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

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.
Therefore, str modifies to “aad”.
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 <bits/stdc++.h>
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

<script>

// Javascript program to implement
// the above approach

// Function to minimize the
    // length of the String by
    // replacing distinct pairs
    // of adjacent characters
    function MinLen( str , K)
    {
        // Stores the length
        // of the given String
        var N = str.length;

        // Stores the index
        // of the given String
        var 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

        var str = "aabc";
        var K = 1;
        document.write(MinLen(str, K));

// This code contributed by gauravrajput1

</script>

Output:

3

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