Given a string S and a character ‘c’, the task is to count the occurrence of the given character in the string.
Examples:
Input : S = “neveropen” and c = ‘e’
Output : 4
Explanation: ‘e’ appears four times in str.Input : S = “abccdefgaa” and c = ‘a’
Output : 3
Explanation: ‘a’ appears three times in str.
Count occurrence of a given character in a string using simple Iteration:
Iterate through the string and during each iteration, check if the current character is equal to the given character c then increments the count variable which stores count of the occurrences of the given character c in the string.
Below is the implementation of above approach:
C++
// C++ program to count occurrences of a given// character#include <iostream>#include <string>using namespace std;// Function that return count of the given // character in the stringint count(string s, char c){ // Count variable int res = 0; for (int i=0;i<s.length();i++) // checking character in string if (s[i] == c) res++; return res;}// Driver codeint main(){ string str= "neveropen"; char c = 'e'; cout << count(str, c) << endl; return 0;} |
Java
// JAVA program to count occurrences// of a characterclass GFG{ // Method that return count of the given // character in the string public static int count(String s, char c) { int res = 0; for (int i=0; i<s.length(); i++) { // checking character in string if (s.charAt(i) == c) res++; } return res; } // Driver method public static void main(String args[]) { String str= "neveropen"; char c = 'e'; System.out.println(count(str, c)); }} |
Python3
# Python program to count occurrences# of a given character# Method that return count of the # given character in the stringdef count(s, c) : # Count variable res = 0 for i in range(len(s)) : # Checking character in string if (s[i] == c): res = res + 1 return res # Driver codestr= "neveropen"c = 'e'print(count(str, c)) # This code is contributed by "rishabh_jain". |
C#
// C# program to count occurrences// of a characterusing System; public class GFG { // Method that return count of the given // character in the string public static int count(string s, char c) { int res = 0; for (int i = 0; i < s.Length; i++) { // checking character in string if (s[i] == c) res++; } return res; } // Driver method public static void Main() { string str = "neveropen"; char c = 'e'; Console.WriteLine(count(str, c)); }}// This code is contributed by Sam007. |
Javascript
<script>// JAVASCRIPT program to count occurrences// of a character // Method that return count of the given // character in the string function count(s, c) { let res = 0; for (let i = 0; i < s.length; i++) { // checking character in string if (s.charAt(i) == c) res++; } return res; } // Driver method let str= "neveropen"; let c = 'e'; document.write(count(str, c)); // This code is contributed by shivanisinghss2110 </script> |
PHP
<?php// PHP program to count // occurrences of a given// character// Function that return count of// the given character in the stringfunction count($s, $c){ // Count variable $res = 0; for ($i = 0; $i < strlen($s); $i++) // checking character in string if ($s[$i] == $c) $res++; return $res;} // Driver Code $str= "neveropen"; $c = 'e'; echo count($str, $c) ; return 0; // This code is contributed by nitin mittal.?> |
Output
4
Time Complexity: O(len), where len is the size of the string given.
Auxiliary Space: O(1)
Count occurrence of a given character in a string using Inbuild Functions:
The idea is to use inbuild method in different programming languages which returns the count of occurences of a character in a string.
Below is the implementation of above approach:
C++
// CPP program to count occurrences of// a character using library#include<bits/stdc++.h>using namespace std;// Driver codeint main(){ string str = "neveropen"; char c = 'e'; // Count returns number of occurrences of // c between two given positions provided // as two iterators. cout << count(str.begin(), str.end(), c); return 0;} |
Java
import java.util.*;public class Main { public static void main(String[] args) { String str = "neveropen"; char c = 'e'; // Count returns number of occurrences of // c between two given positions provided // as two iterators. System.out.println(Collections.frequency( Arrays.asList(str.split("")), String.valueOf(c))); }} |
Python3
# Python program to count occurrences of# a character using library# Driver code to test above functionstr = "neveropen"c = 'e'# Count returns number of occurrences of# c between two given positions provided# as two iterators.print(len(str.split(c)) - 1);# The code is contributed by Arushi Goel. |
C#
using System;using System.Linq;class Program { static void Main(string[] args) { string str = "neveropen"; char c = 'e'; // Count returns number of occurrences of // c between two given positions provided // as two iterators. Console.WriteLine(str.Count(x => x == c)); }} |
Javascript
// JavaScript program to count occurrences of// a character using library// Driver code to test above functionlet str = "neveropen";let c = 'e';// Count returns number of occurrences of// c between two given positions provided// as two iterators.console.log(str.split(c).length - 1); |
Output
4
Time Complexity: O(len), where len is the size of the string given.
Auxiliary Space: O(1)
Count occurrence of a given character in a string using Recursion
Use a recursive approach to count the occurrences of a given character in a string. Checks if the character at the current index matches the target character, increments the count if it does, and then makes a recursive call to check the remaining part of the string. The process continues until the end of the string is reached, and the accumulated count would be the result.
Below is the implementation of above approach:
C++
#include<bits/stdc++.h>using namespace std;int countinString(char ch,int idx, string s){ // base case; if (idx == s.size()) return 0; int count = 0; // checking if the current character of // the given string is that character // or not if (s[idx] == ch) count++; // this will count the occurrence of // given character in the string // from the remaining part of the string. count += countinString(ch,idx+1,s); return count;}int main(){ string str = "neveropen"; char c = 'e'; cout<<(countinString(c,0, str));} |
Java
public class Main { // Function to count occurrences of a character in a string public static int countInString(char ch, int idx, String s) { // Base case: if the index reaches the end of the string, return 0 if (idx == s.length()) return 0; int count = 0; // Check if the current character of the string matches the given character if (s.charAt(idx) == ch) count++; // Recursively count occurrences in the remaining part of the string count += countInString(ch, idx + 1, s); return count; } public static void main(String[] args) { String str = "neveropen"; char c = 'e'; System.out.println(countInString(c, 0, str)); }} |
Python3
def count_in_string(ch, idx, s): # Base case: if the index reaches the end of the string, return 0 if idx == len(s): return 0 count = 0 # Check if the current character of the string matches the given character if s[idx] == ch: count += 1 # Recursively count occurrences in the remaining part of the string count += count_in_string(ch, idx + 1, s) return countif __name__ == "__main__": str = "neveropen" c = 'e' print(count_in_string(c, 0, str)) |
C#
using System;class Program { // Function to count occurrences of a character in a string static int CountInString(char ch, int idx, string s) { // Base case: if the index reaches the end of the string, return 0 if (idx == s.Length) return 0; int count = 0; // Check if the current character of the string matches the given character if (s[idx] == ch) count++; // Recursively count occurrences in the remaining part of the string count += CountInString(ch, idx + 1, s); return count; } static void Main(string[] args) { string str = "neveropen"; char c = 'e'; Console.WriteLine(CountInString(c, 0, str)); }} |
Javascript
function countInString(ch, idx, s) { // Base case: if the index reaches the end of the string, return 0 if (idx === s.length) return 0; let count = 0; // Check if the current character of the string matches the given character if (s[idx] === ch) count++; // Recursively count occurrences in the remaining part of the string count += countInString(ch, idx + 1, s); return count;}const str = "neveropen";const c = 'e';console.log(countInString(c, 0, str)); |
Output
4
Time Complexity: O(len), where len is the size of the string given.
Auxiliary Space: O(len), where len is the size of the string given.
This article is contributed by Sahil Rajput. If you like neveropen and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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