Given two binary strings of equal length, the task is to find the minimum number of swaps to make them equal. Swapping between two characters from two different strings is only allowed, return -1 if strings can’t be made equal.
Examples:
Input: s1 = “0011”, s2 = “1111”
Output: 1
Explanation:
Swap s1[0] and s2[1].After swap
s1 = 1011 and s2 = 1011Input: s1 = “00011”, s2 = “01001”
Output: 2
Explanation:
Swap s1[1] and s2[1]. After swap
s1 = 01011, s2 = 00001
Swap s1[3] and s2[1]. After swap,
s1 = 01001, s2 = 01001
Approach:
- Swapping of s1[i] and s2[j] is allowed, so we need to find in which positions two strings are different. If s1[i] and s2[i] are the same, we do not swap them.
- If s1[i] and s2[i] are not the same then we can find 3 patterns:
- 00 and 11, we can perform a diagonal swap and the result will be 01 01 or 10 10. In the case of the diagonal swap, we need to form pairs to solve the disproportion of this type. Swap required to restore a single pair is 1.
- 11 and 00, we can perform a diagonal swap and the result will be 01 01 or 10 10. In the case of the diagonal swap, we need to form pairs to solve the disproportion of this type. Swap required to restore a single pair is 1.
- 10 and 01, we can perform a vertical swap and the result will be 00 11 or 11 00 and this type will be transformed into type 1 or type 2 problem and another diagonal swap to make them equal. we need to form pair to solve disproportion of this type. Swap required to restore a single pair is 2.
- From the above observation, we can follow the below greedy method:
- Count the positions where s1[i] = 0, and s2[i] = 1 (count0). We need (count0)/2 number of diagonal swaps in each pair of type 1.
- Count the positions where s1[i] =1, and s2[i] = 0 (count1). We need (count1)/2 number of diagonal swaps in each pair of type 2.
- If both count0 and count1 are even we can output the answer = ((count0) + (count1))/2. If both count0 and count1 is odd, we can have one single pair of type 3 disproportion, so we need 2 extra swaps. Output the answer as ((count0) + (count1))/2 + 2. If one of the count0 and count1 is odd, we can not make two strings equal. As in all cases we need to form pairs, odd count means a single position will be left alone making 2 strings unequal.
Below is the implementation of above approach:
C++
// C++ program for// the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate// min swaps to make// binary strings equalint minSwaps(string& s1, string& s2){ int c0 = 0, c1 = 0; for (int i = 0; i < s1.size(); i++) { // Count of zero's if (s1[i] == '0' && s2[i] == '1') { c0++; } // Count of one's else if (s1[i] == '1' && s2[i] == '0') { c1++; } } // As discussed // above int ans = c0 / 2 + c1 / 2; if (c0 % 2 == 0 && c1 % 2 == 0) { return ans; } else if ((c0 + c1) % 2 == 0) { return ans + 2; } else { return -1; }}// Driver codeint main(){ string s1 = "0011", s2 = "1111"; int ans = minSwaps(s1, s2); cout << ans << '\n'; return 0;} |
Java
// Java program for the above approach class GFG { // Function to calculate // min swaps to make // binary strings equal static int minSwaps(String s1, String s2) { int c0 = 0, c1 = 0; for (int i = 0; i < s1.length(); i++) { // Count of zero's if (s1.charAt(i) == '0' && s2.charAt(i) == '1') { c0++; } // Count of one's else if (s1.charAt(i) == '1' && s2.charAt(i) == '0') { c1++; } } // As discussed // above int ans = c0 / 2 + c1 / 2; if (c0 % 2 == 0 && c1 % 2 == 0) { return ans; } else if ((c0 + c1) % 2 == 0) { return ans + 2; } else { return -1; } } // Driver code public static void main (String[] args) { String s1 = "0011", s2 = "1111"; int ans = minSwaps(s1, s2); System.out.println(ans); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 program for # the above approach # Function to calculate # min swaps to make # binary strings equal def minSwaps(s1, s2) : c0 = 0; c1 = 0; for i in range(len(s1)) : # Count of zero's if (s1[i] == '0' and s2[i] == '1') : c0 += 1; # Count of one's elif (s1[i] == '1' and s2[i] == '0') : c1 += 1; # As discussed above ans = c0 // 2 + c1 // 2; if (c0 % 2 == 0 and c1 % 2 == 0) : return ans; elif ((c0 + c1) % 2 == 0) : return ans + 2; else : return -1; # Driver code if __name__ == "__main__" : s1 = "0011"; s2 = "1111"; ans = minSwaps(s1, s2); print(ans); # This code is contributed by AnkitRai01 |
C#
// C# program for the above approach using System;class GFG { // Function to calculate // min swaps to make // binary strings equal static int minSwaps(string s1, string s2) { int c0 = 0, c1 = 0; for (int i = 0; i < s1.Length; i++) { // Count of zero's if (s1[i] == '0' && s2[i] == '1') { c0++; } // Count of one's else if (s1[i] == '1' && s2[i] == '0') { c1++; } } // As discussed // above int ans = c0 / 2 + c1 / 2; if (c0 % 2 == 0 && c1 % 2 == 0) { return ans; } else if ((c0 + c1) % 2 == 0) { return ans + 2; } else { return -1; } } // Driver code public static void Main () { string s1 = "0011", s2 = "1111"; int ans = minSwaps(s1, s2); Console.WriteLine(ans); } }// This code is contributed by AnkitRai01 |
Javascript
// Javascript program for the above approach// Function to calculate// min swaps to make// binary strings equalfunction minSwaps(s1, s2){ let c0 = 0; let c1 = 0; for (let i = 0; i < s1.length; i++) { // Count of zero's if (s1[i] == '0' && s2[i] == '1') { c0++; } // Count of one's else if (s1[i] == '1' && s2[i] == '0') { c1++; } } // As discussed // above let ans = c0 / 2 + c1 / 2; if (c0 % 2 == 0 && c1 % 2 == 0) { return ans; } else if ((c0 + c1) % 2 == 0) { return ans + 2; } else { return -1; }}// Driver codelet s1 = "0011";let s2 = "1111";let ans = minSwaps(s1, s2);console.log(ans);// This code is contributed by Samim Hossain Mondal. |
Output:
1
Time Complexity: O(n)
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Another Approach:
- Initialize two empty vectors pos1 and pos2.
- Initialize an integer variable diff to 0.
- Traverse both strings s1 and s2 from index 0 to n-1, where n is the length of the strings.
- If the characters at index i in s1 and s2 are not equal, increment diff by 1 and do the following:
a. If the character at index i in s1 is ‘0’, append i to the pos1 vector.
b. Else, append i to the pos2 vector. - If diff is odd, return -1.
- Initialize a swaps variable to 0.
- Traverse the pos1 vector from index 0 to pos1.size()-1, incrementing i by 2 in each iteration.
- If the character at index pos1[i] in s2 is ‘1’, increment swaps by 1.
- Traverse the pos2 vector from index 0 to pos2.size()-1, incrementing i by 2 in each iteration.
- If the character at index pos2[i] in s1 is ‘1’, increment swaps by 1.
- Return the swaps variable.
Below is the implementation of the above approach:
C++
// C++ program for// the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate// min swaps to make// binary strings equalint minSwaps(string s1, string s2) { int n = s1.length(); int diff = 0; vector<int> pos1, pos2; for (int i = 0; i < n; i++) { if (s1[i] != s2[i]) { diff++; if (s1[i] == '0') { pos1.push_back(i); } else { pos2.push_back(i); } } } if (diff % 2 != 0) { return -1; } int swaps = 0; for (int i = 0; i < pos1.size(); i += 2) { if (s2[pos1[i]] == '1') { swaps++; } } for (int i = 0; i < pos2.size(); i += 2) { if (s1[pos2[i]] == '1') { swaps++; } } return swaps;}// Driver codeint main(){ string s1 = "0011", s2 = "1111"; int ans = minSwaps(s1, s2); cout << ans << '\n'; return 0;} |
Output
1
Time complexity: O(n)
Auxiliary Space: O(n)
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