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 charactersvoid 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 Codeint main(){ string str = "Hellolllee"; concesStr(str); return 0;} |
Java
// Java code to implement above approachimport 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 charactersdef 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 codeif __name__ == '__main__': consecStr("Hellolllee");# This code is contributed by 29AjayKumar |
C#
// C# program for above approachusing 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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
