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# Minimum number of adjacent swaps to convert a string into its given anagram

Given two strings s1 and s2, the task is to find the minimum number of steps required to convert s1 into s2. The only operation allowed is to swap adjacent elements in the first string. Every swap is counted as a single step.
Examples:

Input: s1 = “abcd”, s2 = “cdab”
Output:
Swap 2nd and 3rd element, abcd => acbd
Swap 1st and 2nd element, acbd => cabd
Swap 3rd and 4th element, cabd => cadb
Swap 2nd and 3rd element, cadb => cdab
Minimum 4 swaps are required.
Input: s1 = “abcfdegji”, s2 = “fjiacbdge”
Output:17

Approach: Use two pointers i and j for first and second strings respectively. Initialise i and j to 0
Iterate over the first string and find the position j such that s1[j] = s2[i] by incrementing the value to j. Keep on swapping the adjacent elements j and j – 1 and decrement j until it is greater than i
Now the ith element of the first string is equal to the second string, hence increment the value of i
This technique will give the minimum number of steps as there are zero unnecessary swaps.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach` `#include ` `using` `namespace` `std;`   `// Function that returns true if s1` `// and s2 are anagrams of each other` `bool` `isAnagram(string s1, string s2)` `{` `    ``sort(s1.begin(), s1.end());` `    ``sort(s2.begin(), s2.end());` `    ``if` `(s1 == s2)` `        ``return` `1;` `    ``return` `0;` `}`   `// Function to return the minimum swaps required` `int` `CountSteps(string s1, string s2, ``int` `size)` `{` `    ``int` `i = 0, j = 0;` `    ``int` `result = 0;`   `    ``// Iterate over the first string and convert` `    ``// every element equal to the second string` `    ``while` `(i < size) {` `        ``j = i;`   `        ``// Find index element of first string which` `        ``// is equal to the ith element of second string` `        ``while` `(s1[j] != s2[i]) {` `            ``j += 1;` `        ``}`   `        ``// Swap adjacent elements in first string so` `        ``// that element at ith position becomes equal` `        ``while` `(i < j) {`   `            ``// Swap elements using temporary variable` `            ``char` `temp = s1[j];` `            ``s1[j] = s1[j - 1];` `            ``s1[j - 1] = temp;` `            ``j -= 1;` `            ``result += 1;` `        ``}` `        ``i += 1;` `    ``}` `    ``return` `result;` `}`   `// Driver code` `int` `main()` `{` `    ``string s1 = ``"abcd"``;` `    ``string s2 = ``"cdab"``;`   `    ``int` `size = s2.size();`   `    ``// If both the strings are anagrams` `    ``// of each other then only they` `    ``// can be made equal` `    ``if` `(isAnagram(s1, s2))` `        ``cout << CountSteps(s1, s2, size);` `    ``else` `        ``cout << -1;`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the above approach` `import` `java.util.*;`   `class` `GFG ` `{`   `// Function that returns true if s1` `// and s2 are anagrams of each other` `static` `boolean` `isAnagram(String s1, String s2)` `{` `    ``s1 = sortString(s1);` `    ``s2 = sortString(s2);` `    ``return` `(s1.equals(s2));` `}`   `// Method to sort a string alphabetically ` `public` `static` `String sortString(String inputString) ` `{ ` `    ``// convert input string to char array ` `    ``char` `tempArray[] = inputString.toCharArray(); ` `    `  `    ``// sort tempArray ` `    ``Arrays.sort(tempArray); ` `    `  `    ``// return new sorted string ` `    ``return` `new` `String(tempArray); ` `} `   `// Function to return the minimum swaps required` `static` `int` `CountSteps(``char` `[]s1, ``char``[] s2, ``int` `size)` `{` `    ``int` `i = ``0``, j = ``0``;` `    ``int` `result = ``0``;`   `    ``// Iterate over the first string and convert` `    ``// every element equal to the second string` `    ``while` `(i < size)` `    ``{` `        ``j = i;`   `        ``// Find index element of first string which` `        ``// is equal to the ith element of second string` `        ``while` `(s1[j] != s2[i])` `        ``{` `            ``j += ``1``;` `        ``}`   `        ``// Swap adjacent elements in first string so` `        ``// that element at ith position becomes equal` `        ``while` `(i < j)` `        ``{`   `            ``// Swap elements using temporary variable` `            ``char` `temp = s1[j];` `            ``s1[j] = s1[j - ``1``];` `            ``s1[j - ``1``] = temp;` `            ``j -= ``1``;` `            ``result += ``1``;` `        ``}` `        ``i += ``1``;` `    ``}` `    ``return` `result;` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``String s1 = ``"abcd"``;` `    ``String s2 = ``"cdab"``;`   `    ``int` `size = s2.length();`   `    ``// If both the strings are anagrams` `    ``// of each other then only they` `    ``// can be made equal` `    ``if` `(isAnagram(s1, s2))` `        ``System.out.println(CountSteps(s1.toCharArray(), s2.toCharArray(), size));` `    ``else` `        ``System.out.println(-``1``);` `}` `}`   `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the above approach `   `# Function that returns true if s1 ` `# and s2 are anagrams of each other ` `def` `isAnagram(s1, s2) : ` `    ``s1 ``=` `list``(s1);` `    ``s2 ``=` `list``(s2);` `    ``s1 ``=` `s1.sort();` `    ``s2 ``=` `s2.sort();` `    `  `    ``if` `(s1 ``=``=` `s2) :` `        ``return` `1``;` `        `  `    ``return` `0``; `   `# Function to return the minimum swaps required ` `def` `CountSteps(s1, s2, size) : ` `    ``s1 ``=` `list``(s1);` `    ``s2 ``=` `list``(s2);` `    `  `    ``i ``=` `0``;` `    ``j ``=` `0``;` `    ``result ``=` `0``;` `    `  `    ``# Iterate over the first string and convert ` `    ``# every element equal to the second string ` `    ``while` `(i < size) :` `        ``j ``=` `i;` `        `  `        ``# Find index element of first string which` `        ``# is equal to the ith element of second string` `        ``while` `(s1[j] !``=` `s2[i]) :` `            ``j ``+``=` `1``;` `            `  `        ``# Swap adjacent elements in first string so` `        ``# that element at ith position becomes equal` `        ``while` `(i < j) :` `            `  `            ``# Swap elements using temporary variable` `            ``temp ``=` `s1[j];` `            ``s1[j] ``=` `s1[j ``-` `1``];` `            ``s1[j ``-` `1``] ``=` `temp; ` `            ``j ``-``=` `1``;` `            ``result ``+``=` `1``; ` `            `  `        ``i ``+``=` `1``;` `        `  `    ``return` `result; `   `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: `   `    ``s1 ``=` `"abcd"``; ` `    ``s2 ``=` `"cdab"``; `   `    ``size ``=` `len``(s2); `   `    ``# If both the strings are anagrams ` `    ``# of each other then only they ` `    ``# can be made equal ` `    ``if` `(isAnagram(s1, s2)) :` `        ``print``(CountSteps(s1, s2, size)); ` `    ``else` `:` `        ``print``(``-``1``); `   `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the above approach` `using` `System;` `using` `System.Linq;`   `class` `GFG ` `{`   `// Function that returns true if s1` `// and s2 are anagrams of each other` `static` `Boolean isAnagram(String s1, String s2)` `{` `    ``s1 = sortString(s1);` `    ``s2 = sortString(s2);` `    ``return` `(s1.Equals(s2));` `}`   `// Method to sort a string alphabetically ` `public` `static` `String sortString(String inputString) ` `{ ` `    ``// convert input string to char array ` `    ``char` `[]tempArray = inputString.ToCharArray(); ` `    `  `    ``// sort tempArray ` `    ``Array.Sort(tempArray); ` `    `  `    ``// return new sorted string ` `    ``return` `new` `String(tempArray); ` `} `   `// Function to return the minimum swaps required` `static` `int` `CountSteps(``char` `[]s1, ``char``[] s2, ``int` `size)` `{` `    ``int` `i = 0, j = 0;` `    ``int` `result = 0;`   `    ``// Iterate over the first string and convert` `    ``// every element equal to the second string` `    ``while` `(i < size)` `    ``{` `        ``j = i;`   `        ``// Find index element of first string which` `        ``// is equal to the ith element of second string` `        ``while` `(s1[j] != s2[i])` `        ``{` `            ``j += 1;` `        ``}`   `        ``// Swap adjacent elements in first string so` `        ``// that element at ith position becomes equal` `        ``while` `(i < j)` `        ``{`   `            ``// Swap elements using temporary variable` `            ``char` `temp = s1[j];` `            ``s1[j] = s1[j - 1];` `            ``s1[j - 1] = temp;` `            ``j -= 1;` `            ``result += 1;` `        ``}` `        ``i += 1;` `    ``}` `    ``return` `result;` `}`   `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `    ``String s1 = ``"abcd"``;` `    ``String s2 = ``"cdab"``;`   `    ``int` `size = s2.Length;`   `    ``// If both the strings are anagrams` `    ``// of each other then only they` `    ``// can be made equal` `    ``if` `(isAnagram(s1, s2))` `        ``Console.WriteLine(CountSteps(s1.ToCharArray(), s2.ToCharArray(), size));` `    ``else` `        ``Console.WriteLine(-1);` `}` `}`   `/* This code is contributed by PrinciRaj1992 */`

## Javascript

 ``

Output:

`4`

Time Complexity: O(N*N), as we are using nested loops for traversing N*N times. Where N is the length of the string.

Auxiliary Space: O(1), as we are not using any extra space.

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