basantCoder's blog

By basantCoder, history, 5 weeks ago, In English

You start with 0 and you have to Find the minimum number of operations to reach 'n'. In one operation you can add or subtract any 2^k to your current number. Where k can be any positive integer.
Constraints:
test cases <200,000;
1 <= | n | <= 1,000,000,000 . Example:
INPUT:
First line is no. of test cases, then a single integer 'n' on each line.
5
1
2
0
6
10
OUTPUT:
1
1
0
2
2
Can anybody help??

 
 
 
 
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5 weeks ago, # |
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Auto comment: topic has been updated by basantCoder (previous revision, new revision, compare).

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5 weeks ago, # |
Rev. 4   Vote: I like it 0 Vote: I do not like it

Why not try to reach from n to 0 by subtracting nearest power of 2(floor)

long long n,ans=0;
cin>>n;
while(n>0){
    n = n - pow(2,(int)__lg(n));
    ans++;
}
cout<<ans<<'\n';

Maybe Like this...Hope its not live Rn

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    5 weeks ago, # ^ |
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    its not always optimal!
    Like for 28 your code will give 3. But you can reach 28 is 2 operations also.

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5 weeks ago, # |
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As any n can be represented in binary form which is nothing but sum of powers of 2. So the answer will be number of 1s in binary notation of n. Example: $$$10 = (1010)_2$$$ Here count of 1s = 2 = answer

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    5 weeks ago, # ^ |
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    for example take n=28, binary form = (11100)2
    According to your logic: answer = 3.
    But correct answer is 2. First operation add 32, second operation substract 4.

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      5 weeks ago, # ^ |
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      Your binary representation of 28 is wrong.

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      5 weeks ago, # ^ |
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      Sorry, I missed this case. I think we can always create consecutive 1's by subtracting

      $$$ 0 - 1 = 10 $$$

      (by taking a carry) in binary form, provided there is a 1 in the left.
      That is we can always replace continuous 1s with 2 operations, putting a 1 in the left and subtracting it by 1. Example: We want 1110

      $$$ 10000 - 10 = 1110 $$$

      So either it is 2 operations or (no of 1 bits) = x operations. We have to take minimum of it and sum all of such operations on consecutive 1s in the binary representation of n to get the answer.

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    5 weeks ago, # ^ |
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    codinginveins Consider n=7 According to your logic it will give 3 , but the correct answer is 2 (Add 8 then subtract 1).

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      5 weeks ago, # ^ |
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      https://codeforces.com/blog/entry/95133?#comment-841488

      I still dont know if I am missing something here.

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        5 weeks ago, # ^ |
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        actually we can get cosecutive 1's and 0's format in 2 operations. Like is the number is (11111110000000)2. But still there can be complicated cases with random 1's and 0's where I am stuck.

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          5 weeks ago, # ^ |
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          $$$ 111000111000111000_2 $$$ can be divided and conquered as sum of answers of $$$ 111000_2 $$$, $$$ 111000000000_2 $$$ and $$$ 111000000000000000_2 $$$. Infact it will also be equivalent to

          $$$ ans(111000_2) + ans(111000_2) + ans(111000_2) = 3 * ans(111_2) $$$

          We just need to process the segments where we get 1s.
          ans(1) = 1 ans(11) = 2 ans(111) = 2 ans(1111) = 2 and so on

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5 weeks ago, # |
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int recur(n){ if(n == 0) return 0; else if((n & (n - 1)) == 0) return 1; int temp 1e9; for(int i = 0; i <= 14; i++){ int val = abs(n - (1 << i)); temp = min(temp, 1 + recur(val)); } return temp; } void solve() { ll n; cin >> n; n = abs(n); cout << recur(n); }

to reduce time complexity you can do memoization

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5 weeks ago, # |
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Let, f(x) = Number of set bit's in binary representation of x. Answer will be min(f(x), f(-x)).

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5 weeks ago, # |
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Generate a vector with all powers of 2 less than equal to 1,000,000,000. Then use while(n > 0) n = n — lower_bound(n); Where lower bound on the generated array will return the power just less than N. This way you'll get the most optimal solution.

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5 weeks ago, # |
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Had a hard time reading the title