Count the minimum number of groups formed in a string

Given a string ‘S’ which comprises two or more uppercase English letters. The task is to count the minimum number of groups required to completely cover the string by replacing consecutive characters with the same single character.

Examples:

Input : S = "TTWWW"
Output : 2
Explanation : 
There are 2 groups formed. One by covering the 2 consecutive T and the other by 3 consecutive W. 

Input : S = "FFMMMF"
Output : 3
Explanation : 
Minimum number of groups formed is 3 that is two F's, three M's and one F .
Note: Three F's were not included in one group because they are not consecutive in the string s.

Approach:

To solve the problem mentioned above the main idea is to compare the adjacent characters in the string ‘S’ one by one. If for instance, the characters are different that is the consecutive letters of the string are not the same then, the counter for the total group formed is incremented by 1 and so on until we reach the length if the string.

Below is the implementation of the above-mentioned approach:

C++

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// C++ implementation to Count the
// minimum number of groups formed in a string
// by replacing consecutive characters
// with same single character
#include<bits/stdc++.h>
  
using namespace std;
  
// Function to count the minimum number
// of groups formed in the given string s
void group_formed(string S){
  
    // Initializing count as one since
    // the string is not NULL
    int count = 1;
  
    for (int i = 0; i < S.size() - 1; i++){
  
        // Comparing adjacent characters
        if ( S[i] != S[i+1])
            count += 1;
          }
  
    cout << (count);
  }
  
// Driver Code
int main(){
    string S = "TTWWW";
  
    group_formed(S);
  }
  
// This code is contributed by mohit kumar 29

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Java

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// Java implementation to Count the 
// minimum number of groups formed in a string 
// by replacing consecutive characters 
// with same single character 
class GFG {
      
    // Function to count the minimum number 
    // of groups formed in the given string s 
    static void group_formed(String S){ 
      
        // Initializing count as one since 
        // the string is not NULL 
        int count = 1
      
        for (int i = 0; i < S.length() - 1; i++){ 
      
            // Comparing adjacent characters 
            if ( S.charAt(i) != S.charAt(i+1)) 
                count += 1
            
      
        System.out.println(count); 
    
      
    // Driver Code 
    public static void main (String[] args) { 
        String S = "TTWWW"
      
        group_formed(S); 
    
}
  
 // This code is contributed by AnkitRai01

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Python3

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# Python3 implementation to Count the
# minimum number of groups formed in a string
# by replacing consecutive characters
# with same single character
  
# Function to count the minimum number
# of groups formed in the given string s
def group_formed(S):
      
    # Initializing count as one since
    # the string is not NULL 
    count = 1
      
    for i in range (len(S)-1):
        a = S[i]
        b = S[i + 1]
          
        # Comparing adjacent characters
        if ( a != b):
              
            count += 1
              
    print (count)
  
# Driver Code 
if __name__ == "__main__"
    S = "TTWWW"
      
    group_formed(S)
  

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C#

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// C# implementation to Count the 
// minimum number of groups formed 
// in a string by replacing consecutive 
// characters with same single character 
using System;
  
class GFG{ 
      
// Function to count the minimum number 
// of groups formed in the given string s 
static void group_formed(String S)
      
    // Initializing count as one since 
    // the string is not NULL 
    int count = 1; 
      
    for(int i = 0; i < S.Length - 1; i++)
    
         
       // Comparing adjacent characters 
       if (S[i] != S[i + 1]) 
           count += 1; 
    }
    Console.Write(count); 
      
// Driver Code 
public static void Main (String[] args)
    String S = "TTWWW"
      
    group_formed(S); 
  
// This code is contributed by Rajnis09 

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Output:

2

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