Given two strings S and T of equal length. Both strings contain only the characters ‘0’ and ‘1’. The task is to find the minimum number of operations to convert string S to T. There are 2 types of operations allowed on string S:
- Swap any two characters of the string.
- Replace a ‘0’ with a ‘1’ or vice versa.
Examples:
Input: S = “011”, T = “101”
Output: 1
Swap the first and second character.Input: S = “010”, T = “101”
Output: 2
Swap the first and second character and replace the third character with ‘1’.
Approach: Find 2 values for the string S, the number of indices that have 0 but should be 1 and the number of indices that have 1 but should be 0. The result would be the maximum of these 2 values since we can use swaps on the minimum of these 2 values and the remaining unmatched characters can be inverted i.e. ‘0’ can be changed to ‘1’ and ‘1’ can be changed to ‘0’.
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 operations // of the given type required to convert // string s to string t int minOperations(string s, string t, int n) { int ct0 = 0, ct1 = 0; for (int i = 0; i < n; i++) { // Characters are already equal if (s[i] == t[i]) continue; // Increment count of 0s if (s[i] == '0') ct0++; // Increment count of 1s else ct1++; } return max(ct0, ct1); } // Driver code int main() { string s = "010", t = "101"; int n = s.length(); cout << minOperations(s, t, n); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG { // Function to return the minimum // operations of the given type required // to convert string s to string t static int minOperations(String s, String t, int n) { int ct0 = 0, ct1 = 0; for (int i = 0; i < n; i++) { // Characters are already equal if (s.charAt(i) == t.charAt(i)) continue; // Increment count of 0s if (s.charAt(i) == '0') ct0++; // Increment count of 1s else ct1++; } return Math.max(ct0, ct1); } // Driver code public static void main(String args[]) { String s = "010", t = "101"; int n = s.length(); System.out.println(minOperations(s, t, n)); } } // This code is contributed by // Surendra_Gangwar |
Python 3
# Python 3 implementation of the approach # Function to return the minimum operations # of the given type required to convert # string s to string t def minOperations(s, t, n): ct0 = 0 ct1 = 0 for i in range(n): # Characters are already equal if (s[i] == t[i]): continue # Increment count of 0s if (s[i] == '0'): ct0 += 1 # Increment count of 1s else: ct1 += 1 return max(ct0, ct1) # Driver code if __name__ == "__main__": s = "010" t = "101" n = len(s) print(minOperations(s, t, n)) # This code is contributed by ita_c |
C#
// C# implementation of the approach using System; class GFG { // Function to return the minimum operations // of the given type required to convert // string s to string t static int minOperations(string s, string t, int n) { int ct0 = 0, ct1 = 0; for (int i = 0; i < n; i++) { // Characters are already equal if (s[i] == t[i]) continue; // Increment count of 0s if (s[i] == '0') ct0++; // Increment count of 1s else ct1++; } return Math.Max(ct0, ct1); } // Driver code public static void Main() { string s = "010", t = "101"; int n = s.Length; Console.Write(minOperations(s, t, n)); } } // This code is contributed // by Akanksha Rai |
PHP
<?php // PHP implementation of the approach // Function to return the minimum operations // of the given type required to convert // string s to string t function minOperations($s, $t, $n) { $ct0 = 0 ; $ct1 = 0; for ($i = 0; $i < $n; $i++) { // Characters are already equal if ($s[$i] == $t[$i]) continue; // Increment count of 0s if ($s[$i] == '0') $ct0++; // Increment count of 1s else $ct1++; } return max($ct0, $ct1); } // Driver code $s = "010"; $t = "101"; $n = strlen($s); echo minOperations($s, $t, $n); // This code is contributed by Ryuga ?> |
2
Time Complexity: O(N)
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