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’.
Brute Force:
To solve this problem, we can use a greedy approach. We can iterate over both strings and keep track of the number of differences between the corresponding characters in both strings. Let diffCount be the count of differences between the corresponding characters in s and t.
We can perform the following operations to reduce the value of diffCount:
- Swap the two different characters in s and t.
- Replace a character in s or t with its complement.
We can choose the operation that reduces diffCount the most at each step. We repeat this process until diffCount becomes 0.
Here’s the step-by-step approach:
- Initialize diffCount to 0.
- Iterate over both strings and count the number of differences between the corresponding characters in both strings. Update diffCount accordingly.
- While diffCount is greater than 0:
- If diffCount is 2 or more, find the two different characters in s and t that occur at the same position and swap them. Update diffCount accordingly.
- f diffCount is 1, find the different character in s and t that occurs at the same position and replace it with its complement. Update diffCount accordingly.
- Repeat steps 2-5 until diffCount becomes 0.
- Return the number of operations performed.
Below is the implementation of the above approach:
C
#include <stdio.h>#include <string.h>// Function to return the minimum operations// of the given type required to convert// string s to string tint minOperations(char s[], char t[], int n){ int diffCount = 0; for (int i = 0; i < n; i++) { if (s[i] != t[i]) { diffCount++; } } int operations = 0; while (diffCount > 0) { if (diffCount >= 2) { // Find the two different characters // in s and t that occur at the same position int pos1 = -1, pos2 = -1; for (int i = 0; i < n; i++) { if (s[i] != t[i]) { if (pos1 == -1) { pos1 = i; } else { pos2 = i; break; } } } // Swap the two different characters char temp = s[pos1]; s[pos1] = s[pos2]; s[pos2] = temp; temp = t[pos1]; t[pos1] = t[pos2]; t[pos2] = temp; // Update the difference count diffCount -= 2; operations++; } else { // Find the different character in s and t // that occurs at the same position int pos = -1; for (int i = 0; i < n; i++) { if (s[i] != t[i]) { pos = i; break; } } // Replace the different character with its // complement s[pos] = (s[pos] == '0') ? '1' : '0'; // Update the difference count diffCount--; operations++; } } return operations;}// Driver codeint main(){ char s[] = "010", t[] = "101"; int n = strlen(s); printf("%d", minOperations(s, t, n)); return 0;} |
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 tint minOperations(string s, string t, int n){ int diffCount = 0; for (int i = 0; i < n; i++) { if (s[i] != t[i]) { diffCount++; } } int operations = 0; while (diffCount > 0) { if (diffCount >= 2) { // Find the two different characters // in s and t that occur at the same position int pos1 = -1, pos2 = -1; for (int i = 0; i < n; i++) { if (s[i] != t[i]) { if (pos1 == -1) { pos1 = i; } else { pos2 = i; break; } } } // Swap the two different characters swap(s[pos1], s[pos2]); swap(t[pos1], t[pos2]); // Update the difference count diffCount -= 2; operations++; } else { // Find the different character in s and t // that occurs at the same position int pos = -1; for (int i = 0; i < n; i++) { if (s[i] != t[i]) { pos = i; break; } } // Replace the different character with its complement s[pos] = (s[pos] == '0') ? '1' : '0'; // Update the difference count diffCount--; operations++; } } return operations;}// Driver codeint main(){ string s = "010", t = "101"; int n = s.length(); cout << minOperations(s, t, n); return 0;} |
Java
import java.util.*;public class Main { // 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 diffCount = 0; for (int i = 0; i < n; i++) { if (s.charAt(i) != t.charAt(i)) { diffCount++; } } int operations = 0; while (diffCount > 0) { if (diffCount >= 2) { // Find the two different characters // in s and t that occur at the same // position int pos1 = -1, pos2 = -1; for (int i = 0; i < n; i++) { if (s.charAt(i) != t.charAt(i)) { if (pos1 == -1) { pos1 = i; } else { pos2 = i; break; } } } // Swap the two different characters char[] sChars = s.toCharArray(); char[] tChars = t.toCharArray(); char temp = sChars[pos1]; sChars[pos1] = sChars[pos2]; sChars[pos2] = temp; temp = tChars[pos1]; tChars[pos1] = tChars[pos2]; tChars[pos2] = temp; s = new String(sChars); t = new String(tChars); // Update the difference count diffCount -= 2; operations++; } else { // Find the different character in s and t // that occurs at the same position int pos = -1; for (int i = 0; i < n; i++) { if (s.charAt(i) != t.charAt(i)) { pos = i; break; } } // Replace the different character with its // complement char[] sChars = s.toCharArray(); if (s.charAt(pos) == '0') { sChars[pos] = '1'; } else { sChars[pos] = '0'; } s = new String(sChars); // Update the difference count diffCount--; operations++; } } return operations; } // Driver code public static void main(String[] args) { String s = "010", t = "101"; int n = s.length(); System.out.println(minOperations(s, t, n)); }} |
Python3
# Python implementation of the approach# Function to return the minimum operations# of the given type required to convert# string s to string tdef minOperations(s, t, n): diffCount = 0 for i in range(n): if s[i] != t[i]: diffCount += 1 operations = 0 while diffCount > 0: if diffCount >= 2: # Find the two different characters # in s and t that occur at the same position pos1, pos2 = -1, -1 for i in range(n): if s[i] != t[i]: if pos1 == -1: pos1 = i else: pos2 = i break # Swap the two different characters s = s[:pos1] + t[pos1] + s[pos1+1:pos2] + t[pos2] + s[pos2+1:] t = t[:pos1] + s[pos1] + t[pos1+1:pos2] + s[pos2] + t[pos2+1:] # Update the difference count diffCount -= 2 operations += 1 else: # Find the different character in s and t # that occurs at the same position pos = -1 for i in range(n): if s[i] != t[i]: pos = i break # Replace the different character with its complement s = s[:pos] + ('1' if s[pos] == '0' else '0') + s[pos+1:] # Update the difference count diffCount -= 1 operations += 1 return operations# Driver codes = "010"t = "101"n = len(s)print(minOperations(s, t, n)) |
C#
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 diffCount = 0; for (int i = 0; i < n; i++) { if (s[i] != t[i]) { diffCount++; } } int operations = 0; while (diffCount > 0) { if (diffCount >= 2) { // Find the two different characters // in s and t that occur at the same position int pos1 = -1, pos2 = -1; for (int i = 0; i < n; i++) { if (s[i] != t[i]) { if (pos1 == -1) { pos1 = i; } else { pos2 = i; break; } } } // Swap the two different characters char[] sArr = s.ToCharArray(); char[] tArr = t.ToCharArray(); char temp = sArr[pos1]; sArr[pos1] = sArr[pos2]; sArr[pos2] = temp; s = new string(sArr); temp = tArr[pos1]; tArr[pos1] = tArr[pos2]; tArr[pos2] = temp; t = new string(tArr); // Update the difference count diffCount -= 2; operations++; } else { // Find the different character in s and t // that occurs at the same position int pos = -1; for (int i = 0; i < n; i++) { if (s[i] != t[i]) { pos = i; break; } } // Replace the different character with its complement char[] sArr = s.ToCharArray(); sArr[pos] = sArr[pos] == '0' ? '1' : '0'; s = new string(sArr); // Update the difference count diffCount--; operations++; } } return operations; } // Driver code static public void Main() { string s = "010", t = "101"; int n = s.Length; Console.WriteLine(minOperations(s, t, n)); }}// This code is contributed by sarojmcy2e |
Javascript
// JavaScript implementation of the approach// Function to return the minimum operations// of the given type required to convert// string s to string tfunction minOperations(s, t, n) {let diffCount = 0;for (let i = 0; i < n; i++) {if (s[i] !== t[i]) {diffCount += 1;}}let operations = 0;while (diffCount > 0) { if (diffCount >= 2) { // Find the two different characters // in s and t that occur at the same position let pos1 = -1, pos2 = -1; for (let i = 0; i < n; i++) { if (s[i] !== t[i]) { if (pos1 === -1) { pos1 = i; } else { pos2 = i; break; } } } // Swap the two different characters s = s.substring(0, pos1) + t.charAt(pos1) + s.substring(pos1+1, pos2) + t.charAt(pos2) + s.substring(pos2+1); t = t.substring(0, pos1) + s.charAt(pos1) + t.substring(pos1+1, pos2) + s.charAt(pos2) + t.substring(pos2+1); // Update the difference count diffCount -= 2; operations += 1; } else { // Find the different character in s and t // that occurs at the same position let pos = -1; for (let i = 0; i < n; i++) { if (s[i] !== t[i]) { pos = i; break; } } // Replace the different character with its complement s = s.substring(0, pos) + (s.charAt(pos) === '0' ? '1' : '0') + s.substring(pos+1); // Update the difference count diffCount -= 1; operations += 1; }}return operations;}// Driver codelet s = "010";let t = "101";let n = s.length;console.log(minOperations(s, t, n)); |
Output
2
Time Complexity: O(n^2) due to the nested loops used to generate all possible swaps and replacements.
Auxiliary Space: O(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
#include <stdio.h>#include <string.h>// Function to return the minimum operations// of the given type required to convert// string s to string tint minOperations(char* s, char* 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 (ct0 > ct1) ? ct0 : ct1;}// Driver codeint main(){ char s[] = "010"; char t[] = "101"; int n = strlen(s); printf("%d", minOperations(s, t, n)); return 0;} |
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 tint 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 codeint main(){ string s = "010", t = "101"; int n = s.length(); cout << minOperations(s, t, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{// Function to return the minimum // operations of the given type required // to convert string s to string tstatic 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 codepublic 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 |
Python3
# Python 3 implementation of the approach# Function to return the minimum operations# of the given type required to convert# string s to string tdef 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 codeif __name__ == "__main__": s = "010" t = "101" n = len(s) print(minOperations(s, t, n))# This code is contributed by ita_c |
Javascript
<script>// JavaScript implementation of the approach// Function to return the minimum operations// of the given type required to convert// string s to string tfunction minOperations(s, t, n) { var ct0 = 0, ct1 = 0; for(var 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 codevar s = "010",t = "101";var n = s.length;document.write(minOperations(s, t, n));// This code is contributed by rdtank</script> |
C#
// C# implementation of the approachusing System;class GFG{// Function to return the minimum operations// of the given type required to convert// string s to string tstatic 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 codepublic 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?> |
Output
2
Time Complexity: O(N)
Auxiliary Space: O(1) it is using constant space for variables
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