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Minimum swaps required to make a binary string alternating
• Difficulty Level : Hard
• Last Updated : 28 Dec, 2018

You are given a binary string of even length and equal number of 0’s and 1’s. What is the minimum number of swaps to make the string alternating. A binary string is alternating if no two consecutive elements are equal.

Examples:

```Input : 000111
Output : 1
Explanation : Swap index 2 and index 5 to get 010101

Input : 1010
Output : 0
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We may either get a 1 at first position or a zero at first position. We consider two cases and find the minimum of two cases. Note that it is given that there are equal of numbers of 1’s and 0’s in the string and string is of even length.

1. Count number of zeroes at odd position and even position of the string. Let their count be odd_0 and even_0 respectively.
2. Count number of ones at odd position and even position of the string. Let their count be odd_1 and even_1 respectively.
3. We will always swap a 1 with a 0 (never a 1 with a 1 or a 0 with a 0). So we just check if our alternating string starts with 0 then the number of swaps is min(even_0, odd_1) and if our alternating string starts with 1 then the number of swaps is min(even_1, odd_0). The answer is min of these two .

Below is the implementation of above approach:

## C++

 `// CPP implementation of the approach``#include``using` `namespace` `std;`` ` `    ``// returns the minimum number of swaps``    ``// of a binary string``    ``// passed as the argument``    ``// to make it alternating``    ``int` `countMinSwaps(string st)``    ``{`` ` `        ``int` `min_swaps = 0;`` ` `        ``// counts number of zeroes at odd ``        ``// and even positions``        ``int` `odd_0 = 0, even_0 = 0;`` ` `        ``// counts number of ones at odd ``        ``// and even positions``        ``int` `odd_1 = 0, even_1 = 0;`` ` `        ``int` `n = st.length();``        ``for` `(``int` `i = 0; i < n; i++) {``            ``if` `(i % 2 == 0) { ``                ``if` `(st[i] == ``'1'``) ``                    ``even_1++;``                ``else``                    ``even_0++;``            ``}``            ``else` `{ ``                ``if` `(st[i] == ``'1'``) ``                    ``odd_1++;``                ``else``                    ``odd_0++;``            ``}``        ``}`` ` `        ``// alternating string starts with 0``        ``int` `cnt_swaps_1 = min(even_0, odd_1); `` ` `        ``// alternating string starts with 1``        ``int` `cnt_swaps_2 = min(even_1, odd_0); `` ` `        ``// calculates the minimum number of swaps``        ``return` `min(cnt_swaps_1, cnt_swaps_2);``    ``}`` ` `    ``// Driver code``    ``int` `main()``    ``{``        ``string st = ``"000111"``;``        ``cout<

## Java

 `// Java implementation of the approach`` ` `class` `GFG {`` ` `    ``// returns the minimum number of swaps``    ``// of a binary string``    ``// passed as the argument``    ``// to make it alternating``    ``static` `int` `countMinSwaps(String st)``    ``{`` ` `        ``int` `min_swaps = ``0``;`` ` `        ``// counts number of zeroes at odd ``        ``// and even positions``        ``int` `odd_0 = ``0``, even_0 = ``0``;`` ` `        ``// counts number of ones at odd ``        ``// and even positions``        ``int` `odd_1 = ``0``, even_1 = ``0``;`` ` `        ``int` `n = st.length();``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(i % ``2` `== ``0``) { ``                ``if` `(st.charAt(i) == ``'1'``) ``                    ``even_1++;``                ``else` `                    ``even_0++;``            ``}``            ``else` `{ ``                ``if` `(st.charAt(i) == ``'1'``) ``                    ``odd_1++;``                ``else` `                    ``odd_0++;``            ``}``        ``}`` ` `        ``// alternating string starts with 0``        ``int` `cnt_swaps_1 = Math.min(even_0, odd_1); `` ` `        ``// alternating string starts with 1``        ``int` `cnt_swaps_2 = Math.min(even_1, odd_0); `` ` `        ``// calculates the minimum number of swaps``        ``return` `Math.min(cnt_swaps_1, cnt_swaps_2);``    ``}`` ` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String st = ``"000111"``;``        ``System.out.println(countMinSwaps(st));``    ``}``}`

## Python 3

 `# Python3 implementation of the ``# above approach`` ` `# returns the minimum number of swaps ``# of a binary string ``# passed as the argument ``# to make it alternating ``def` `countMinSwaps(st) :`` ` `    ``min_swaps ``=` `0`` ` `    ``# counts number of zeroes at odd ``    ``# and even positions ``    ``odd_0, even_0 ``=` `0``, ``0`` ` `    ``# counts number of ones at odd ``    ``# and even positions ``    ``odd_1, even_1 ``=` `0``, ``0`` ` `    ``n ``=` `len``(st)`` ` `    ``for` `i ``in` `range``(``0``, n) :`` ` `        ``if` `i ``%` `2` `=``=` `0` `:`` ` `            ``if` `st[i] ``=``=` `"1"` `:``                ``even_1 ``+``=` `1``            ``else` `:``                ``even_0 ``+``=` `1``                 ` `        ``else` `:``            ``if` `st[i] ``=``=` `"1"` `:``                ``odd_1 ``+``=` `1``            ``else` `:``                ``odd_0 ``+``=` `1`` ` `    ``# alternating string starts with 0 ``    ``cnt_swaps_1 ``=` `min``(even_0, odd_1)`` ` `    ``# alternating string starts with 1 ``    ``cnt_swaps_2 ``=` `min``(even_1, odd_0)`` ` `    ``# calculates the minimum number of swaps ``    ``return` `min``(cnt_swaps_1, cnt_swaps_2)`` ` `# Driver code     ``if` `__name__ ``=``=` `"__main__"` `:`` ` `    ``st ``=` `"000111"`` ` `    ``# Function call``    ``print``(countMinSwaps(st))`` ` `# This code is contributed by ``# ANKITRAI1`

## C#

 `// C# implementation of the approach``using` `System;`` ` `public` `class` `GFG``{`` ` `    ``// returns the minimum number of swaps ``    ``// of a binary string ``    ``// passed as the argument ``    ``// to make it alternating ``    ``public` `static` `int` `countMinSwaps(``string` `st)``    ``{`` ` `        ``int` `min_swaps = 0;`` ` `        ``// counts number of zeroes at odd  ``        ``// and even positions ``        ``int` `odd_0 = 0, even_0 = 0;`` ` `        ``// counts number of ones at odd  ``        ``// and even positions ``        ``int` `odd_1 = 0, even_1 = 0;`` ` `        ``int` `n = st.Length;``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``if` `(i % 2 == 0)``            ``{``                ``if` `(st[i] == ``'1'``)``                ``{``                    ``even_1++;``                ``}``                ``else``                ``{``                    ``even_0++;``                ``}``            ``}``            ``else``            ``{``                ``if` `(st[i] == ``'1'``)``                ``{``                    ``odd_1++;``                ``}``                ``else``                ``{``                    ``odd_0++;``                ``}``            ``}``        ``}`` ` `        ``// alternating string starts with 0 ``        ``int` `cnt_swaps_1 = Math.Min(even_0, odd_1);`` ` `        ``// alternating string starts with 1 ``        ``int` `cnt_swaps_2 = Math.Min(even_1, odd_0);`` ` `        ``// calculates the minimum number of swaps ``        ``return` `Math.Min(cnt_swaps_1, cnt_swaps_2);``    ``}`` ` `    ``// Driver code ``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``string` `st = ``"000111"``;``        ``Console.WriteLine(countMinSwaps(st));``    ``}``}`` ` `// This code is contributed by Shrikant13`

## PHP

 ``
Output:
```1
```

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