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 = 2
Output: 1
Explanation: 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=3
Output: 1
Naive 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 i
- Initialize 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++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find minimum Hamming// Distance after atmost one operationint minimumHammingDistance(string S, int K){ // Store the size of the string int n = S.size(); // Store the prefix sum of 1s int pref[n]; // Create Prefix Sum array pref[0] = S[0] - '0'; for (int i = 1; i < n; i++) pref[i] = pref[i - 1] + (S[i] - '0'); // Initialize cnt as number of ones // in string S int cnt = pref[n - 1]; // Store the required result int ans = cnt; // Traverse the string, S for (int i = 0; i < n - K; i++) { // Store the number of 1s in the // substring S[i, i+K-1] int value = pref[i + K - 1] - (i - 1 >= 0 ? pref[i - 1] : 0); // Update the answer ans = min(ans, cnt - value + (K - value)); } // Return the result return ans;}// Driver Codeint main(){ // Given Input string s = "101"; int K = 2; // Function Call cout << minimumHammingDistance(s, K); return 0;} |
Java
// Java program for the above approachpublic class GFG {// Function to find minimum Hamming// Distance after atmost one operationstatic int minimumHammingDistance(String S, int K){ // Store the size of the string int n = S.length(); // Store the prefix sum of 1s int []pref = new int [n]; // Create Prefix Sum array pref[0] = S.charAt(0) - '0'; for (int i = 1; i < n; i++) pref[i] = pref[i - 1] + (S.charAt(i) - '0'); // Initialize cnt as number of ones // in string S int cnt = pref[n - 1]; // Store the required result int ans = cnt; // Traverse the string, S for (int i = 0; i < n - K; i++) { // Store the number of 1s in the // substring S[i, i+K-1] int value = pref[i + K - 1] - (i - 1 >= 0 ? pref[i - 1] : 0); // Update the answer ans = Math.min(ans, cnt - value + (K - value)); } // Return the result return ans;}// Driver Codepublic static void main(String args[]){ // Given Input String s = "101"; int K = 2; // Function Call System.out.println(minimumHammingDistance(s, K)); }}// This code is contributed by SoumikMondal |
Python3
# Py program for the above approach# Function to find minimum Hamming# Distance after atmost one operationdef minimumHammingDistance(S, K): # Store the size of the string n = len(S) # Store the prefix sum of 1s pref = [0] * n # Create Prefix Sum array pref[0] = ord(S[0]) - ord('0') for i in range(1,n): pref[i] = pref[i - 1] + (ord(S[i]) - ord('0')) # Initialize cnt as number of ones # in string S cnt = pref[n - 1] # Store the required result ans = cnt # Traverse the string, S for i in range(n - K): # Store the number of 1s in the # substring S[i, i+K-1] value = pref[i + K - 1] - (pref[i-1] if (i - 1) >= 0 else 0) # Update the answer ans = min(ans, cnt - value + (K - value)) # Return the result return ans# Driver Codeif __name__ == '__main__': # Given Input s = "101" K = 2 # Function Call print (minimumHammingDistance(s, K))# This code is contributed by mohit kumar 29. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG {// Function to find minimum Hamming// Distance after atmost one operationstatic int minimumHammingDistance(string S, int K){ // Store the size of the string int n = S.Length; // Store the prefix sum of 1s int []pref = new int [n]; // Create Prefix Sum array pref[0] = (int)S[0] - 48; for (int i = 1; i < n; i++) pref[i] = pref[i - 1] + ((int)S[i] - 48); // Initialize cnt as number of ones // in string S int cnt = pref[n - 1]; // Store the required result int ans = cnt; // Traverse the string, S for (int i = 0; i < n - K; i++) { // Store the number of 1s in the // substring S[i, i+K-1] int value = pref[i + K - 1] - (i - 1 >= 0 ? pref[i - 1] : 0); // Update the answer ans = Math.Min(ans, cnt - value + (K - value)); } // Return the result return ans;}// Driver Codepublic static void Main(){ // Given Input string s = "101"; int K = 2; // Function Call Console.Write(minimumHammingDistance(s, K)); }}// This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script> // JavaScript program for the above approach// Function to find minimum Hamming// Distance after atmost one operationfunction minimumHammingDistance(S, K){ // Store the size of the string let n = S.length; // Store the prefix sum of 1s let pref = new Array(n); // Create Prefix Sum array pref[0] = S[0] - '0'; for (let i = 1; i < n; i++) pref[i] = pref[i - 1] + (S[i] - '0'); // Initialize cnt as number of ones // in string S let cnt = pref[n - 1]; // Store the required result let ans = cnt; // Traverse the string, S for (let i = 0; i < n - K; i++) { // Store the number of 1s in the // substring S[i, i+K-1] let value = pref[i + K - 1] - (i - 1 >= 0 ? pref[i - 1] : 0); // Update the answer ans = Math.min(ans, cnt - value + (K - value)); } // Return the result return ans;}// Driver Code // Given Input let s = "101"; let K = 2; // Function Call document.write(minimumHammingDistance(s, K)); </script> |
Output
2
Time Complexity: O(N)
Auxiliary Space: O(N)
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