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# Minimize length of string by replacing K pairs of distinct adjacent characters

• Difficulty Level : Medium
• Last Updated : 07 May, 2021

Given string str of length N, the task is to find the minimum length to which the given string can be reduced by replacing any pair of non-equal adjacent characters with a single character at most K times.

Examples:

Input: str = “aabc”, K =1
Output: 3
Explanation:
Replace “bc” with a single character “d” and decrement K by 1.
Since K = 0, the minimized length of the string is 3.

Input: str= “aabc”, K = 2
Output: 2

Naive Approach: The simplest approach is to traverse the string K times and during each traversal, check if any pair of adjacent characters of the given string are distinct or not. If found to be true, then replace both the characters with a single character that is not equal to its adjacent characters. Finally, print the minimum length of the string.

Time Complexity: O(N * K)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to traverse the given string and check if any pair of adjacent characters of the given string are distinct or not. If found to be true, then print the value of max(1, (N – K)). Otherwise, print N. Follow the steps below to solve the problem:

1. Traverse the given string str and check if any pair of adjacent characters are distinct or not.
2. If found to be true, then print max(1, (N – K)) as the required answer.
3. Otherwise, print N as the required answer.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to minimize the length``// of the string by replacing distinct``// pairs of adjacent characters``int` `MinLen(string str, ``int` `K)``{``    ``// Stores the length``    ``// of the given string``    ``int` `N = str.length();` `    ``// Stores the index``    ``// of the given string``    ``int` `i = 0;` `    ``// Traverse the given string``    ``while` `(i < N - 1) {` `        ``// If two adjacent``        ``// characters are distinct``        ``if` `(str[i] != str[i + 1]) {``            ``break``;``        ``}``        ``i++;``    ``}` `    ``// If all characters``    ``// are equal``    ``if` `(i == N - 1) {``        ``return` `N;``    ``}` `    ``// If all characters``    ``// are distinct``    ``return` `max(1, N - K);``}` `// Driver Code``int` `main()``{``    ``string str = ``"aabc"``;``    ``int` `K = 1;``    ``cout << MinLen(str, K);``}`

## Java

 `// Java program to implement``// the above approach``class` `GFG{` `// Function to minimize the``// length of the String by``// replacing distinct pairs``// of adjacent characters``static` `int` `MinLen(String str,``                  ``int` `K)``{``  ``// Stores the length``  ``// of the given String``  ``int` `N = str.length();` `  ``// Stores the index``  ``// of the given String``  ``int` `i = ``0``;` `  ``// Traverse the given String``  ``while` `(i < N - ``1``)``  ``{``    ``// If two adjacent``    ``// characters are distinct``    ``if` `(str.charAt(i) !=``        ``str.charAt(i + ``1``))``    ``{``      ``break``;``    ``}``    ``i++;``  ``}` `  ``// If all characters``  ``// are equal``  ``if` `(i == N - ``1``)``  ``{``    ``return` `N;``  ``}` `  ``// If all characters``  ``// are distinct``  ``return` `Math.max(``1``, N - K);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ``String str = ``"aabc"``;``  ``int` `K = ``1``;``  ``System.out.print(MinLen(str, K));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to minimize the length``# of the by replacing distinct``# pairs of adjacent characters``def` `MinLen(``str``, K):``    ` `    ``# Stores the length``    ``# of the given string``    ``N ``=` `len``(``str``)` `    ``# Stores the index``    ``# of the given string``    ``i ``=` `0` `    ``# Traverse the given string``    ``while` `(i < N ``-` `1``):` `        ``# If two adjacent``        ``# haracters are distinct``        ``if` `(``str``[i] !``=` `str``[i ``+` `1``]):``            ``break``        ` `        ``i ``+``=` `1` `    ``# If all characters``    ``# are equal``    ``if` `(i ``=``=` `N ``-` `1``):``        ``return` `N` `    ``# If all characters``    ``# are distinct``    ``return` `max``(``1``, N ``-` `K)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``str` `=` `"aabc"``    ``K ``=` `1``    ` `    ``print``(MinLen(``str``, K))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to implement``// the above approach``using` `System;``class` `GFG{` `// Function to minimize the``// length of the String by``// replacing distinct pairs``// of adjacent characters``static` `int` `MinLen(String str,``                  ``int` `K)``{``  ``// Stores the length``  ``// of the given String``  ``int` `N = str.Length;` `  ``// Stores the index``  ``// of the given String``  ``int` `i = 0;` `  ``// Traverse the given String``  ``while` `(i < N - 1)``  ``{``    ``// If two adjacent``    ``// characters are distinct``    ``if` `(str[i] != str[i + 1])``    ``{``      ``break``;``    ``}``    ``i++;``  ``}` `  ``// If all characters``  ``// are equal``  ``if` `(i == N - 1)``  ``{``    ``return` `N;``  ``}` `  ``// If all characters``  ``// are distinct``  ``return` `Math.Max(1, N - K);``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ``String str = ``"aabc"``;``  ``int` `K = 1;``  ``Console.Write(MinLen(str, K));``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`3`

Time Complexity: O(N)
Auxiliary Space: O(1)

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