I'm using bottom-up implementations and pull dp when possible. Pull dp is when we calculate each dp entry as a function of previously calculated dp entries. This is the way used in recursion / memoization. The other alternative would be push dp, where we update future dp entries using the current dp entry.

I think CSES is a nice collection of important CP problems, and would like it to have editorials. Without editorials users will get stuck on problems, and give up without learning the solution. I think this slows down learning significantly compared to solving problems with editorials. Therefore, I encourage others who want to contribute, to write editorials for other sections of CSES.

Feel free to point out mistakes.

## Dice Combinations (1633)

`dp[x] = number of ways to make sum x using numbers from 1 to 6.`

Sum over the last number used to create x, it was some number between 1 and 6. For example, the number of ways to make sum x ending with a 3 is dp[x-3]. Summing over the possibilities gives dp[x] = dp[x-1] + dp[x-2] + dp[x-3] + dp[x-4] + dp[x-5] + dp[x-6].

We initialize by dp[0] = 1, saying there is one way with sum zero (the empty set).

The complexity is $$$O(n)$$$.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int mod = 1e9+7;
int n;
cin >> n;
vector<int> dp(n+1,0);
dp[0] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= 6 && i-j >= 0; j++) {
(dp[i] += dp[i-j]) %= mod;
}
}
cout << dp[n] << endl;
}
```

## Minimizing Coins (1634)

This is a classical problem called the **unbounded knapsack problem**.

`dp[x] = minimum number of coins with sum x.`

We look at the last coin added to get sum x, say it has value v. We need dp[x-v] coins to get value x-v, and 1 coin for value v. Therefore we need dp[x-v]+1 coins if we are to use a coin with value v. Checking all possibilities for v must include the optimal choice of last coin.

As an implementation detail, we use dp[x] = 1e9 = $$$10^9 \approx \infty$$$ to signify that it is not possible to make value x with the given coins.

The complexity is $$$O(n\cdot target)$$$.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, target;
cin >> n >> target;
vector<int> c(n);
for (int&v : c) cin >> v;
vector<int> dp(target+1,1e9);
dp[0] = 0;
for (int i = 1; i <= target; i++) {
for (int j = 0; j < n; j++) {
if (i-c[j] >= 0) {
dp[i] = min(dp[i], dp[i-c[j]]+1);
}
}
}
cout << (dp[target] == 1e9 ? -1 : dp[target]) << endl;
}
```

## Coin Combinations I (1635)

This problem has a very similar implementation to the previous problem.

`dp[x] = number of ways to make value x.`

We initialize dp[0] = 1, saying the empty set is the only way to make 0.

Like in "Minimizing Coins", we loop over the possibilities for last coin added. There are dp[x-v] ways to make x, when adding a coin with value v last. This is since we can choose any combination for the first coins to sum to x-v, but need to choose v as the last coin. Summing over all the possibilities for v gives dp[x].

The complexity is $$$O(n\cdot target)$$$.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int mod = 1e9+7;
int n, target;
cin >> n >> target;
vector<int> c(n);
for (int&v : c) cin >> v;
vector<int> dp(target+1,0);
dp[0] = 1;
for (int i = 1; i <= target; i++) {
for (int j = 0; j < n; j++) {
if (i-c[j] >= 0) {
(dp[i] += dp[i-c[j]]) %= mod;
}
}
}
cout << dp[target] << endl;
}
```

## Coin Combinations II (1636)

`dp[i][x] = number of ways to pick coins with sum x, using the first i coins.`

Initially, we say we have dp[0][0] = 1, i.e we have the empty set with sum zero.

When calculating dp[i][x], we consider the i'th coin. Either we didn't pick the coin, then there are dp[i-1][x] possibilities. Otherwise, we picked the coin. Since we are allowed to pick it again, there are dp[i][x — *<value of i'th coin>*] possibilities (not dp[i-1][x — *<value of i'th coin>*] possibilities).

Because we consider the coins in order, we will only count one order of coins. This is unlike the previous task, where we considered every coin at all times.

The complexity is $$$O(n\cdot target)$$$.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int mod = 1e9+7;
int n, target;
cin >> n >> target;
vector<int> x(n);
for (int&v : x) cin >> v;
vector<vector<int>> dp(n+1,vector<int>(target+1,0));
dp[0][0] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= target; j++) {
dp[i][j] = dp[i-1][j];
int left = j-x[i-1];
if (left >= 0) {
(dp[i][j] += dp[i][left]) %= mod;
}
}
}
cout << dp[n][target] << endl;
}
```

## Removing Digits (1637)

`dp[x] = minimum number of operations to go from x to zero.`

When considering a number x, for each digit in the decimal representation of x, we can try to remove it. The transition is therefore: dp[x] = $$$\min_{d \in digits(x)}$$$ dp[x-d].

We initialize dp[0] = 0.

The complexity is $$$O(n)$$$.

Note that the greedy solution of always subtracting the maximum digit is also correct, but we are practicing DP :)

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> dp(n+1,1e9);
dp[0] = 0;
for (int i = 0; i <= n; i++) {
for (char c : to_string(i)) {
dp[i] = min(dp[i], dp[i-(c-'0')]+1);
}
}
cout << dp[n] << endl;
}
```

## Grid Paths (1638)

`dp[r][c] = number of ways to reach row r, column c.`

We say there is one way to reach (0,0), dp[0][0] = 1.

When we are at some position with a `.`

, we came either from the left or top. So the number of ways to get to there is the number of ways to get to the position above, plus the number of ways to get to the position to the left. We also need to make sure that the number of ways to get to any position with a `#`

is 0.

The complexity is $$$O(n^2)$$$, so linear in the number of cells of input.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int mod = 1e9+7;
int n;
cin >> n;
vector<vector<int>> dp(n, vector<int>(n, 0));
dp[0][0] = 1;
for (int i = 0; i < n; i++) {
string row;
cin >> row;
for (int j = 0; j < n; j++) {
if (row[j] == '.') {
if (i > 0) {
(dp[i][j] += dp[i-1][j]) %= mod;
}
if (j > 0) {
(dp[i][j] += dp[i][j-1]) %= mod;
}
} else {
dp[i][j] = 0;
}
}
}
cout << dp[n-1][n-1] << endl;
}
```

## Book Shop (1158)

This is a case of the classical problem called **0-1 knapsack**.

`dp[i][x] = maximum number of pages we can get for price at most x, only buying among the first i books.`

Initially dp[0][x] = 0 for all x, as we can't get any pages without any books.

When calculating dp[i][x], we look at the last considered book, the i'th book. We either didn't buy it, leaving x money for the first i-1 books, giving dp[i-1][x] pages. Or we bought it, leaving x-price[i-1] money for the other i-1 books, and giving pages[i-1] extra pages from the bought book. Thus, buying the i'th book gives `dp[i-1][x-price[i-1]] + pages[i-1]`

pages.

The complexity is $$$O(n\cdot x)$$$.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, x;
cin >> n >> x;
vector<int> price(n), pages(n);
for (int&v : price) cin >> v;
for (int&v : pages) cin >> v;
vector<vector<int>> dp(n+1,vector<int>(x+1,0));
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= x; j++) {
dp[i][j] = dp[i-1][j];
int left = j-price[i-1];
if (left >= 0) {
dp[i][j] = max(dp[i][j], dp[i-1][left]+pages[i-1]);
}
}
}
cout << dp[n][x] << endl;
}
```

## Array Description (1746)

`dp[i][v] = number of ways to fill the array up to index i, if x[i] = v.`

We treat i = 0 separately. Either x[0] = 0, so we can replace it by anything (i.e dp[0][v] = 1 for all v). Otherwise x[0] = v $$$\ne$$$ 0, so that dp[0][v] = 1 is the only allowed value.

Now to the other indices i > 0. If x[i] = 0, we can replace it by any value. However, if we replace it by v, the previous value must be either v-1, v or v+1. Thus the number of ways to fill the array up to i, is the sum of the previous value being v-1, v and v+1. If x[i] = v from the input, only dp[i][v] is allowed (i.e dp[i][j] = 0 if j $$$\ne$$$ v). Still dp[i][v] = dp[i-1][v-1] + dp[i-1][v] + dp[i-1][v+1].

The complexity is $$$O(n\cdot m)$$$ with worst-case when x is all zeros.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int mod = 1e9+7;
int n, m;
cin >> n >> m;
vector<vector<int>> dp(n,vector<int>(m+1,0));
int x0;
cin >> x0;
if (x0 == 0) {
fill(dp[0].begin(), dp[0].end(), 1);
} else {
dp[0][x0] = 1;
}
for (int i = 1; i < n; i++) {
int x;
cin >> x;
if (x == 0) {
for (int j = 1; j <= m; j++) {
for (int k : {j-1,j,j+1}) {
if (k >= 1 && k <= m) {
(dp[i][j] += dp[i-1][k]) %= mod;
}
}
}
} else {
for (int k : {x-1,x,x+1}) {
if (k >= 1 && k <= m) {
(dp[i][x] += dp[i-1][k]) %= mod;
}
}
}
}
int ans = 0;
for (int j = 1; j <= m; j++) {
(ans += dp[n-1][j]) %= mod;
}
cout << ans << endl;
}
```

## Edit Distance (1639)

This is a classic problem called **edit distance**.

We call the input strings a and b, and refer to the first i characters of a by a[:i].

`dp[i][k] = minimum number of moves to change a[:i] to b[:k].`

When we calculate dp[i][k], there are four possibilities to consider for the rightmost operation. We check all of them and take the cheapest one.

1. We deleted character a[i-1]. This took one operation, and we still need to change a[:i-1] to b[:k]. So this costs 1 + dp[i-1][k] operations.

2. We added character b[k-1] to the end of a[:i]. This took one operation, and we still need to change a[:i] to b[:k-1]. So this costs 1 + dp[i][k-1] operations.

3. We replaced a[i-1] with b[k-1]. This took one operation, and we still need to change a[:i-1] to b[:k-1]. So this costs 1 + dp[i-1][k-1] operations.

4. a[i-1] was already equal to b[k-1], so we just need to change a[:i-1] to b[:k-1]. That takes dp[i-1][k-1] operations. This possibility can be viewed as a replace operation where we don't actually need to replace a[i-1].

The complexity is $$$O(|a|\cdot |b|)$$$.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
string a, b;
cin >> a >> b;
int na = a.size(), nb = b.size();
vector<vector<int>> dp(na+1, vector<int>(nb+1,1e9));
dp[0][0] = 0;
for (int i = 0; i <= na; i++) {
for (int j = 0; j <= nb; j++) {
if (i) {
dp[i][j] = min(dp[i][j], dp[i-1][j]+1);
}
if (j) {
dp[i][j] = min(dp[i][j], dp[i][j-1]+1);
}
if (i && j) {
dp[i][j] = min(dp[i][j], dp[i-1][j-1]+(a[i-1] != b[j-1]));
}
}
}
cout << dp[na][nb] << endl;
}
```

## Rectangle Cutting (1744)

`dp[w][h] = minimum number of cuts needed to cut a w x h piece into squares.`

Consider a $$$w \times h$$$ piece. If it is already square (w = h), we need 0 cuts. Otherwise, we need to make the first cut either horizontally or vertically. Say we make it horizontally, then we can cut at any position 1,2,..,h-1. If we cut at position k, then we are left with two pieces of sizes $$$w \times k$$$ and $$$w \times h-k$$$. We can look up the number of moves to reduce these to squares in the dp array. We loop over all possibilities k and take the best one. Similarly for vertical cuts.

The complexity is $$$O(a^2\cdot b + a\cdot b^2)$$$.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int w, h;
cin >> w >> h;
vector<vector<int>> dp(w+1,vector<int>(h+1));
for (int i = 0; i <= w; i++) {
for (int j = 0; j <= h; j++) {
if (i == j) {
dp[i][j] = 0;
} else {
dp[i][j] = 1e9;
for (int k = 1; k < i; k++) {
dp[i][j] = min(dp[i][j], dp[k][j]+dp[i-k][j]+1);
}
for (int k = 1; k < j; k++) {
dp[i][j] = min(dp[i][j], dp[i][k]+dp[i][j-k]+1);
}
}
}
}
cout << dp[w][h] << endl;
}
```

## Money Sums (1745)

This is a case of the classical problem called **0-1 knapsack**.

`dp[i][x] = true if it is possible to make x using the first i coins, false otherwise.`

It is possible to make x with the first i coins, if either it was possible with the first i-1 coins, or we chose the i'th coin, and it was possible to make x — *<value of i'th coin>* using the first i-1 coins.

Note that we only need to consider sums up to 1000 $$$\cdot$$$ n, since we can't make more than that using n coins of value $$$\le$$$ 1000.

The complexity is $$$O(n^2\cdot \max x_i)$$$.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int max_sum = n*1000;
vector<int> x(n);
for (int&v : x) cin >> v;
vector<vector<bool>> dp(n+1,vector<bool>(max_sum+1,false));
dp[0][0] = true;
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= max_sum; j++) {
dp[i][j] = dp[i-1][j];
int left = j-x[i-1];
if (left >= 0 && dp[i-1][left]) {
dp[i][j] = true;
}
}
}
vector<int> possible;
for (int j = 1; j <= max_sum; j++) {
if (dp[n][j]) {
possible.push_back(j);
}
}
cout << possible.size() << endl;
for (int v : possible) {
cout << v << ' ';
}
cout << endl;
}
```

## Removal Game (1097)

The trick here is to see that since the sum of the two players' scores is the sum of the input list, player 1 tries to maximize $$$score_1-score_2$$$, while player 2 tries to minimize it.

`dp[l][r] = difference`

$$$score_1-score_2$$$`if considering the game played only on interval [l, r].`

If the interval contains only one element (l = r), then the first player must take that element. So dp[i][i] = x[i].

Otherwise, player 1 can choose to take the first element or the last element. If he takes the first element, he gets x[l] points, and we are left with the interval [l+1,r], but with player 2 starting. $$$score_1-score_2$$$ on interval [l+1,r] is just dp[l+1][r] if player 1 starts. Since player 2 starts, it is -dp[l+1][r]. Thus, the difference of scores will be x[l]-dp[l+1][r] if player 1 chooses the first element. Similarly, it will be x[r]-dp[l][r-1] if he chooses the last element. He always chooses the maximum of those, so `dp[l][r] = max(x[l]-dp[l+1][r], x[r]-dp[l][r-1])`

.

In this problem dp[l][r] depends on dp[l+1][r], and therefore we need to compute larger l before smaller l. We do it by looping through l from high to low. r still needs to go from low to high, since we depend only on smaller r (dp[l][r] depends on dp[l][r-1]). Note that in all the other problems in this editorial, dp only depends on smaller indices (like dp[x] depending on dp[x-v], or dp[i][x] depending on dp[i-1][x]), which means looping through indices in increasing order is correct.

We can reconstruct the score of player 1 as the mean of, the sum of all input values, and $$$score_1-score_2$$$.

The complexity is $$$O(n^2)$$$.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> x(n);
long long sum = 0;
for (int&v : x) {
cin >> v;
sum += v;
}
vector<vector<long long>> dp(n,vector<long long>(n));
for (int l = n-1; l >= 0; l--) {
for (int r = l; r < n; r++) {
if (l == r) {
dp[l][r] = x[l];
} else {
dp[l][r] = max(x[l]-dp[l+1][r],
x[r]-dp[l][r-1]);
}
}
}
cout << (sum+dp[0][n-1])/2 << endl;
}
```

## Two Sets II (1093)

This is a 0-1 knapsack in disguise. If we are to have two subsets of equal sum, they must sum to half the total sum each. This means if the total sum $$$\frac{n(n+1)}{2}$$$ is odd, the answer is zero (no possibilities). Otherwise we run 0-1 knapsack to get the number of ways to reach $$$\frac{n(n+1)}{4}$$$ using subsets of the numbers 1..n-1. Why n-1? Because by only considering numbers up to n-1, we always put n in the second set, and therefore only count each pair of sets once (otherwise we count every pair of sets twice).

`dp[i][x] = number of ways to make sum x using subsets of the numbers 1..i .`

We say there is one way (the empty set) to make sum 0, so dp[0][0] = 1;

For counting number of ways to make sum x using values up to i, we consider the number i. Either we didn't include it, then there are dp[i-1][x] possibilities, or we included it, and there are dp[i-1][x-i] possibilities. So `dp[i][x] = dp[i-1][x] + dp[i-1][x-i]`

.

The complexity is $$$O(n^3)$$$.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int mod = 1e9+7;
int n;
cin >> n;
int target = n*(n+1)/2;
if (target%2) {
cout << 0 << endl;
return 0;
}
target /= 2;
vector<vector<int>> dp(n,vector<int>(target+1,0));
dp[0][0] = 1;
for (int i = 1; i < n; i++) {
for (int j = 0; j <= target; j++) {
dp[i][j] = dp[i-1][j];
int left = j-i;
if (left >= 0) {
(dp[i][j] += dp[i-1][left]) %= mod;
}
}
}
cout << dp[n-1][target] << endl;
}
```

## Increasing Subsequence (1145)

This is a classical problem called **Longest Increasing Subsequence** or LIS for short.

`dp[x] = minimum ending value of an increasing subsequence of length x+1, using the elements considered so far.`

We add elements one by one from left to right. Say we want to add a new value v. For this to be part of an increasing subsequence, the previous value in the subsequence must be lower than v. We might as well take the maximum length subsequence leading up to v, as the values don't matter for the continuation to the right of v. Therefore we need to extend the current longest increasing subsequence ending in a value less than v. This means we want to find the rightmost element in the dp array (as the position corresponds to the length of the subsequence), with value less than v. Say it is at position x. We can put v as a new candidate for ending value at position x+1 (since we have an increasing subsequence of length x+1 + 1, which ends on v). Note that since x was the rightmost position with value less than v, changing dp[x+1] to v can only make the value smaller (better), so we can always set dp[x+1] = v without checking if it is an improvement first.

Naively locating the position x with a for loop gives complexity $$$O(n^2)$$$. However, dp is always an increasing array. So we can locate x position by binary search (std::lower_bound in C++ directly gives position x+1).

The final answer is the length of the dp array after considering all elements.

The complexity is $$$O(n\cdot \log n)$$$.

In this task we were asked to find the longest strictly increasing subsequence. To find the longest increasing subsequence where we allow consecutive equal values (for example 1,2,2,3), change lower_bound to upper_bound.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> dp;
for (int i = 0; i < n; i++) {
int x;
cin >> x;
auto it = lower_bound(dp.begin(), dp.end(), x);
if (it == dp.end()) {
dp.push_back(x);
} else {
*it = x;
}
}
cout << dp.size() << endl;
}
```

## Projects (1140)

Even though days can go up to $$$10^9$$$, we only care about days where we either start or just finished a project. So before doing anything else, we compress all days to their index among all interesting days (i.e days corresponding to $$$a_i$$$ or $$$b_i+1$$$ for some i). This way, days range from 0 to less than $$$2 n \le 4\cdot 10^5$$$.

`dp[i] = maximum amount of money we can earn before day i.`

On day i, maybe we just did nothing, so we earn what we earned on day i-1, i.e dp[i-1]. Otherwise, we just finished some project. We earned some money doing the project, and use dp[start of project] to know how much money we could have earned before starting the project. Loop through all projects finishing just before day i, and take the best one.

The complexity is $$$O(n\cdot \log n)$$$, log comes from day compression.

**Code**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
map<int,int> compress;
vector<int> a(n),b(n),p(n);
for (int i = 0; i < n; i++) {
cin >> a[i] >> b[i] >> p[i];
b[i]++;
compress[a[i]], compress[b[i]];
}
int coords = 0;
for (auto&v : compress) {
v.second = coords++;
}
vector<vector<pair<int,int>>> project(coords);
for (int i = 0; i < n; i++) {
project[ compress[b[i]] ].emplace_back( compress[a[i]], p[i] );
}
vector<long long> dp(coords, 0);
for (int i = 0; i < coords; i++) {
if (i > 0) {
dp[i] = dp[i-1];
}
for (auto p : project[i]) {
dp[i] = max(dp[i], dp[p.first]+p.second);
}
}
cout << dp[coords-1] << endl;
}
```

Hello. Thanks for editorial. Do you have links to above problems or where I can submit them. I want to try all of them as they look good from IPOV.

https://cses.fi/problemset/

Thanks a lot.

Cool, thanks. I already recommend CSES problem set to some people and I will now link to this blog for dp. Just next time consider using names like

`ways`

or`best_score`

in the future instead of`dp`

in every single problem — if you want to then publish your codes to help others.In my opinion, open editorials to CSES problems (with code!) goes against the spirit of the platform. Not providing solutions or access to accepted submissions must have been a conscious decision by the maintainers.

I thought about that, and therefore asked pllk for permission before writing the editorial. He encourages editorials for the problems. However, he mentioned that it might cause some people to read the editorial without thinking about the problem.

meooow I think you are wrong here. I agree one should think before rushing to editorials but having no editorials forces one to leave the question without any further learning and development.

Thanks to icecuber who invested his precious time in writing some editorials and pllk for allowing it.

Having no editorial is not enough reason to leave a problem. There are many avenues to ask for help once you have truly exhausted all approaches you can think of.

However, since pllk approves of this, my point of complaint isn't valid.

Can you kindly suggest some of the many avenues you mention of? And are they more effective and efficient than having the editorial at your disposal?

Asking for help FAQ

Is the answer you get by asking for help going to be better than an editorial? Maybe not.

Is the absence of an editorial going to make you work harder to solve the problem? Absolutely yes.

Please do not think that I am preaching for a world with no editorials. It is just that the design of the CSES problemset implies that it is meant to solved by putting in maximum individual effort.

Well, I believe that every problem should have an editorial. Let's say I put in 2 days worth of effort solving a problem without making any real progress then an editorial guarantees me closure for my efforts. On the other hand, asking help from others doesn't provide me that certainty. Especially for the guys like me who don't have coding circles where we can ask for help from friends. Cyans(or lower rating) people asking for help from a random person usually gets ignored. That's just what I have observed and my opinion.

shoya right. exactly right.

Idk, I usually answer people that ask nicely. Show the problem source and more people might help.

Thanks for this. I love the work here. CSES needs more editorials like this; sometimes even just stalking people's code isn't enough to understand what's going on. I hope you keep going with this too.

How to see other peoples submission in CSES.fi ?

I meant looking at people’s GitHubs or something like that since you can’t view other solutions until you solve it yourself

msf_sheet you can see others submission after you solved the problem in

hackingtab.Thanks to this blog, I go back to my account on CSES problem set. I found out that I didn't solve 2 problems in DP section (Got

`TLE`

several testcases). I solved it and AC in first submit, but the logic is the same as I have solved it before. Lmao =)))I've gotten two more questions about TLE in Coin Combinations II, so I guess it deserves a comment. Since we are doing $$$10^8$$$ operations in one second, we need those operations to be efficient. This means we can't have many cache misses.

You get cache misses by accessing array entries that are far away from each other. My implementation loops through

`i`

, then`j`

. It accesses`dp[i][j]`

,`dp[i-1][j]`

and`dp[i][j-x[i-1]]`

.If you order your loops differently (

`j`

, then`i`

), or use`dp[j][i]`

instead of`dp[i][j]`

(so you swapped the meaning of the dimensions), you will likely get TLE.In terms of rules of thumb, we see that the dimension containing

`j-x[i-1]`

varies most, and therefore put it as the inner dimension of the dp array. And we loop through the dimensions the same order as the dimensions of the array. Below are some more detailed explanations.If we loop through

`i`

,`j-x[i-1]`

varies a lot, this means cache misses. Therefore we need to loop through`i`

in the outer loop. Looping through`j`

gives contiguous memory accesses, so we get few cache misses by having`j`

as the inner loop.If you define your array as

`dp[j][i]`

instead of`dp[i][j]`

, then`dp[j-x[i-1]][i]`

goes far from`dp[j][i]`

, compared to`dp[i][j]`

and`dp[i][j-x[i-1]]`

. This is because the outer dimension gives smaller distance in memory than the inner dimension (you can think of`dp[i][j]`

as accessing index $$$10^6 i + j$$$ in a linear array, if the second dimension has size $$$10^6$$$).Is it the same reason (that O(n * sum) solution requires 10

^{8}ops, in worst case) and hence top-down is not feasible due to extra overhead of recursion ?actually if u notice then in case of top down we dont visit each and every cell of n*sum dp table... but in bottom up we also go to the states which wont be occuring if we use the TOP down method..so some how Top down is better .Thats my logic behind it.. Any new logics are highly appreciated!! Do comment if u feel something wrong in it!

Thanks,i didn't think caching might be a problem in a program although it makes quite a sense now. however, we can do the program in

O(x)space complexity,right? That could avoid the problem of caching." If you order your loops differently (j, then i), or use dp[j][i] instead of dp[i][j] (so you swapped the meaning of the dimensions), you will likely get TLE." this was reason i was getting TLE, thanks for your comment

## include<bits/stdc++.h>

using namespace std;

## define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

## define pb push_back

## define ll long long int

ll mod = 1e9+7; const int N = 1e5+7;

int dp[101][1000001];

int main(){ fast; int n , val;

}

This Code is still giving TLE ..Can you please let me know where its going wrong? Its the same code as yours!!

What's essentialy the difference between coins combinations

IandII, why the added dimension and why we reversed the order of loops?Next Graph theory, please...

Thanks, Due to no editorial novice programmers couldn't try those problems or either leave it after sometime. Please make next one on graph and tree. icecuber

In

Book Shop, if I buy the`i'th`

book then shouldn't`x-price[i]`

money be left for the remaining`i-1`

books , so I can't see whymoney is left for the remainingx-price[i-1]`i-1`

books as that would imply that the money left is after buying the`(i-1)'th`

book. So if icecuber or anyone else could please explain this it would be a great help. Thanks.The array "price" is 0-indexed, so the price of the i'th book is price[i-1].

I feel so dumb for asking such a question as now it is obvious what you were trying to say. And I am sorry for asking something like this. I am thankful to you for answering and also for the editorial. Thanks icecuber.

For Coin Combinations II (1636), my solution getting TLE in 3 TCs. Unlike editorial, I'm trying top-down approach. Can someone help please?

TLE on 100 1000000? I don't know but I think top-down will be slower when the table is filled-up

I think there might be 2 reasons for that :

If you are using a 2D vector for dp like 'vector<vector> dp(n+1,vector(sum+1,0));'(ll -> long long), then you might want to change the data type from long long int to int like 'vector<vector> dp(n+1,vector(sum+1,0));'

while doing the MOD operation like '_dp[i][j] = ((dp[i][j-v[i-1]]%MOD) + (dp[i-1][j]%MOD))%MOD;_', you might want to do something like '(dp[i][j] += dp[i][left]) %= MOD;

This worked for me.

In Two Sets II problem,

`Because by only considering numbers up to n-1, we always put n in the second set, and therefore only count each pair of sets once (otherwise we count every pair of sets twice).`

Why putting n always in second set gurantees unique sets? How to prove above statement.

Since you aren't having 'n' in your first set, you'll never have a first set that consists of a combination involving 'n' i.e. all the times a combination requires 'n' you are simply not counting it... thus you end up counting things only once.

Please do the same for "Graph Algorithms" section. Thanks.

please provide editorial for graphs problem. Thanks:)

It takes a long time to write up an editorial like this, and the graph section has like a billion problems. If icecuber provides an editorial that's great, but there exists code on github for all the problems so if you're stuck you shouldn't rely on him making an editorial, as he may not want to. Notice how people asked 7 months ago (comment) and he still has not, so I doubt it'll happen smh. No need for people to keep asking.

Check out the AC codes for all CSES dp problems here:

https://github.com/pratikkundnani/cses-dp-solutions/tree/master

I will keep updating the repository! I will be solving graph problems too.

Star it if you liked it.

Thank you!

Can you write an editorial for the graph section too...it will be a great help..

I am not sure about the editorials but I will be solving the graph section and I will upload all AC codes of the graph section on my Github. I will update you all!

Why my code for Minimizing Coins gives TLE for some test cases, even it's complexity is also O(n*target)?

CodeThe

`i%arr[j]`

is causing the TLE I believe. I removed it, since it isn't required as`dp[i-arr[j]]`

will handle it.This is because the constraints are already too tight and modulo operation is time costly operation. Also, sorting

`arr`

also improved the time a bit, but major contribution was made by remove the module operation.Hope this helps.

CodeSorry for the poor formatting, I am not able to figure out a better way.

Yeah, Got it!!

You can also use something like this:

if (value >= MOD) value -= MOD;

This is much much faster than the costly operation of modulus operator!

what about values twice or thrice bigger than mod

In case of addition, the biggest possible result would be MOD — 1 + MOD — 1 = 2*MOD — 2. Otherwise, it has already been drawn down.

can you also provide editorials for the graph section as well

You can also refer this competitive programming book by cses

Hello coders!! I'm listing out my AC solutions of CSES DP questions. It's not fully updated. I'm working on it. you can have a visit to look into the AC codes. Do star it if you like it.

Thank you. Suggestions are welcome.https://github.com/noob-hu-yaar/CSES-Problem-set-solutions-DP-?files=1

Can anyone explain the matrix expo solution to Dice Combinations? It has much lower complexity ($$$6^3 \log n$$$ afaict). This is the code, for example, which used the matrix

raises it to the power $$$n$$$, and then prints

`dp[0][0]`

as the answer. Can someone explain how do we come up with this matrix in the first place?Check CP handbook's matrix chapter's linear recurrences section.

Why does this code give a runtime error? It works fine on my computer for the same inputs.

CodeI figured it out myself, the problem with the code was that I was using long long for dp array which was exceeding the memory limit, changing it to int passed all the test cases. I just wrote it here in case someone faces the same problem.

Thanks

Thanks a lot icecuber! I found your editorial super useful.

It was hard for me to understand the Removal Game (1097) solution at first. I even found another version here, and struggled comprehending it as well :-) Like, I saw how the code could possibly work, of course, but the trail of thought leading to that code was eluding me. So I found my own trail then, which may be somewhat more beginner-friendly. So, here it is:

Given the input numbers

`xs`

, I started with two arrays:`A[l][r]`

,`B[l][r]`

-- the scores for each player, when they have the first move on the range of`[l, r]`

. Now, say, I'm the player A, and I pull from the front of the range -- the`xs[l]`

. Let`t`

be the`range_sum(l, r)`

. Then,`t = xs[l] + B[l + 1][r] + alpha`

, which leads to`alpha = t - xs[l] - B[l + 1][r]`

. Thus,`A[l][r] = xs[l] + alpha = t - B[l + 1][r]`

. Similarly, if I pull from the back of the range,`A[l][r] = t - B[l][r - 1]`

. The optimal strategy is to take the largest of the two numbers.Next, we make an observation, that the

`A`

will be identical to`B`

, and we only need one DP matrix.The codeWhat is the alpha here?

and can u explain how A and B and identical?

Oh, right. I didn't explain that

`alpha`

, did I?`alpha`

is whatever remains from`A[l][r]`

after`xs[l]`

is taken:`alpha := A[l][r] - xs[l]`

. It's just a temporary variable that helps in deriving the recurrence.Now, regarding the sameness of

`A`

and`B`

. As defined,`A[l][r]`

is the top score for a player, when they have the first move on the range of`[l, r]`

.And,whether that player 1st or 2nd is, is fully determined by the parity (even/odd) of the interval's length:`r - l + 1`

. Thus, designating the`A`

from the`B`

is actually superfluous.That's clear now

thanks

In the edit distance problem, it says -- "When we calculate dp[i][k], there are four possibilities to consider for the rightmost operation. We check all of them and take the cheapest one."

Why do we only need to consider the rightmost operation? I feel this is worth discussing (either in the post or as a reply to my comment).

In Coin Combinations 2, you don't need extra dimension/state. dp[x + 1] = {0} would work fine. dp[0] = 1. dp[i] represents no. of ways to reach value 'i' with coins considered so far. So loop through coins in any order, and update values of future states of dp by looping through previous states of dp from left to right i.e. from 0 to x. That is dp[current_coin + state] += dp[state];

Spoiler - CodeSpace: O(target)

Time: O(target * n)

Can you please clarify your approach?

Can anyone explain me the array description question? Or just explain me the output of this test case: 4 5 2 0 3 0 I thought it would be 6 but it turns out to be 2. Any help would be appreciated.

Hope it's still relevant: https://ideone.com/ATMahc

Thanks, to icecuber for writing an amazing editorial. Without this I wouldn't have thought of even touching single dp problem and someone can explain the array description problem.

In Coin Combinations II(1636) shouldn't the vector x be sorted?

Two Sets II (1093) problem we can also do it in normal way (that is including n too) then finally divide by 2 to avoid repetition. NOTE:- use modulo inverse while performing division.

CODEIn the problem two set (1093) u said that if we count subset till the last element than we end up repeating the same set twice.so we can get our desired answer by just dividing it by 2. but here it fails .

you cant divide it by two because you did mod operation ,all you can do is multiply it by multiplicative inverse of 2 and mod .

you can watch this https://www.youtube.com/watch?v=TOsD3BkIKoQ&feature=youtu.be he was having the same problem as yours

please consider making one editorial on graph section also icecuber

In problem Projects(1140), I'm not able to get the compressing the days part. Can anyone explain with an example?

I am making detailed video explanations for some of the problems of this section, I will keep updating this comment whenever I add a new video. Along with the explanation I live code the solution at the end of the video and submit the code on the platform.

Interested may check the following videos.

Dice Combinations: https://youtu.be/5gd5jptXWAM

Coin Combinations: https://youtu.be/-pXjopzMVrE

Grid Paths: https://youtu.be/V64F4wlodUM

Book Shop: https://youtu.be/qpNy2nuSuaI

Also added on my channel videos for coin combinations 2, array Description, removing digits, edit distance, Rectangle cutting and projects.

Hopefully people will find this helfpul.

Links to Video solution for all problems can be found here: https://codeforces.com/blog/entry/80064

Editorials like this should be appreciated!

Thanks for this :p

I have been trying the problem Book Shop for a very long time now. I am using the same logic as given by you. Still, it is giving me TLE. Is there any other approach for this question or some optimization that I can do on my code:

Here is my code:

Incase you missed it, I have replied to your comment on my youtube video: "Honestly it's sad that the time limit on the platform is very tight. Try changing the loops, make outer loop for n and inner loop for x and it should pass. Try it and tell me if it works. I'll give you a plausible explanation why this works."

Yes, I just did that and it worked. But, I have no idea why this happened. The time complexity for both of them is same but just interchanging the loops is reducing the time it is taking for the program to run. Looking forward to your explanation on this. And yes, Thanks a lot.

I am no expert but here is what I can tell you, There is something known as locality of reference, if your program uses a part of the memory which is located in contiguous memory addresses(or nearby memory addresses) it will execute faster.

Why is that so?

Caching. The OS(again I am no expert, but hopefully someone will put more light on this) will cache some of the regions of the data stored in main memory, accessing cache is much faster than accessing main memory.

Ever Thought why iterating an array is much faster than iterating a link list of same size even though asymptotically both of them are O(N) operations.

Try laying out your 2d array and think which of two approaches will make more compact memory region accesses.

Hope this helps!

Thanks for the reply again but correct me if I am wrong, If I have 2D array of N*X and another of X*N and in both of them I iterate wouldnt it take the same time. I understood the part that contiguous memory will be faster to iterate on but aren't both the loops doing the same thing. In both cases we are iterating on contiguous memory locations. In the first case, we will do like X iterations N times and in the second we will do N itetations X times. So, does it affect the running tume considering X is greater than N. Sorry for the trouble but it would really help me if you van explain a little by taking an example. How about you explain this when N is 2 and X is 5. This will really clear the doubt.

keeping in mind caching and locality of reference think which of the two recurrences will likely evaluate faster:

1. dp[i] = f(dp[i-1])

2. dp[i] = f(dp[i-some_random_number(1,i)])

In main memory 2d, 3d, 4d all arrays are flattened and stored as 1d arrays. So flatten N.X array into N arrays of size X each and X.N array into X arrays of size N each.

Take 5mins and then check if this pic helps you a bit:

Even I would appreciate a response from someone who is an intermediate or expert at COA and OS concepts but till then this is the best explanation I can come up with.

Ohhh I think now I am having a much better understanding of this thing. The picture was quite explanatory. I looked at both of my codes and realised this is actually quite sensible. When I am doing dp[i-1][j- some random number] it is like going 1 time back in 10^3 array and then it is about 10^5 places to move in the end to reach j-some random number. But in case I do dp[j-some random number][i-1] it will go back in 10^5 array which has 10^3 size sub array and this will happen arbitrary number of times so it becomes 10^8. I guess this is only what you are trying to say. Thanks for the wonderful explanation, really appreciate your effort to explain this. I have understood it.

Yes that is what I was trying to say, I am glad it helped :)

Bro, could you please share your code, i am kinda stuck here i thought the correct way yet it gives TLE due to tight limits, i even tried kartik's way maybe i didnt do it rightly

Another way is to just do a 1D dp, which is possible since the transitions only ever access [j-1].

Sharing a 1D dp approach

codeThanks a lot for this! Best problems for beginner in DP and editorial is perfect for beginner to understand dp!

https://pastebin.com/mVN1rXrz Can Someone tell me the issue with my code of

edit distance.It passes all the test cases but fails in one and I have been trying to debug for over an hour but couldn't find the bug.How to see other peoples submission in CSES.fi ?

No you can't because there is no option to see other's code.

You can go to hacking section after submitting your code and you can hack other's code where you can see others code This is only the way to see other's code Hope this help

This is my code for (Two sets II) problem. My approach:-

`dp[i]= number of ways to make sum using subset 1..i.`

Can anyone help me find my mistake?

you can't make dp[sum]/=2 after module

you need to use modular inverse for the divison problem

Can we go for Removing Digits like this. Errichto and icecuber sir give me some shine

i'm just giving my openion. Thankyou

in minimizing coins: O(n.target)->could be 1e8 operation

how it works in 1 second

hey did anyone else get a runtime error (like heap space error or something) on coin combination 2 or is it just my code?

Can someone explain why the solution for increasing subsequence works? Specifically I am confused about how we can replace an element in the dp array with an element later found. Doesn't this not make dp a subsequence?

i know its not a big thing you we don't need a two state dp for problem combinations2 one is enough.

Figured it.

for array description can anyone tell me where i have gone wrong https://ideone.com/FsF43S

i didn't get about removal game can anyone make me understand why we have to take the mean at the end

As stated, score1 + score2 = sum. With our dp, we calculate max(score1 — score2).

Add both of these, sum + (score1-score2). = score1 + score2 + score1 — score2 = 2 * score1

Divide by 2 to get max score1.

Hey! Can someone give the formal proof for the Greedy Solution of 'Removing Digits'?

I tried to formulate a proof, but no luck :(

for projects problem can anyone check it is giving wrong verdict only for 2 cases even though code looks correct can anyone check?

Can someone give a top down recursive dp implementation of problem** Array description**? It does not have to pass in the given TL due to being recursion but the recursive dp should be accurate.

Hope it's still relevant: https://ideone.com/ATMahc

Thanks a lot.

Thanks for the great editorial. I solved some problems on my own. If I didn't got the logic for some, I read the editorial and got the idea. Some problems involved some good tricks which I learnt. Thanks a lot.

Why does taking the mean like this for problem "removal game" work?

what is the top-down approach for coin combination I and II?

Thanks for the editorial.

Two Sets II (1093)for this problem why the recursive solution works, even if we take n instead of n-1.

When current = 1 and sum = 1 when you go to solve(0,0) you returns 0 because of condition current<1 returns 0

your code is not taking 1 in the first set instead of n

The problem Projects — 1140 can also be solved

withoutusing the compression technique.ApproachThe problem can be treated as a 0-1 Knapsack optimisation problem with one extra constraint that the events performed should not overlap. For evaluating this constraint, sort the events by their start times alone, so that we can use upper bound of an event E to find the index of an event which can be performed after an event E.

See the implementation below for more details.

CodeWhy does CSES problem Set give TLE in some problems (in one or two test cases), even if my asymptotic time complexity is the same as that of the intended solution?

In

Rectangle cutting, one can exploit symmetry and stop both the innermost loops at i/2 and j/2 respectively.yes, you can.

For the problem

Removing Digits, I found another solution that works in almost the same time.I'll just add a space optimized DP solution for Money Sums (Can be used for most knapsack problems)

Thank you so much! This is really useful :D

Hey I am stuck in problem of Counting towers of DP section CSES problem set can someone plz help !! Thanks in advance..

i found this useful: youtube video solution

For Two Sets 2 Problem , I have used a similar approach as editorial.However I I have counted every pair of sets twice as I have considered numbers till n and returned dp[n][sum/2]. However my solution still got accepted.Any help in why this is so?

A different approach using vector of pairs, sorting and binary search (nlogn) https://ide.geeksforgeeks.org/hCuD7Po9XT

Code for new problem counting towers.

If someone is looking for memoization based solution for Removing Digits, then this might help.

It would be great if everyone could enclose their code into a spoiler, it creates unnecessary mess sometimes while navigating.

I was trying to solve these problems using recursive dp, I was facing an issue in the Book Shop problem, I was getting runtime error on large test cases though I feel I am using the same space as required in iterative dp. Link to my solution.