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# Count minimum swap to make string palindrome

• Difficulty Level : Medium
• Last Updated : 04 Oct, 2021

Given a string s, the task is to find out the minimum no of adjacent swaps required to make string s palindrome. If it is not possible, then return -1.

Examples:

Input: aabcb
Output:
Explanation:
After 1st swap: abacb
After 2nd swap: abcab
After 3rd swap: abcba

Output: -1

Approach
The following are detailed steps to solve this problem.

1. Take two-pointer where the first pointer track from the left side of a string and second pointer keep track from the right side of a string.
2. Till the time we find the same character, keep moving the right pointer to one step left.
4. If the same character found then swap the right pointer’s character towards the right until it is not placed at its correct position in a string.
5. Increase left pointer and repeat step 2.

Below is the implementation of the above approach:

## C++

 `// C++ program to Count``// minimum swap to make``// string palindrome``#include ``using` `namespace` `std;`` ` `// Function to Count minimum swap``int` `countSwap(string s)``{``    ``// calculate length of string as n``    ``int` `n = s.length();`` ` `    ``// counter to count minimum swap``    ``int` `count = 0;`` ` `    ``// A loop which run till mid of``    ``// string``    ``for` `(``int` `i = 0; i < n / 2; i++) {``        ``// Left pointer``        ``int` `left = i;`` ` `        ``// Right pointer``        ``int` `right = n - left - 1;`` ` `        ``// A loop which run from right``        ``// pointer towards left pointer``        ``while` `(left < right) {``            ``// if both char same then``            ``// break the loop.``            ``// If not, then we have to``            ``// move right pointer to one``            ``// position left``            ``if` `(s[left] == s[right]) {``                ``break``;``            ``}``            ``else` `{``                ``right--;``            ``}``        ``}`` ` `        ``// If both pointers are at same``        ``// position, it denotes that we``        ``// don't have sufficient characters``        ``// to make palindrome string``        ``if` `(left == right) {``            ``return` `-1;``        ``}`` ` `        ``// else swap and increase the count``        ``for` `(``int` `j = right; j < n - left - 1; j++) {``            ``swap(s[j], s[j + 1]);``            ``count++;``        ``}``    ``}`` ` `    ``return` `count;``}`` ` `// Driver code``int` `main()``{``    ``string s = ``"geeksfgeeks"``;`` ` `    ``// Function calling``    ``int` `ans1 = countSwap(s);`` ` `    ``reverse(s.begin(), s.end());``    ``int` `ans2 = countSwap(s);`` ` `    ``cout << max(ans1, ans2);`` ` `    ``return` `0;``}`

## Python3

 `''' Python3 program to Count ``    ``minimum swap to make``    ``string palindrome'''`` ` `# Function to Count minimum swap`` ` ` ` `def` `CountSwap(s, n):``    ``s ``=` `list``(s)`` ` `    ``# Counter to count minimum swap``    ``count ``=` `0``    ``ans ``=` `True`` ` `    ``# A loop which run in half string``    ``# from starting``    ``for` `i ``in` `range``(n ``/``/` `2``):`` ` `        ``# Left pointer``        ``left ``=` `i`` ` `        ``# Right pointer``        ``right ``=` `n ``-` `left ``-` `1`` ` `        ``# A loop which run from right pointer``        ``# to left pointer``        ``while` `left < right:`` ` `            ``# if both char same then``            ``# break the loop if not``            ``# same then we have to move``            ``# right pointer to one step left``            ``if` `s[left] ``=``=` `s[right]:``                ``break``            ``else``:``                ``right ``-``=` `1`` ` `        ``# it denotes both pointer at``        ``# same position and we don't``        ``# have sufficient char to make``        ``# palindrome string``        ``if` `left ``=``=` `right:``            ``ans ``=` `False``            ``break``        ``else``:``            ``for` `j ``in` `range``(right, n ``-` `left ``-` `1``):``                ``(s[j], s[j ``+` `1``]) ``=` `(s[j ``+` `1``], s[j])``                ``count ``+``=` `1``    ``if` `ans:``        ``return` `(count)``    ``else``:``        ``return` `-``1`` ` ` ` `# Driver Code``s ``=` `'geeksfgeeks'`` ` `# Length of string``n ``=` `len``(s)`` ` `# Function calling``ans1 ``=` `CountSwap(s, n)``ans2 ``=` `CountSwap(s[::``-``1``], n)``print``(``max``(ans1, ans2))`

## C#

 `// C# program to Count``// minimum swap to make``// string palindrome``using` `System;`` ` `class` `GFG {`` ` `    ``// Function to Count minimum swap``    ``static` `int` `countSwap(String str)``    ``{`` ` `        ``// Length of string``        ``int` `n = str.Length;`` ` `        ``// it will convert string to``        ``// char array``        ``char``[] s = str.ToCharArray();`` ` `        ``// Counter to count minimum``        ``// swap``        ``int` `count = 0;`` ` `        ``// A loop which run in half``        ``// string from starting``        ``for` `(``int` `i = 0; i < n / 2; i++) {`` ` `            ``// Left pointer``            ``int` `left = i;`` ` `            ``// Right pointer``            ``int` `right = n - left - 1;`` ` `            ``// A loop which run from``            ``// right pointer to left``            ``// pointer``            ``while` `(left < right) {`` ` `                ``// if both char same``                ``// then break the loop``                ``// if not same then we``                ``// have to move right``                ``// pointer to one step``                ``// left``                ``if` `(s[left] == s[right]) {``                    ``break``;``                ``}``                ``else` `{``                    ``right--;``                ``}``            ``}`` ` `            ``// it denotes both pointer at``            ``// same position and we don't``            ``// have sufficient char to make``            ``// palindrome string``            ``if` `(left == right) {``                ``return` `-1;``            ``}``            ``else` `{``                ``for` `(``int` `j = right; j < n - left - 1; j++) {``                    ``char` `t = s[j];``                    ``s[j] = s[j + 1];``                    ``s[j + 1] = t;``                    ``count++;``                ``}``            ``}``        ``}`` ` `        ``return` `count;``    ``}`` ` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String s = ``"geeksfgeeks"``;`` ` `        ``// Function calling``        ``int` `ans1 = countSwap(s);`` ` `        ``char``[] charArray = s.ToCharArray();``        ``Array.Reverse(charArray);``        ``s = ``new` `string``(charArray);`` ` `        ``int` `ans2 = countSwap(s);`` ` `        ``if` `(ans1 > ans2)``            ``Console.WriteLine(ans1);``        ``else``            ``Console.WriteLine(ans2);``    ``}``}`` ` `// This code is contributed by sapnasingh4991`
Output
`9`

Complexity Analysis
Time Complexity: Since we are running two nested loops on the length of string, the time complexity is O(n2)
Auxiliary Space: Since we aren’t using any extra space, Therefore Auxiliary space used is O(1)

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