Minimum number of swaps to make two binary string equal

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:1Explanation:Swap s1[0] and s2[1].After swap s1 = 1011 and s2 = 1011Input:s1 = "00011", s2 = "01001"Output:2 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:**

- Following observation can be found:
- 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 equal` `int` `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 code` `int` `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` |

**Output:**

1

**Time Complexity:** O(n)

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