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Minimum operations required to make the string satisfy the given condition
• Last Updated : 26 Nov, 2019

Given a string str, the task is to make the string start and end at the same character with the minimum number of given operations. In a single operation, any character of the string can be removed. Note that the length of the resultant string must be greater than 1 and it is not possible then print -1.

Examples:

Input: str = “geeksforgeeks”
Output: 3
Remove the first and the last two characters
and the string becomes “eeksforgee”

Input: str = “abcda”
Output: 0

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

Approach: If the string needs to start and end at a character say ch then an optimal way is to remove all the characters before the first occurrence of ch and all the characters after the last occurrence of ch. Find the number of characters that need to be removed for every possible value of ch starting from ‘a’ to ‘z’ and choose the one with the minimum number of delete operations.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;`` ` `const` `int` `MAX = 26;`` ` `// Function to return the minimum``// operations required``int` `minOperation(string str, ``int` `len)``{`` ` `    ``// To store the first and the last``    ``// occurrence of all the characters``    ``int` `first[MAX], last[MAX];`` ` `    ``// Set the first and the last occurrence``    ``// of all the characters to -1``    ``for` `(``int` `i = 0; i < MAX; i++) {``        ``first[i] = -1;``        ``last[i] = -1;``    ``}`` ` `    ``// Update the occurrences of the characters``    ``for` `(``int` `i = 0; i < len; i++) {`` ` `        ``int` `index = (str[i] - ``'a'``);`` ` `        ``// Only set the first occurrence if``        ``// it hasn't already been set``        ``if` `(first[index] == -1)``            ``first[index] = i;`` ` `        ``last[index] = i;``    ``}`` ` `    ``// To store the minimum operations``    ``int` `minOp = -1;`` ` `    ``for` `(``int` `i = 0; i < MAX; i++) {`` ` `        ``// If the frequency of the current``        ``// character in the string``        ``// is less than 2``        ``if` `(first[i] == -1 || first[i] == last[i])``            ``continue``;`` ` `        ``// Count of characters to be``        ``// removed so that the string``        ``// starts and ends at the``        ``// current character``        ``int` `cnt = len - (last[i] - first[i] + 1);`` ` `        ``if` `(minOp == -1 || cnt < minOp)``            ``minOp = cnt;``    ``}`` ` `    ``return` `minOp;``}`` ` `// Driver code``int` `main()``{``    ``string str = ``"abcda"``;``    ``int` `len = str.length();`` ` `    ``cout << minOperation(str, len);`` ` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach ``class` `GFG``{``    ``final` `static` `int` `MAX = ``26``; ``     ` `    ``// Function to return the minimum ``    ``// operations required ``    ``static` `int` `minOperation(String str, ``int` `len) ``    ``{ ``     ` `        ``// To store the first and the last ``        ``// occurrence of all the characters ``        ``int` `first[] = ``new` `int``[MAX];``        ``int` `last[] = ``new` `int``[MAX]; ``     ` `        ``// Set the first and the last occurrence ``        ``// of all the characters to -1 ``        ``for` `(``int` `i = ``0``; i < MAX; i++)``        ``{ ``            ``first[i] = -``1``; ``            ``last[i] = -``1``; ``        ``} ``     ` `        ``// Update the occurrences of the characters ``        ``for` `(``int` `i = ``0``; i < len; i++)``        ``{ ``            ``int` `index = (str.charAt(i) - ``'a'``); ``     ` `            ``// Only set the first occurrence if ``            ``// it hasn't already been set ``            ``if` `(first[index] == -``1``) ``                ``first[index] = i; ``     ` `            ``last[index] = i; ``        ``} ``     ` `        ``// To store the minimum operations ``        ``int` `minOp = -``1``; ``     ` `        ``for` `(``int` `i = ``0``; i < MAX; i++) ``        ``{ ``     ` `            ``// If the frequency of the current ``            ``// character in the string ``            ``// is less than 2 ``            ``if` `(first[i] == -``1` `|| ``                ``first[i] == last[i]) ``                ``continue``; ``     ` `            ``// Count of characters to be ``            ``// removed so that the string ``            ``// starts and ends at the ``            ``// current character ``            ``int` `cnt = len - (last[i] - first[i] + ``1``); ``     ` `            ``if` `(minOp == -``1` `|| cnt < minOp) ``                ``minOp = cnt; ``        ``} ``        ``return` `minOp; ``    ``} ``     ` `    ``// Driver code ``    ``public` `static` `void` `main (String[] args) ``    ``{ ``        ``String str = ``"abcda"``; ``        ``int` `len = str.length(); ``     ` `        ``System.out.println(minOperation(str, len)); ``    ``} ``}`` ` `// This code is contributed by AnkitRai01`

## Python3

 `# Python implementation of the approach``MAX` `=` `26``;`` ` `# Function to return the minimum``# operations required``def` `minOperation(``str``, ``len``):`` ` `    ``# To store the first and the last``    ``# occurrence of all the characters``    ``first, last ``=` `[``0``] ``*` `MAX``, [``0``] ``*` `MAX``;`` ` `    ``# Set the first and the last occurrence``    ``# of all the characters to -1``    ``for` `i ``in` `range``(``MAX``):``        ``first[i] ``=` `-``1``;``        ``last[i] ``=` `-``1``;`` ` `    ``# Update the occurrences of the characters``    ``for` `i ``in` `range``(``len``):`` ` `        ``index ``=` `(``ord``(``str``[i]) ``-` `ord``(``'a'``));`` ` `        ``# Only set the first occurrence if``        ``# it hasn't already been set``        ``if` `(first[index] ``=``=` `-``1``):``            ``first[index] ``=` `i;`` ` `        ``last[index] ``=` `i;`` ` `    ``# To store the minimum operations``    ``minOp ``=` `-``1``;`` ` `    ``for` `i ``in` `range``(``MAX``):`` ` `        ``# If the frequency of the current``        ``# character in the string``        ``# is less than 2``        ``if` `(first[i] ``=``=` `-``1` `or` `first[i] ``=``=` `last[i]):``            ``continue``;`` ` `        ``# Count of characters to be``        ``# removed so that the string``        ``# starts and ends at the``        ``# current character``        ``cnt ``=` `len` `-` `(last[i] ``-` `first[i] ``+` `1``);`` ` `        ``if` `(minOp ``=``=` `-``1` `or` `cnt < minOp):``            ``minOp ``=` `cnt;``    ``return` `minOp;`` ` `# Driver code``str` `=` `"abcda"``;``len` `=` `len``(``str``);`` ` `print``( minOperation(``str``, ``len``));`` ` `# This code is contributed by 29AjayKumar`

## C#

 `// C# implementation of the approach ``using` `System;``     ` `class` `GFG``{``    ``readonly` `static` `int` `MAX = 26; ``     ` `    ``// Function to return the minimum ``    ``// operations required ``    ``static` `int` `minOperation(String str, ``int` `len) ``    ``{ ``     ` `        ``// To store the first and the last ``        ``// occurrence of all the characters ``        ``int` `[]first = ``new` `int``[MAX];``        ``int` `[]last = ``new` `int``[MAX]; ``     ` `        ``// Set the first and the last occurrence ``        ``// of all the characters to -1 ``        ``for` `(``int` `i = 0; i < MAX; i++)``        ``{ ``            ``first[i] = -1; ``            ``last[i] = -1; ``        ``} ``     ` `        ``// Update the occurrences of the characters ``        ``for` `(``int` `i = 0; i < len; i++)``        ``{ ``            ``int` `index = (str[i] - ``'a'``); ``     ` `            ``// Only set the first occurrence if ``            ``// it hasn't already been set ``            ``if` `(first[index] == -1) ``                ``first[index] = i; ``     ` `            ``last[index] = i; ``        ``} ``     ` `        ``// To store the minimum operations ``        ``int` `minOp = -1; ``     ` `        ``for` `(``int` `i = 0; i < MAX; i++) ``        ``{ ``     ` `            ``// If the frequency of the current ``            ``// character in the string ``            ``// is less than 2 ``            ``if` `(first[i] == -1 || ``                ``first[i] == last[i]) ``                ``continue``; ``     ` `            ``// Count of characters to be ``            ``// removed so that the string ``            ``// starts and ends at the ``            ``// current character ``            ``int` `cnt = len - (last[i] - first[i] + 1); ``     ` `            ``if` `(minOp == -1 || cnt < minOp) ``                ``minOp = cnt; ``        ``} ``        ``return` `minOp; ``    ``} ``     ` `    ``// Driver code ``    ``public` `static` `void` `Main (String[] args) ``    ``{ ``        ``String str = ``"abcda"``; ``        ``int` `len = str.Length; ``     ` `        ``Console.WriteLine(minOperation(str, len)); ``    ``} ``}`` ` `// This code is contributed by 29AjayKumar`
Output:
```0
```

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