Minimize hamming distance in Binary String by setting only one K size substring bits

Minimize hamming distance in Binary String by setting only one K size substring bits Given two binary strings S and T of length N and a positive integer K. Initially, all characters of T are ‘0’. The task is to find the minimum Hamming distance after choosing a substring of size K and making all elements of string T as ‘1’ only once.Examples:Input: S = “101”, K = 2Output: 1Explanation: Initially string T = “000”, one possible way is to change all 0s in range [0, 1] to 1. Thus string T becomes “110” and the hamming distance between S and T is 2 which is the minimum possible.Input: S = “1100”, K=3Output: 1Naive Approach: The simplest approach is to consider every substring of size K and make all the elements as 1 and then check the hamming distance with string, S. After checking all the substrings, print the minimum hamming distance.Time Complexity: O(N×K)Auxiliary Space: O(1)Approach: This problem can be solved by creating a prefix array sum which stores the prefix sum of the count of ones in the string S. Follow the steps below to solve the problem:Create a prefix sum array pref[] of string S by initializing pref[0] as 0 updating pref[i] as pref[i-1] +(S[i] – ‘0’) for every index i.Store the total count of ones in the string, S in a variable cnt.Initialize a variable ans as cnt to store the required result.Iterate in the range [0, N-K] using the variable iInitialize a variable val as pref[i+K-1] – pref[i-1] to store the count of ones in the substring S[i, i+K-1].Create two variables A and B to store the hamming distance outside the current substring and the hamming distance inside the current substring and initialize A with cnt – K and B with K – val.Update the value of ans with the minimum of ans and (A + B).Print the value of ans as the result.Below is the implementation of the above approach:C++#include using namespace std;  int minimumHammingDistance(string S, int K){        int n = S.size();          int pref[n];          pref[0] = S[0] – ‘0’;    for (int i = 1; i < n; i++)        pref[i] = pref[i - 1] + (S[i] - '0');              int cnt = pref[n - 1];          int ans = cnt;          for (int i = 0; i < n - K; i++) {                          int value = pref[i + K - 1]                    - (i - 1 >= 0 ? pref[i – 1] : 0);                  ans = min(ans, cnt – value + (K – value));    }          return ans;}  int main(){          string s = “101”;    int K = 2;          cout