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# Minimum adjacent swaps required to Sort Binary array

Given a binary array, task is to sort this binary array using minimum swaps. We are allowed to swap only adjacent elements

Examples:

```Input : [0, 0, 1, 0, 1, 0, 1, 1]
Output : 3
1st swap : [0, 0, 1, 0, 0, 1, 1, 1]
2nd swap : [0, 0, 0, 1, 0, 1, 1, 1]
3rd swap : [0, 0, 0, 0, 1, 1, 1, 1]

Input : Array = [0, 1, 0, 1, 0]
Output : 3```

Approach: This can be done by finding number of zeroes to the right side of every 1 and add them. In order to sort the array every one always has to perform a swap operation with every zero on its right side. So the total number of swap operations for a particular 1 in array is the number of zeroes on its right hand side. Find the number of zeroes on right side for every one i.e. the number of swaps and add them all to obtain the total number of swaps.

Steps to solve this problem:

1. Declare an integer array noOfZeroes of size n and initialize it with zeros using memset function.

2. Initialize a variable count to keep track of the total number of swaps required.

3. Initialize another variable i to traverse the array from the end.

4. Calculate the number of zeroes on the right side of every one and store it in the noOfZeroes array. This is done using a loop which starts from the second last element of the input array and goes till the first element. For every iteration, the value of noOfZeroes[i] is updated with the value of noOfZeroes[i+1]. If the current element of the input array is 0, the value of noOfZeroes[i] is incremented.

5. A loop is run over the input array to count the total number of swaps required. For every iteration, if the current element of the input array is 1, the value of count is incremented by noOfZeroes[i].

6. The function returns the value of count, which represents the minimum number of swaps required to make all ones together.

Implementation :

## C++

 `// C++ code to find minimum number of``// swaps to sort a binary array``#include ` `using` `namespace` `std;` `// Function to find minimum swaps to``// sort an array of 0s and 1s.``int` `findMinSwaps(``int` `arr[], ``int` `n)``{``    ``// Array to store count of zeroes``    ``int` `noOfZeroes[n];``    ``memset``(noOfZeroes, 0, ``sizeof``(noOfZeroes));` `    ``int` `i, count = 0;` `    ``// Count number of zeroes``    ``// on right side of every one.``    ``noOfZeroes[n - 1] = 1 - arr[n - 1];``    ``for` `(i = n - 2; i >= 0; i--) {``        ``noOfZeroes[i] = noOfZeroes[i + 1];``        ``if` `(arr[i] == 0)``            ``noOfZeroes[i]++;``    ``}` `    ``// Count total number of swaps by adding number``    ``// of zeroes on right side of every one.``    ``for` `(i = 0; i < n; i++) {``        ``if` `(arr[i] == 1)``            ``count += noOfZeroes[i];``    ``}` `    ``return` `count;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 0, 0, 1, 0, 1, 0, 1, 1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << findMinSwaps(arr, n);``    ``return` `0;``}`

## Java

 `// Java code to find minimum number of``// swaps to sort a binary array``class` `gfg {``    ` `    ``static` `int` `findMinSwaps(``int` `arr[], ``int` `n)``    ``{``        ``// Array to store count of zeroes``        ``int` `noOfZeroes[] = ``new` `int``[n];``        ``int` `i, count = ``0``;` `        ``// Count number of zeroes``        ``// on right side of every one.``        ``noOfZeroes[n - ``1``] = ``1` `- arr[n - ``1``];``        ``for` `(i = n - ``2``; i >= ``0``; i--)``        ``{``            ``noOfZeroes[i] = noOfZeroes[i + ``1``];``            ``if` `(arr[i] == ``0``)``                ``noOfZeroes[i]++;``        ``}` `        ``// Count total number of swaps by adding number``        ``// of zeroes on right side of every one.``        ``for` `(i = ``0``; i < n; i++)``        ``{``            ``if` `(arr[i] == ``1``)``                ``count += noOfZeroes[i];``        ``}``        ``return` `count;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `ar[] = { ``0``, ``0``, ``1``, ``0``, ``1``, ``0``, ``1``, ``1` `};``        ``System.out.println(findMinSwaps(ar, ar.length));``    ``}``}` `// This code is contributed by Niraj_Pandey.`

## Python3

 `# Python3 code to find minimum number of``# swaps to sort a binary array` `# Function to find minimum swaps to``# sort an array of 0s and 1s.``def` `findMinSwaps(arr, n) :``    ``# Array to store count of zeroes``    ``noOfZeroes ``=` `[``0``] ``*` `n``    ``count ``=` `0``    ` `    ``# Count number of zeroes``    ``# on right side of every one.``    ``noOfZeroes[n ``-` `1``] ``=` `1` `-` `arr[n ``-` `1``]``    ``for` `i ``in` `range``(n``-``2``, ``-``1``, ``-``1``) :``        ``noOfZeroes[i] ``=` `noOfZeroes[i ``+` `1``]``        ``if` `(arr[i] ``=``=` `0``) :``            ``noOfZeroes[i] ``=` `noOfZeroes[i] ``+` `1` `    ``# Count total number of swaps by adding``    ``# number of zeroes on right side of``    ``# every one.``    ``for` `i ``in` `range``(``0``, n) :``        ``if` `(arr[i] ``=``=` `1``) :``            ``count ``=` `count ``+` `noOfZeroes[i]` `    ``return` `count` `# Driver code``arr ``=` `[ ``0``, ``0``, ``1``, ``0``, ``1``, ``0``, ``1``, ``1` `]``n ``=` `len``(arr)``print` `(findMinSwaps(arr, n))` `# This code is contributed by Manish Shaw``# (manishshaw1)`

## C#

 `// C# code to find minimum number of``// swaps to sort a binary array``using` `System;` `class` `GFG {``    ` `    ``static` `int` `findMinSwaps(``int` `[]arr, ``int` `n)``    ``{``        ` `        ``// Array to store count of zeroes``        ``int` `[]noOfZeroes = ``new` `int``[n];``        ``int` `i, count = 0;` `        ``// Count number of zeroes``        ``// on right side of every one.``        ``noOfZeroes[n - 1] = 1 - arr[n - 1];``        ``for` `(i = n - 2; i >= 0; i--)``        ``{``            ``noOfZeroes[i] = noOfZeroes[i + 1];``            ``if` `(arr[i] == 0)``                ``noOfZeroes[i]++;``        ``}` `        ``// Count total number of swaps by``        ``// adding number of zeroes on right``        ``// side of every one.``        ``for` `(i = 0; i < n; i++)``        ``{``            ``if` `(arr[i] == 1)``                ``count += noOfZeroes[i];``        ``}``        ` `        ``return` `count;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]ar = { 0, 0, 1, 0, 1,``                                ``0, 1, 1 };``                                ` `        ``Console.WriteLine(``              ``findMinSwaps(ar, ar.Length));``    ``}``}` `// This code is contributed by vt_m.`

## PHP

 `= 0; ``\$i``--)``    ``{``        ``\$noOfZeroes``[``\$i``] = ``\$noOfZeroes``[``\$i` `+ 1];``        ``if` `(``\$arr``[``\$i``] == 0)``            ``\$noOfZeroes``[``\$i``]++;``    ``}` `    ``// Count total number of swaps by adding``    ``// number of zeroes on right side of every one.``    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)``    ``{``        ``if` `(``\$arr``[``\$i``] == 1)``            ``\$count` `+= ``\$noOfZeroes``[``\$i``];``    ``}` `    ``return` `\$count``;``}` `// Driver code``\$arr` `= ``array``( 0, 0, 1, 0, 1, 0, 1, 1 );``\$n` `= sizeof(``\$arr``);``echo` `findMinSwaps(``\$arr``, ``\$n``);` `// This code is contributed by Sach_code``?>`

## Javascript

 ``

Output

`3`

Time Complexity: O(n)
Auxiliary Space: O(n)

Space Optimized Solution: An auxiliary space is not needed. We just need to start reading the list from the back and keep track of number of zeros we encounter. If we encounter a 1 the number of zeros is the number of swaps needed to put the 1 in correct place.

Implementation:

## C++

 `#include ``using` `namespace` `std;` `int` `minswaps(``int` `arr[], ``int` `n)``{``    ``int` `count = 0;``    ``int` `num_unplaced_zeros = 0;``    ` `    ``for``(``int` `index=n-1;index>=0;index--)``    ``{``        ``if``(arr[index] == 0)``            ``num_unplaced_zeros += 1;``        ``if``(arr[index] == 1)``            ``count += num_unplaced_zeros;``    ``}``    ``return` `count;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = {0, 0, 1, 0, 1, 0, 1, 1};``    ``cout<

## Java

 `import` `java.io.*;` `class` `GFG {``    ``public` `static` `int` `minswaps(``int` `arr[], ``int` `n)``    ``{``        ``int` `count = ``0``;``        ``int` `num_unplaced_zeros = ``0``;` `        ``for` `(``int` `index = n - ``1``; index >= ``0``; index--)``        ``{``            ``if` `(arr[index] == ``0``)``                ``num_unplaced_zeros += ``1``;``            ``else``                ``count += num_unplaced_zeros;``        ``}``        ``return` `count;``    ``}``  ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = { ``0``, ``0``, ``1``, ``0``, ``1``, ``0``, ``1``, ``1` `};``        ``System.out.println(minswaps(arr, ``8``));``    ``}``}` `// this code is contributed by Manu Pathria`

## Python3

 `def` `minswaps(arr):``    ``count ``=` `0``    ``num_unplaced_zeros ``=` `0` `    ``for` `index ``in` `range``(``len``(arr)``-``1``, ``-``1``, ``-``1``):``        ``if` `arr[index] ``=``=` `0``:``            ``num_unplaced_zeros ``+``=` `1``        ``else``:``            ``count ``+``=` `num_unplaced_zeros``    ``return` `count`  `arr ``=` `[``0``, ``0``, ``1``, ``0``, ``1``, ``0``, ``1``, ``1``]``print``(minswaps(arr))`

## C#

 `using` `System;``class` `GFG {``    ` `    ``static` `int` `minswaps(``int``[] arr, ``int` `n)``    ``{``        ``int` `count = 0;``        ``int` `num_unplaced_zeros = 0;`` ` `        ``for` `(``int` `index = n - 2; index >= 0; index--)``        ``{``            ``if` `(arr[index] == 0)``                ``num_unplaced_zeros += 1;``            ``else``                ``count += num_unplaced_zeros;``        ``}``        ``return` `count;``    ``}``    ` `  ``static` `void` `Main() {``    ``int``[] arr = { 0, 0, 1, 0, 1, 0, 1, 1 };``    ``Console.WriteLine(minswaps(arr, 8));``  ``}``}` `// This code is contributed by divyeshrabadiya07.`

## Javascript

 ``

Output

`3`

Time Complexity: O(n)
Auxiliary Space: O(1)