Minimum number of insertions in given String to remove adjacent duplicates

Given a string str of size N, the task is to find the minimum number of additions in the string such that no two consecutive elements are the same.

Examples:

Input: str=”rrg” 
Output: 1
Explanation: Add an element between two r’s

Input: str=”rrrrr”
Output: 4

Approach: The above problem can be solved using the below given steps:

1. Declare variable min_steps and initialise it by 0.
2. Traverse the string using a for-loop from i=0 to i<N.
3. Check if the current character is the same as the character before it.
4. If yes, then increment min_steps by 1.
5. Else, continue the loop.
6. Print min_steps as the answer.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <iostream>
using namespace std;

// Function to find the minimum
// number of additions such that
// no two elements are the same
int minAdditions(string str)
{
    // Storing length of the string
    int len = str.size();

    // Variable to store
    // the number of steps needed
    int min_steps = 0;

    int i;

    // For loop to check
    // all colours in the string
    for (i = 1; i < len; i++) {
        if (str[i] == str[i - 1])
            min_steps++;
    }

    // Returning the number of additions
    return min_steps;
}

// Driver Code
int main()
{
    string str = "RRG";
    cout << minAdditions(str);
    return 0;
}

Java

// Java program for above approach
import java.util.*;
public class GFG {

  // Function to find the minimum
  // number of additions such that
  // no two elements are the same
  static int minAdditions(String str)
  {

    // Storing length of the string
    int len = str.length();

    // Variable to store
    // the number of steps needed
    int min_steps = 0;

    int i;

    // For loop to check
    // all colours in the string
    for (i = 1; i < len; i++) {
      if (str.charAt(i) == str.charAt(i - 1))
        min_steps++;
    }

    // Returning the number of additions
    return min_steps;
  }

  // Driver Code
  public static void main(String args[])
  {
    String str = "RRG";

    System.out.println(minAdditions(str));
  }
}

// This code is contributed by Samim Hossain Mondal.

Python3

# python3 program for the above approach

# Function to find the minimum
# number of additions such that
# no two elements are the same
def minAdditions(str):

    # Storing length of the string
    le = len(str)

    # Variable to store
    # the number of steps needed
    min_steps = 0

    # For loop to check
    # all colours in the string
    for i in range(1, le):
        if (str[i] == str[i - 1]):
            min_steps += 1

    # Returning the number of additions
    return min_steps


# Driver Code
if __name__ == "__main__":

    str = "RRG"
    print(minAdditions(str))

# This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
class GFG {

  // Function to find the minimum
  // number of additions such that
  // no two elements are the same
  static int minAdditions(string str)
  {

    // Storing length of the string
    int len = str.Length;

    // Variable to store
    // the number of steps needed
    int min_steps = 0;

    int i;

    // For loop to check
    // all colours in the string
    for (i = 1; i < len; i++) {
      if (str[i] == str[i - 1])
        min_steps++;
    }

    // Returning the number of additions
    return min_steps;
  }

  // Driver Code
  public static void Main()
  {
    string str = "RRG";
    Console.Write(minAdditions(str));
  }
}

// This code is contributed by ukasp.

Javascript

<script>
// Javascript program for the above approach

// Function to find the minimum
// number of additions such that
// no two elements are the same
function minAdditions(str)
{

    // Storing length of the string
    let len = str.length;

    // Variable to store
    // the number of steps needed
    let min_steps = 0;

    let i;

    // For loop to check
    // all colours in the string
    for (i = 1; i < len; i++) {
        if (str[i] == str[i - 1])
            min_steps++;
    }

    // Returning the number of additions
    return min_steps;
}

// Driver Code

let str = "RRG";
document.write(minAdditions(str))

// This code is contributed by gfgking.
</script>

Output

1

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