**PROBLEM STATEMENT :**

You have given 4 numbers A, B, C, D and you have to perform an algorithm:-

int sum=0;

for(int i=A;i<=B;i++) {

for(int j=C;j<=D;j++) sum+=i^j;

}

As the sum could be very large, compute it modulo 1000000007 (10e9+7).

**Constraints :** 1<=A,B,C,D<=1000000000

Time Limit — 1 sec

INPUT :- 4 integers A,B,C,D is given.

OUTPUT :- Print the sum after performing the algorithm.

EXAMPLE INPUT:- 1 2 3 4

EXAMPLE OUTPUT:- 14

EXPLANATION :- 1^3+1^4+2^3+2^4

2 + 5 + 1 + 6 = 14

This problem was asked in my recent coding round of Uber. Can anyone help me with its approach?

It should be sufficient to just count contribution of each bit into the answer separately.

But we cannot traverse from A to B or from C to D, then how will we count then contribution of each bit?...could you elaborate your answer more?

for example i has 1 at certain places(let say it has 1 at position x) so when you will exor it with j you will add 2^x if j has 0 at that place or 0 otherwise.

Therefore for every bit from 0 to 63rd bit you have to

2^x*(count(numbers with 1 at x in a to b)*count(numbers with 0 at x in c to d)+count(numbers with 0 at x in a to b )*count(numbers with 1 at x in c to d))

Understood!!

a similar question came in july lunchtime/cook-off .

u may see it's editorial for better visualization.

If u did understood , can u explain me pls, I agree we'll have to find the number of set bits at some position x from 'a' to 'b' and number of unset bits from 'c' to 'd' and same vice versa ...

But even in that we would have to iterate over entire range from a to b and from c to d , which would still in worst case take O(n) time and that won't work here.

Pls if possible explain me , I know I am wrong somewhere

Hi Everyone, please I want a verification for my method or point out if I am wrong. let $$$A_1,A_2...A_k$$$ are the numbers in range $$$[A,B]$$$ and similarly let $$$C_1,C_2...C_m$$$ are the numbers in range $$$[C,D]$$$ so basically I have to calculate $$$\displaystyle \sum\limits_{i = 1}^k\displaystyle \sum\limits_{j = 1}^m A_i \,xor\,C_j $$$. but we know that $$$A_i \,xor\,C_j=(A_i +C_j)-2(A_i\,and\,C_j)$$$ also we know that $$$A\,and\,B+A\,and\,B=A\,and\,(B+C)$$$. Now let $$$\displaystyle \sum\limits_{j = 1}^mC_j=S_1$$$. So our inner summation $$$\displaystyle \sum\limits_{j = 1}^m A_i \,xor\,C_j $$$ becomes $$$mA_i+S_1-2(A_i\,and\,S_1)$$$. Now let $$$\displaystyle \sum\limits_{i= 1}^kA_i=S_2$$$. So $$$\displaystyle \sum\limits_{i = 1}^k (mA_i+S_1-2(A_i\,and\,S_1))$$$ is our final answer and it equals $$$m*S_2+k*S_1-2*(S_1\,and\,S_2).$$$ Since we can find $$$S_1,S_2$$$ in constant time so time complexity is $$$O(1).$$$ Please point out anyone if I am wrong.

aah,bitwise and is not commutative over addition

i meant distributive**

Geometric progression sum with binary exponentiation

Not understanding the approach to it, can you explain your approach more clearly?

Hey, I think you misunderstood the question (and I also).

`i^j`

is`i xor j`

not`i raised to j`

.Well it was mentioned there that '^' means xor...I forgot to mention that in the blog

just off topic: It was on-campus or off-campus?

It was DTU on campus

Hey have placement tests started at DTU? or are these internship tests?

I am not studying there, but yes acc to my knowledge both are going!

Just count number of set bits for each position(bit) in A to B and same for C to D. Then ans is just contribution of each bit ie. For each bit contribution is ((no. of set bits in A-B)*(no. of unset C-D)+(no. of set bits in C-D)*(no. of unset A-B))*(2^position). Hope this helps.

fucking genius!!!!

But bro we cannot traverse even from A to B or from C to D so how will we count the set and unset bits?

Hint : For the ith bit, there's a pattern and pattern length is 2^(i+1).

Ohh...okay okay..understood how the concept will work..Thanks!

Hint : For the ith bit, there's a pattern and pattern length is 2^(i+1).Can you elaborate the hint?Hi can you explain about the pattern? I cannot understand ur statement.

For every ith bit there is a pattern of they change....for ex- 0-7

0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111,

0th bit repeats pattern at 2 , 1st bit at 4 , 2nd bit at 8 and so on...

Hope you understand!

Thanks. Do u have a working code? If u have share the code please.

Use Digit Dp

pastebin link code if u want, (not thoroughly tested).

Did you calculated each bit contribution for A to B and C to D with this code?

I calculated until R at ith position bit,

Observation ::

1] at ith position zeros are 2^i — 1, we remove that from R

2] Now there alternating pattern of 1 and 0 with each length of 2^(i+1) , you can calculate that by quotient and remainder dividing by 2^(i+1)

Can you please explain why did you subtract 2^i-1 ?

If we start from 0, then we can say that the pattern at ith position is (0000..2^i times) and then (1111....2^i times).

We calculate number of zeros in [L,R] is = zeros[0,R] — zeros[0,L] . Number of ones is (R-L+1)-(no of zeros).

Why do we need to substract 2^i-1 from R ?

I think I understand why we subtract 2^i — 1 from R. You will get it too. Just start writing numbers in binary form but start from 1 instead from 0. The code link also calculates set bit in ith position for the numbers 1 to R and not 0 to R. This should help.

wow i didnt know that taxi drivers should be good at cp nowadays, what a great time we live in!

lol

What if Uber's algorithm fails some day and they have to find shortest path ;)

This problem ultimately reduces to "Find number of setbits of ith bit for numbers between range x to y"

To find this there is a pattern, run this prog you will understand https://ideone.com/vMaCNs

Yes..I understood it after several explanations in the comments, thank you so much!

its damn easy.i did a similar problem.just store the number of ones and zeros in both ranges.then just iterate over the 42 bits and pick zero from one set(a-b)and one from another and vice-versa.then multiply with 2^i(0-based index)using fast expo and sum it.and you get your answer

Coding round for interships or coding round for full-time ?? @D_Coder_03

Mine was for Internship

Could you share rest questions also in brief

It will be helpful

There were 3 problems

1st problem

Check whether the string is good. A string is good if all its odd length substring are palindromes.

Constraints 1<=|s|<=40

Example

INPUT- aaaaa OUTPUT- YES

INPUT- xyz OUTPUT- NO

2nd problem

A number N is given. Find out the minimum cost to make it a good number.

A good number is the sum of two numbers which are positive power of a single digit odd prime number.

In one operation:

You can add 1 to n with x cost

You can subtract 1 from n with y cost

You can do the operations any number of times.

Contraints: 1<=N,x,y<=1e9

Example

INPUT- 4 5 6 OUTPUT- 24 Explanation- good no. is 3^1+5^1=8->(8-4)*6 = 24

3rd problem was in the blog itself

HOPE IT HELPS

Thanks man

In problem 2 , for the given input 4,5,6 don't you think that the output should be 0 as N=4 is already a good no. coz it can be represented as 1^1+3^1.

I forgot to add prime there..edited it though.

Just a try at this..Will this logic work? Bruteforce for all powers of 3 , 5 and 7 and just check which combination gives the smallest answer becoz we know that the number of elements in each of their power lists cannot be more than 64.?

An also how many questions were required to be solved for clearing this round ? 2 out of 3 ?

I guess all 3 will be required because first 2 were easy and the 3rd one was the only question which had extremely less fully accepted submissions.

I thought so XD. Too difficult to crack now it seems. Were there any partial points for the 3rd problem. XD

Yes partial pts were available for every question

Just precompute the sums of all the powers of 3 5 and 7 taking 2 at a time less than or equal to 1e9 then check between which 2 nos the N lies.

Yes this is also another bruteforce logic which can be used.

how to precompute the sums of all the powers of 3,5,7 taking 2 at a time ? I mean like for example we choose 3 and 5, then we can store all combinations of 3^x . 5^y in sorted order or store in a set but how to do it ? just like a loop

please do we need to do something like this ?

Hey,Can you help me with the second question Edit:Got it Thanks

In the second question, Is both prime numbers can be equal? Ex- 6 4 5

the answer is zero because 3^1+3^1 = 6.

No, the nos are different

just curious. In q1. can we just find if whole string contain 1 char only or it is alternate with 2 char. Is this the logic??

If the string contains the condition a[i]!=a[i+2] then the ans is false otherwise true

(1<<i)-1no of zeroes at ith place before the start of the patternLength of the pattern is (1<<(i+1))for every placetotal_no_of_complete_pattern (N)= (15 — ((1<<i) — 1) ) / ( 1<<i+1 )now we can calculate the 1 part for 2 part we havea partial pattern of length (L) = ( 15 — ((1<<i) — 1) ) % ( 1<<i+1 )now we can find no. of zeroes from it with simple if elseHi Everyone, please I want a verification for my method or point out if I am wrong. let $$$A_1,A_2...A_k$$$ are the numbers in range $$$[A,B]$$$ and similarly let $$$C_1,C_2...C_m$$$ are the numbers in range $$$[C,D]$$$ so basically I have to calculate $$$\displaystyle \sum\limits_{i = 1}^k\displaystyle \sum\limits_{j = 1}^m A_i \,xor\,C_j $$$. but we know that $$$A_i \,xor\,C_j=(A_i +C_j)-2(A_i\,and\,C_j)$$$ also we know that $$$A\,and\,B+A\,and\,B=A\,and\,(B+C)$$$. Now let $$$\displaystyle \sum\limits_{j = 1}^mC_j=S_1$$$. So our inner summation $$$\displaystyle \sum\limits_{j = 1}^m A_i \,xor\,C_j $$$ becomes $$$mA_i+S_1-2(A_i\,and\,S_1)$$$. Now let $$$\displaystyle \sum\limits_{i= 1}^kA_i=S_2$$$. So $$$\displaystyle \sum\limits_{i = 1}^k (mA_i+S_1-2(A_i\,and\,S_1))$$$ is our final answer and it equals $$$m*S_2+k*S_1-2*(S_1\,and\,S_2).$$$ Since we can find $$$S_1,S_2$$$ in constant time so time complexity is $$$O(1).$$$ Please point out anyone if I am wrong.

okay:

a ^ b = a + b — 2 * (a & b) [it's true]

you are writing: A and B + A and B = A and (B + C) [it is not true]

maybe you meant it?

A and B + A and C = A and (B + C)

however for A = B = C = 1

=> A and B + A and C = 1 + 1 = 2, A and (B + C) = 1 and 2 = 01 and 10 = 0, 2 != 0

similar properties are very beautiful, if you wanted to write something like that, please share. That's very beautiful

I am so sorry. Thanks for pointing. This mistake (A and B + A and C = A and (B + C)) came because of Electrical stuff because there only LEDs are either on/off(0/1 bit) and for this that expression is true. But here we have Integers. Sorry, my method is useless then.

Sorry,I didn't see the operator.Its exor I thought its addition. Pls don't read Try using prefix sum.Time complexity=O(n). Create a array of length=max_sum you can achieve+1 say the name of array is arr Initialize it with zeros Then run a loop from A to B: arr[i+C]+=1 arr[i+D+1]-=1 Then run a loop on array and perform arr[i]+=arr[i-1] Create a new var ans=0 now again iterate arr: and do ans+=arr[i]*i

Well O(n) will give TLE bro...constraints are upto 10^9.

CodeIf there is any mistake feel free to correct me :)

For each integer in range, a to b and range c to d find the number of integers having i-th (0 <= i <= 30) bit set and unset.

Let c11 denoted the count of integers in range a to b having i-th bit set. Then c12 = (b-a+1-c11) integers are having bits unset in this range.

Similarly, let c21 denoted the count of integers in range c to d having i-th bit set. Then c22 = (d-c+1-c21) integers are having bits unset in this range.

Now for each i-th bit in range (0,30). Add to answer the value equals ((c11*c22 + c12*c21) % mod) *(1LL<<i)) % mod where mod = 1e9+7.

Pseudo codeReference GFG article: Count integers up to N having K-th bit set in O(1)

understood the solution ! Is there some derivation on how

`((n>>(k+1))<<k)`

gives count of numbers in range [1 to n-1] having kth bit setIt's all about finding "how many times the pattern is repeating?". Pls, try to convince yourself by doing it with small numbers up to 50 or so.

Hello! Is there any way to check the code approach to this problem?

Note to me. My super weird approach: 122420626

Here is my approach for the problem:

Note considering the MOD part in the code (that can be included if required)

is this what they ask to be a driver?

so you do coding to get a driver job?

i guess