# Minimum move to end operations to make all strings equal

Given n strings that are permutations of each other. We need to make all strings same with an operation that takes front character of any string and moves it to the end.**Examples:**

Input : n = 2 arr[] = {"molzv", "lzvmo"} Output : 2 Explanation: In first string, we remove first element("m") from first string and append it end. Then we move second character of first string and move it to end. So after 2 operations, both strings become same. Input : n = 3 arr[] = {"kc", "kc", "kc"} Output : 0 Explanation: already all strings are equal.

The move to end operation is basically left rotation. We use the approach discussed in check if strings are rotations of each other or not to count a number of move to front operations required to make two strings the same. We one by one consider every string as the target string. We count rotations required to make all other strings the same as the current target and finally return a minimum of all counts.

Below is the implementation of the above approach.

## C++

`// CPP program to make all strings same using` `// move to end operations.` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Returns minimum number of moves to end` `// operations to make all strings same.` `int` `minimunMoves(string arr[], ` `int` `n)` `{` ` ` `int` `ans = INT_MAX;` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `{` ` ` `int` `curr_count = 0; ` ` ` `// Consider s[i] as target string and` ` ` `// count rotations required to make` ` ` `// all other strings same as str[i].` ` ` `for` `(` `int` `j = 0; j < n; j++) {` ` ` `string tmp = arr[j] + arr[j];` ` ` `// find function returns the index where we` ` ` `// found arr[i] which is actually count of` ` ` `// move-to-front operations.` ` ` `int` `index = tmp.find(arr[i]);` ` ` `// If any two strings are not rotations of` ` ` `// each other, we can't make them same. ` ` ` `if` `(index == string::npos)` ` ` `return` `-1;` ` ` `curr_count += index;` ` ` `}` ` ` `ans = min(curr_count, ans);` ` ` `}` ` ` `return` `ans;` `}` `// driver code for above function.` `int` `main()` `{` ` ` `string arr[] = {` `"xzzwo"` `, ` `"zwoxz"` `, ` `"zzwox"` `, ` `"xzzwo"` `}; ` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);` ` ` `cout << minimunMoves(arr, n);` ` ` `return` `0;` `}` |

## Java

`// Java program to make all` `// strings same using move` `// to end operations.` `import` `java.util.*;` `class` `GFG` `{` `// Returns minimum number of` `// moves to end operations` `// to make all strings same.` `static` `int` `minimunMoves(String arr[], ` `int` `n)` `{` ` ` `int` `ans = Integer.MAX_VALUE;` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `{` ` ` `int` `curr_count = ` `0` `;` ` ` `// Consider s[i] as target` ` ` `// string and count rotations` ` ` `// required to make all other` ` ` `// strings same as str[i].` ` ` `String tmp = ` `""` `;` ` ` `for` `(` `int` `j = ` `0` `; j < n; j++)` ` ` `{` ` ` `tmp = arr[j] + arr[j];` ` ` `// find function returns the` ` ` `// index where we found arr[i]` ` ` `// which is actually count of` ` ` `// move-to-front operations.` ` ` `int` `index = tmp.indexOf(arr[i]);` ` ` `// If any two strings are not` ` ` `// rotations of each other,` ` ` `// we can't make them same.` ` ` `if` `(index != -` `1` `)` ` ` `curr_count += index;` ` ` `else` ` ` `curr_count = -` `1` `; ` ` ` `}` ` ` `ans = Math.min(curr_count, ans);` ` ` `}` ` ` `return` `ans;` `}` `// Driver code` `public` `static` `void` `main(String args[])` `{` ` ` `String arr[] = {` `"xzzwo"` `, ` `"zwoxz"` `,` ` ` `"zzwox"` `, ` `"xzzwo"` `};` ` ` `int` `n = arr.length;` ` ` `System.out.println(minimunMoves(arr, n));` `}` `}` `// This code is contributed` `// by Kirti_Mangal` |

## Python 3

`# Python 3 program to make all strings` `# same using move to end operations.` `import` `sys` `# Returns minimum number of moves to end` `# operations to make all strings same.` `def` `minimunMoves(arr, n):` ` ` `ans ` `=` `sys.maxsize` ` ` `for` `i ` `in` `range` `(n):` ` ` `curr_count ` `=` `0` ` ` `# Consider s[i] as target string and` ` ` `# count rotations required to make` ` ` `# all other strings same as str[i].` ` ` `for` `j ` `in` `range` `(n):` ` ` `tmp ` `=` `arr[j] ` `+` `arr[j]` ` ` `# find function returns the index where` ` ` `# we found arr[i] which is actually` ` ` `# count of move-to-front operations.` ` ` `index ` `=` `tmp.find(arr[i])` ` ` `# If any two strings are not rotations of` ` ` `# each other, we can't make them same.` ` ` `if` `(index ` `=` `=` `len` `(arr[i])):` ` ` `return` `-` `1` ` ` `curr_count ` `+` `=` `index` ` ` `ans ` `=` `min` `(curr_count, ans)` ` ` `return` `ans` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` ` ` `arr ` `=` `[` `"xzzwo"` `, ` `"zwoxz"` `, ` `"zzwox"` `, ` `"xzzwo"` `]` ` ` `n ` `=` `len` `(arr)` ` ` `print` `( minimunMoves(arr, n))` `# This code is contributed by ita_c` |

## C#

`using` `System;` `// C# program to make all` `// strings same using move` `// to end operations.` `public` `class` `GFG` `{` `// Returns minimum number of` `// moves to end operations` `// to make all strings same.` `public` `static` `int` `minimunMoves(` `string` `[] arr, ` `int` `n)` `{` ` ` `int` `ans = ` `int` `.MaxValue;` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `{` ` ` `int` `curr_count = 0;` ` ` `// Consider s[i] as target` ` ` `// string and count rotations` ` ` `// required to make all other` ` ` `// strings same as str[i].` ` ` `string` `tmp = ` `""` `;` ` ` `for` `(` `int` `j = 0; j < n; j++)` ` ` `{` ` ` `tmp = arr[j] + arr[j];` ` ` `// find function returns the` ` ` `// index where we found arr[i]` ` ` `// which is actually count of` ` ` `// move-to-front operations.` ` ` `int` `index = tmp.IndexOf(arr[i], StringComparison.Ordinal);` ` ` `// If any two strings are not` ` ` `// rotations of each other,` ` ` `// we can't make them same.` ` ` `if` `(index == arr[i].Length)` ` ` `{` ` ` `return` `-1;` ` ` `}` ` ` `curr_count += index;` ` ` `}` ` ` `ans = Math.Min(curr_count, ans);` ` ` `}` ` ` `return` `ans;` `}` `// Driver code` `public` `static` `void` `Main(` `string` `[] args)` `{` ` ` `string` `[] arr = ` `new` `string` `[] {` `"xzzwo"` `, ` `"zwoxz"` `, ` `"zzwox"` `, ` `"xzzwo"` `};` ` ` `int` `n = arr.Length;` ` ` `Console.WriteLine(minimunMoves(arr, n));` `}` `}` `// This code is contributed by Shrikant13` |

## Javascript

`<script>` `// Javascript program to make all` `// strings same using move` `// to end operations.` ` ` ` ` `// Returns minimum number of` ` ` `// moves to end operations` ` ` `// to make all strings same.` ` ` `function` `minimunMoves(arr,n)` ` ` `{` ` ` `let ans = Number.MAX_VALUE;` ` ` `for` `(let i = 0; i < n; i++)` ` ` `{` ` ` `let curr_count = 0;` ` ` ` ` `// Consider s[i] as target` ` ` `// string and count rotations` ` ` `// required to make all other` ` ` `// strings same as str[i].` ` ` `let tmp = ` `""` `;` ` ` `for` `(let j = 0; j < n; j++)` ` ` `{` ` ` `tmp = arr[j] + arr[j];` ` ` ` ` `// find function returns the` ` ` `// index where we found arr[i]` ` ` `// which is actually count of` ` ` `// move-to-front operations.` ` ` `let index = tmp.indexOf(arr[i]);` ` ` ` ` `// If any two strings are not` ` ` `// rotations of each other,` ` ` `// we can't make them same.` ` ` `if` `(index == arr[i].length)` ` ` `return` `-1;` ` ` ` ` `curr_count += index;` ` ` `}` ` ` ` ` `ans = Math.min(curr_count, ans);` ` ` `}` ` ` ` ` `return` `ans;` ` ` `}` ` ` ` ` `// Driver code` ` ` `let arr=[` `"xzzwo"` `, ` `"zwoxz"` `,` ` ` `"zzwox"` `, ` `"xzzwo"` `];` ` ` `let n = arr.length;` ` ` `document.write(minimunMoves(arr, n));` ` ` ` ` ` ` `// This code is contributed by avanitrachhadiya2155` ` ` `</script>` |

**Output:**

5

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