Given a binary string str of length N and two integers A and B such that 0 ? A < B < n. The task is to count the minimum number of operations on the string such that it gives 10A as remainder when divided by 10B. An operation means changing 1 to 0 or 0 to 1.
Examples:
Input: str = “1001011001”, A = 3, B = 6
Output: 2
The string after 2 operations is 1001001000.
1001001000 % 106 = 103Input: str = “11010100101”, A = 1, B = 5
Output: 3
Approach: In order for the number to give 10A as remainder when divided by 10B, the last B digits of the string has to be 0 except the digit at (A + 1)th position from the last which should be 1. Therefore, check the last B digits of the string for the above condition and increase the count by 1 for each mismatch of digit.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the minimum number// of operations on a binary string such that// it gives 10^A as remainder when divided by 10^Bint findCount(string s, int n, int a, int b){ // Initialize result int res = 0; // Loop through last b digits for (int i = 0; i < b; i++) { if (i == a) res += (s[n - i - 1] != '1'); else res += (s[n - i - 1] != '0'); } return res;}// Driver codeint main(){ string str = "1001011001"; int N = str.size(); int A = 3, B = 6; cout << findCount(str, N, A, B); return 0;} |
Java
// Java implementation of the approach import java.util.*;class GFG{ // Function to return the minimum number // of operations on a binary string such that // it gives 10^A as remainder when divided by 10^B static int findCount(String s, int n, int a, int b) { // Initialize result int res = 0; char []s1 = s.toCharArray(); // Loop through last b digits for (int i = 0; i < b; i++) { if (i == a) { if (s1[n - i - 1] != '1') res += 1; } else { if (s1[n - i - 1] != '0') res += 1 ; } } return res; } // Driver code static public void main (String []args) { String str = "1001011001"; int N = str.length() ; int A = 3, B = 6; System.out.println(findCount(str, N, A, B)); }}// This code is contributed by ChitraNayal |
Python3
# Python 3 implementation of the approach# Function to return the minimum number# of operations on a binary string such that# it gives 10^A as remainder when divided by 10^Bdef findCount(s, n, a, b): # Initialize result res = 0 # Loop through last b digits for i in range(b): if (i == a): res += (s[n - i - 1] != '1') else: res += (s[n - i - 1] != '0') return res# Driver codeif __name__ == '__main__': str = "1001011001" N = len(str) A = 3 B = 6 print(findCount(str, N, A, B))# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approach using System;class GFG{ // Function to return the minimum number // of operations on a binary string such that // it gives 10^A as remainder when divided by 10^B static int findCount(string s, int n, int a, int b) { // Initialize result int res = 0; // Loop through last b digits for (int i = 0; i < b; i++) { if (i == a) { if (s[n - i - 1] != '1') res += 1; } else { if (s[n - i - 1] != '0') res += 1 ; } } return res; } // Driver code static public void Main () { string str = "1001011001"; int N = str.Length ; int A = 3, B = 6; Console.WriteLine(findCount(str, N, A, B)); }}// This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript implementation of the approach// Function to return the minimum number// of operations on a binary string such that// it gives 10^A as remainder when divided by 10^Bfunction findCount(s, n, a, b){ // Initialize result var res = 0; // Loop through last b digits for(var i = 0; i < b; i++) { if (i == a) res += (s[n - i - 1] != '1'); else res += (s[n - i - 1] != '0'); } return res;}// Driver codevar str = "1001011001";var N = str.length;var A = 3, B = 6;document.write(findCount(str, N, A, B));// This code is contributed by itsok</script> |
Output:
2
Time Complexity: O(N )
Auxiliary Space: O(1)
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