Print the frequency of adjacent repeating characters in given string
Last Updated :
31 Jan, 2022
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 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");
}
}
# 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# 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.
<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>
Time Complexity: O(N2)
Auxiliary Space: O(1)
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