Print the frequency of adjacent repeating characters in given string

Given a string str of length N. The task is to print the frequency of adjacent repeating characters. 

Examples:

Input: str = “Hello”
Output: l: 2
Explanation: Consecutive Repeating Character from the given string is “l” and its frequency is 2.

Input: str = “Hellolllee”
Output: l: 2
              l: 3
              e: 2
Explanation: Consecutive Repeating Characters from the given string are “l, “, “l” and “e” 
and its frequencies are as follows: 2, 3, 2.

Approach: This problem can be solved simply by traversing and keeping track of adjacent repeating characters. Follow the steps below to solve the given problem.

• Iterate from i = 0 till string length.
• Maintain a counter.
• Again iterate via the next loop from i+1 till string length.
• The counter will increment until the next character is different.
• For characters having more than 2 frequencies increment i so that the count will remain intact.
• If the counter is greater than 1 then print.

Below is the implementation of the above approach:

C++

// C++ program for above approach
#include <iostream>
using namespace std;

// Function to find frequency
// of repeating characters
void concesStr(string str)
{

    // Length of string
    int lenStr = str.length();

    // Iterate through 1st pointer
    for (int i = 0; i < lenStr; i++) {

        // Keep a counter
        int curr_count = 1;

        // Iterate through 2nd pointer
        for (int j = i + 1; j < lenStr;
             j++) {

            // if next element is different
            // then break
            if (str[i] != str[j]) {
                break;
            }
            curr_count++;

            // Example: if count is 3
            // then move the first
            // pointer so that
            // count remains intact
            if (curr_count > 2) {
                i++;
            }
        }

        // Condition for print more than 1
        // count characters
        if (curr_count > 1) {
            cout << str[i] << ": "
                 << curr_count << endl;
        }
    }
}

// Driver Code
int main()
{
    string str = "Hellolllee";

    concesStr(str);
    return 0;
}

Java

// Java code to implement above approach
import java.io.*;

class GFG {

    // Function to find frequency
    // of repeating characters
    public static void consecStr(String str)
    {
        int lenStr = str.length();

        // Iterate through 1st pointer
        for (int i = 0; i < lenStr; i++) {

            // keep a counter
            int curr_count = 1;

            // Iterate through 2nd pointer
            for (int j = i + 1; j < lenStr;
                 j++) {

                // if next element is different
                // then break
                if (str.charAt(i) != str.charAt(j)) {
                    break;
                }
                curr_count++;

                // Example: if count is 3 then
                // move the first pointer
                // so that count remains intact
                if (curr_count > 2) {
                    i++;
                }
            }

            // Condition for print
            // more than 1 count characters
            if (curr_count > 1) {
                System.out.print(str.charAt(i)
                                 + ": " + curr_count
                                 + "\n");
            }
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        consecStr("Hellolllee");
    }
}

Python3

# Python code to implement above approach

# Function to find frequency
# of repeating characters
def consecStr(str):
    lenStr = len(str);
    i = 0;

    # Iterate through 1st pointer
    for k in range(lenStr):

        # keep a counter
        curr_count = 1;

        # Iterate through 2nd pointer
        for j in range(i+1,lenStr):

            # if next element is different
            # then break
            if (str[i] != str[j]):
                break;

            curr_count += 1;

            # Example: if count is 3 then
            # move the first pointer
            # so that count remains intact
            if (curr_count > 2):
                i += 1;

        # Condition for print
        # more than 1 count characters
        if (curr_count > 1):
            print(str[i] , ": " , curr_count , "");
        i += 1;

# Driver code
if __name__ == '__main__':
    consecStr("Hellolllee");

# This code is contributed by 29AjayKumar

C#

// C# program for above approach
using System;

class GFG
{
  // Function to find frequency
  // of repeating characters
  static void concesStr(string str)
  {

    // Length of string
    int lenStr = str.Length;

    // Iterate through 1st pointer
    for (int i = 0; i < lenStr; i++)
    {

      // Keep a counter
      int curr_count = 1;

      // Iterate through 2nd pointer
      for (int j = i + 1; j < lenStr;
           j++)
      {

        // if next element is different
        // then break
        if (str[i] != str[j])
        {
          break;
        }
        curr_count++;

        // Example: if count is 3
        // then move the first
        // pointer so that
        // count remains intact
        if (curr_count > 2)
        {
          i++;
        }
      }

      // Condition for print more than 1
      // count characters
      if (curr_count > 1)
      {
        Console.WriteLine(str[i] + ": " + curr_count);
      }
    }
  }

  // Driver Code
  public static void Main()
  {
    string str = "Hellolllee";

    concesStr(str);
  }
}

// This code is contributed by gfgking.

Javascript

<script>
        // JavaScript code for the above approach

        // Function to find frequency
        // of repeating characters
        function concesStr(str) {

            // Length of string
            let lenStr = str.length;

            // Iterate through 1st pointer
            for (let i = 0; i < lenStr; i++) {

                // Keep a counter
                let curr_count = 1;

                // Iterate through 2nd pointer
                for (let j = i + 1; j < lenStr;
                    j++) {

                    // if next element is different
                    // then break
                    if (str[i] != str[j]) {
                        break;
                    }
                    curr_count++;

                    // Example: if count is 3
                    // then move the first
                    // pointer so that
                    // count remains intact
                    if (curr_count > 2) {
                        i++;
                    }
                }

                // Condition for print more than 1
                // count characters
                if (curr_count > 1) {
                    document.write(str[i] + ": "
                        + curr_count + '<br>');
                }
            }
        }

        // Driver Code
        let str = "Hellolllee";
        concesStr(str);

       // This code is contributed by Potta Lokesh
    </script>

Output

l: 2
l: 3
e: 2

Time Complexity: O(N²)
Auxiliary Space: O(1)