Finding the minimum number of swaps required for a substring to contain exactly K ones is a common problem in computer science and programming. In this article, we will delve into this problem and provide a C solution for it. This question has applications in various fields, including string manipulation, data structure optimization, and coding challenges in interviews.
Given a binary string and a number K, the task is to find the minimum number of swaps required to ensure that each substring of the string has exactly K 1's.
To solve this problem, we can use the two-pointer method and sliding window technology. The basic idea is to maintain a window of size K and calculate the number of exchanges required for all 1s in the window.
This is a C function that implements the above method -
#include<bits/stdc++.h> using namespace std; int minSwaps(string s, int K) { int n = s.length(); vector<int> onesPrefix(n, 0); if(s[0] == '1') onesPrefix[0] = 1; for(int i = 1; i < n; i++) { onesPrefix[i] = onesPrefix[i-1]; if(s[i] == '1') onesPrefix[i]++; } int ans = INT_MAX; for(int i = 0; i <= n - K; i++) { int j = i + K - 1; int ones = onesPrefix[j] - ((i == 0) ? 0 : onesPrefix[i - 1]); ans = min(ans, K - ones); } return ans; } int main() { string s = "10010110"; int K = 3; cout << "Minimum number of swaps = " << minSwaps(s, K) << endl; return 0; }
Minimum number of swaps = 1
Assume the string is "10010110", K = 3.
In the initial binary string "10010110", we want each substring of size 3 to have exactly 3 1's. For example, the substring "100" requires 2 exchanges to become "111". Similarly, the substring "001" also requires 2 exchanges. By iterating over the string, we find that the minimum number of swaps required for the substring "101" is 1.
This question is a great example of how to combine algorithms, data structures, and an understanding of the C language to solve complex problems. Understanding and implementing such questions can be very beneficial for software engineers, especially in coding interviews and competitive programming.
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