# Minimize replacements by previous or next alphabet required to make all characters of a string the same

Given a string S of length N consisting of lowercase alphabets, the task is to find the minimum number of operations required to make all the characters of the string S the same. In each operation, choose any character and replace it with its next or previous alphabet.

Note: The alphabets are considered to be cyclic i.e., Next character of z is considered to be a and previous character of a is considered to be z.

Examples:

Input: S = “abc”
Output: 2
Explanation:
To minimize the number of operation change all characters of the strings to ‘b’.
Operation 1: Change a to b.
Operation 2: Change c to b.

Input: S = “zzza”
Output: 1
Explanation:
To minimize the number of operation change all characters of the strings to ‘z’.
Operation 1: Change a to z.

Approach: To solve the problem, the idea is to calculate the cost of making all the characters equal to each alphabets, ‘a’ to ‘z’, one by one and print the minimum cost required for any of the conversions. Follow the steps below to solve the problem:

• Initialize the variable min with a large value which will store the minimum answer.
• Traverse over the range [0, 25] where i represent (i + 1)thalphabet from ‘a’ to ‘z’ and perform the following steps:
• Initialize variable cnt initialized as 0 which will store the answer to convert all character of string same.
• Traverse the given string from j = 0 to (N – 1) and add min(abs(i + ‘a’ – S[j]), 26 – abs(i + ‘a’ – S[j])) into cnt.
• After the above step, update min to a minimum of min and cnt.
• After traversing the string for each character, print the value of min as the minimum count of operations.

Below is the implementation of the above approach:

 `// C++ program for the above approach`   `#include `   `using` `namespace` `std;`   `// Function to find the minimum count` `// of operations to make all characters` `// of the string same` `int` `minCost(string s, ``int` `n)` `{`   `    ``// Set min to some large value` `    ``int` `minValue = 100000000;`   `    ``// Find minimum operations for` `    ``// each character` `    ``for` `(``int` `i = 0; i <= 25; i++) {`   `        ``// Initialize cnt` `        ``int` `cnt = 0;`   `        ``for` `(``int` `j = 0; j < n; j++) {`   `            ``// Add the value to cnt` `            ``cnt += min(``abs``(i - (s[j] - ``'a'``)),` `                       ``26 - ``abs``(i - (s[j] - ``'a'``)));` `        ``}`   `        ``// Update minValue` `        ``minValue = min(minValue, cnt);` `    ``}`   `    ``// Return minValue` `    ``return` `minValue;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given string str` `    ``string str = ``"geeksforgeeks"``;`   `    ``int` `N = str.length();`   `    ``// Function Call` `    ``cout << minCost(str, N);`   `    ``return` `0;` `}`

 `// Java program for the above approach` `import` `java.io.*;`   `class` `GFG{` `    `  `// Function to find the minimum count` `// of operations to make all characters` `// of the String same` `static` `int` `minCost(String s, ``int` `n)` `{` `    `  `    ``// Set min to some large value` `    ``int` `minValue = ``100000000``;` ` `  `    ``// Find minimum operations for` `    ``// each character` `    ``for``(``int` `i = ``0``; i <= ``25``; i++) ` `    ``{` `        `  `        ``// Initialize cnt` `        ``int` `cnt = ``0``;` ` `  `        ``for``(``int` `j = ``0``; j < n; j++) ` `        ``{` `            `  `            ``// Add the value to cnt` `            ``cnt += Math.min(Math.abs(i - (s.charAt(j) - ``'a'``)),` `                       ``26` `- Math.abs(i - (s.charAt(j) - ``'a'``)));` `        ``}` ` `  `        ``// Update minValue` `        ``minValue = Math.min(minValue, cnt);` `    ``}` ` `  `    ``// Return minValue` `    ``return` `minValue;` `}` ` `  `// Driver Code` `public` `static` `void` `main (String[] args) ` `{` `    `  `    ``// Given String str` `    ``String str = ``"geeksforgeeks"``;` ` `  `    ``int` `N = str.length();` ` `  `    ``// Function call` `    ``System.out.println(minCost(str, N));` `}` `}`   `// This code is contributed by sanjoy_62`

 `# Python3 program for the ` `# above approach`   `# Function to find the minimum ` `# count of operations to make ` `# all characters of the string same` `def` `minCost(s, n):`   `    ``# Set min to some ` `    ``# large value` `    ``minValue ``=` `100000000`   `    ``# Find minimum operations ` `    ``# for each character` `    ``for` `i ``in` `range``(``26``):`   `        ``# Initialize cnt` `        ``cnt ``=` `0`   `        ``for` `j ``in` `range``(n):`   `            ``# Add the value to cnt` `            ``cnt ``+``=` `min``(``abs``(i ``-` `(``ord``(s[j]) ``-` `                                ``ord``(``'a'``))),` `                       ``26` `-` `abs``(i ``-` `(``ord``(s[j]) ``-` `                                     ``ord``(``'a'``))))`   `        ``# Update minValue` `        ``minValue ``=` `min``(minValue, cnt)`   `    ``# Return minValue` `    ``return` `minValue`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``# Given string str` `    ``st ``=` `"geeksforgeeks"`   `    ``N ``=` `len``(st)`   `    ``# Function Call` `    ``print``(minCost(st, N))`   `# This code is contributed by Chitranayal`

 `// C# program for the above approach` `using` `System;`   `class` `GFG{` `     `  `// Function to find the minimum count` `// of operations to make all characters` `// of the String same` `static` `int` `minCost(``string` `s, ``int` `n)` `{` `    `  `    ``// Set min to some large value` `    ``int` `minValue = 100000000;` `  `  `    ``// Find minimum operations for` `    ``// each character` `    ``for``(``int` `i = 0; i <= 25; i++) ` `    ``{` `        `  `        ``// Initialize cnt` `        ``int` `cnt = 0;` `  `  `        ``for``(``int` `j = 0; j < n; j++) ` `        ``{` `             `  `            ``// Add the value to cnt` `            ``cnt += Math.Min(Math.Abs(i - (s[j] - ``'a'``)),` `                       ``26 - Math.Abs(i - (s[j] - ``'a'``)));` `        ``}` `  `  `        ``// Update minValue` `        ``minValue = Math.Min(minValue, cnt);` `    ``}` `  `  `    ``// Return minValue` `    ``return` `minValue;` `}` `  `  `// Driver code` `public` `static` `void` `Main()` `{` `    `  `    ``// Given String str` `    ``string` `str = ``"geeksforgeeks"``;` `  `  `    ``int` `N = str.Length;` `  `  `    ``// Function call` `    ``Console.WriteLine(minCost(str, N));` `}` `}`   `// This code is contributed by code_hunt`

Output
`60`

Time Complexity: O(N * 26)
Auxiliary Space: O(N)

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Improved By : sanjoy_62, code_hunt, chitranayal

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