Find maximum occurring character in a string Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice Given string str. The task is to find the maximum occurring character in the string str.Examples:Input: geeksforgeeksOutput: eExplanation: 'e' occurs 4 times in the stringInput: testOutput: tExplanation: 't' occurs 2 times in the stringReturn the maximum occurring character in an input string using Hashing:Naive approach : ( using unordered_map ) In this approach we simply use the unordered_map from STL to store the frequency of every character and while adding characters to map we take a variable count to determine the element having highest frequency.Implementation : C++ // C++ program for the above approach #include <bits/stdc++.h> using namespace std; // function that return maximum occurring character char getMaxOccurringChar(string str) { // create unordered_map to store frequency of every character unordered_map<char,int>mp; // to store length of string int n = str.length(); // to store answer char ans; // to check count of answer character is less or greater // than another elements count int cnt=0; // traverse the string for(int i=0 ;i<n ; i++){ // push element into map and increase its frequency mp[str[i]]++; // update answer and count if(cnt < mp[str[i]]){ ans = str[i]; cnt = mp[str[i]]; } } return ans; } // Driver Code int main() { string str = "sample string"; cout << "Max occurring character is: " << getMaxOccurringChar(str); } // this code is contributed by bhardwajji C #include <stdio.h> #include <stdlib.h> #include <string.h> // Function to find the maximum occurring character char getMaxOccurringChar(char str[]) { // Create a hash table (unordered_map) to store the // frequency of each character int count[256] = { 0 }; // Traverse the string and update the frequency of each // character int length = strlen(str); for (int i = 0; i < length; i++) count[(int)str[i]]++; // Find the character with the maximum frequency char maxChar; int maxCount = 0; for (int i = 0; i < length; i++) { if (count[(int)str[i]] > maxCount) { maxCount = count[(int)str[i]]; maxChar = str[i]; } } return maxChar; } // Driver Code int main() { char str[] = "sample string"; printf("Max occurring character is: %c\n", getMaxOccurringChar(str)); return 0; } Java import java.util.*; public class Main { // function that returns maximum occurring character static char getMaxOccurringChar(String str) { // create HashMap to store frequency of every character HashMap<Character, Integer> mp = new HashMap<>(); // to store length of string int n = str.length(); // to store answer char ans = 0; // to check count of answer character is less or greater // than another elements count int cnt = 0; // traverse the string for(int i = 0; i < n; i++) { // push element into map and increase its frequency char c = str.charAt(i); mp.put(c, mp.getOrDefault(c, 0) + 1); // update answer and count if(cnt < mp.get(c)) { ans = c; cnt = mp.get(c); } } return ans; } // Driver Code public static void main(String[] args) { String str = "sample string"; System.out.println("Max occurring character is: " + getMaxOccurringChar(str)); } } // This code is contributed by kalyanbef Python # function that return maximum occurring character def getMaxOccurringChar(str): # create dictionary to store frequency of every character mp = {} # to store length of string n = len(str) # to store answer ans = '' # to check count of answer character is less or greater # than another elements count cnt = 0 # traverse the string for i in range(n): # push element into dictionary and increase its frequency if str[i] in mp: mp[str[i]] += 1 else: mp[str[i]] = 1 # update answer and count if cnt < mp[str[i]]: ans = str[i] cnt = mp[str[i]] return ans # Driver Code str = "sample string" print("Max occurring character is:", getMaxOccurringChar(str)) C# // C# program for the above approach using System; using System.Collections.Generic; class MainClass { public static void Main(string[] args) { string str = "sample string"; Console.WriteLine("Max occurring character is: " + getMaxOccurringChar(str)); } // function that return maximum occurring character static char getMaxOccurringChar(string str) { // create dictionary to store frequency of every // character Dictionary<char, int> mp = new Dictionary<char, int>(); // to store length of string int n = str.Length; // to store answer char ans = '\0'; // to check count of answer character is less or // greater than another elements count int cnt = 0; // traverse the string for (int i = 0; i < n; i++) { // push element into map and increase its // frequency if (mp.ContainsKey(str[i])) { mp[str[i]]++; } else { mp.Add(str[i], 1); } // update answer and count if (cnt < mp[str[i]]) { ans = str[i]; cnt = mp[str[i]]; } } return ans; } } JavaScript // JavaScript program for the above approach // function that return maximum occurring character function getMaxOccurringChar(str) { // create map to store frequency of every character let mp = new Map(); // to store length of string let n = str.length; // to store answer let ans; // to check count of answer character is less or greater // than another elements count let cnt=0; // traverse the string for(let i=0 ;i<n ; i++){ // push element into map and increase its frequency mp.set(str[i], (mp.get(str[i]) || 0) + 1); // update answer and count if(cnt < mp.get(str[i])){ ans = str[i]; cnt = mp.get(str[i]); } } return ans; } // Driver Code let str = "sample string"; console.log("Max occurring character is: " + getMaxOccurringChar(str)); // This code is contributed by rutikbhosale OutputMax occurring character is: sTime Complexity: O(N), Traversing the string of length N one time.Auxiliary Space: O(N), where N is the size of the stringThe idea is to store the frequency of every character in the array and return the character with maximum count.Follow the steps to solve the problem:Create a count array of size 256 to store the frequency of every character of the stringMaintain a max variable to store the maximum frequency so far whenever encounter a frequency more than the max then update the maxAnd update that character in our result variable.Below is the implementation of the above approach: C++ #include <iostream> #include <unordered_map> #include <algorithm> std::unordered_map<char, int> findMaxCharacterCount(const std::string& str) { std::unordered_map<char, int> countMap; // Count occurrences of each character for (char ch : str) { countMap[ch]++; } // Find the character with the maximum count char maxChar = '\0'; int maxCount = 0; for (const auto& entry : countMap) { if (entry.second > maxCount) { maxChar = entry.first; maxCount = entry.second; } } std::unordered_map<char, int> result; result[maxChar] = maxCount; return result; } int main() { std::string str = "geeksforgeeks"; // Call the function and print the result std::unordered_map<char, int> result = findMaxCharacterCount(str); std::cout << "Character: " << result.begin()->first << ", Count: " << result.begin()->second << std::endl; return 0; } Java import java.util.HashMap; import java.util.Map; public class Main { public static Map<Character, Integer> findMaxCharacterCount(String str) { // Initialize max with the first character and its count Map<Character, Integer> max = new HashMap<>(); max.put(str.charAt(0), str.length() - str.replace(String.valueOf(str.charAt(0)), "").length()); // Iterate through the string to find the character with the maximum count for (char i : str.toCharArray()) { int count = str.length() - str.replace(String.valueOf(i), "").length(); if (count > max.get(max.keySet().iterator().next())) { // Update max if the count is higher max.clear(); max.put(i, count); } } return max; } public static void main(String[] args) { String str = "geeksforgeeks"; // Call the function and print the result Map<Character, Integer> result = findMaxCharacterCount(str); System.out.println("Character: " + result.keySet().iterator().next() + ", Count: " + result.get(result.keySet().iterator().next())); } } Python from collections import defaultdict def find_max_character_count(s): count_map = defaultdict(int) # Count occurrences of each character for ch in s: count_map[ch] += 1 # Find the character with the maximum count max_char = '' max_count = 0 for char, count in count_map.items(): if count > max_count: max_char = char max_count = count return {max_char: max_count} # Main function to test if __name__ == "__main__": string = "geeksforgeeks" # Call the function and print the result result = find_max_character_count(string) for char, count in result.items(): # Using .format() for string formatting compatibility with older Python versions print("Character: {}, Count: {}".format(char, count)) C# using System; using System.Collections.Generic; using System.Linq; class Program { static Dictionary<char, int> FindMaxCharacterCount(string str) { Dictionary<char, int> countMap = new Dictionary<char, int>(); // Count occurrences of each character foreach (char ch in str) { if (countMap.ContainsKey(ch)) { countMap[ch]++; } else { countMap[ch] = 1; } } // Find the character with the maximum count char maxChar = '\0'; int maxCount = 0; foreach (var entry in countMap) { if (entry.Value > maxCount) { maxChar = entry.Key; maxCount = entry.Value; } } Dictionary<char, int> result = new Dictionary<char, int>(); result[maxChar] = maxCount; return result; } static void Main(string[] args) { string str = "geeksforgeeks"; // Call the function and print the result Dictionary<char, int> result = FindMaxCharacterCount(str); Console.WriteLine($"Character: {result.First().Key}, Count: {result.First().Value}"); } } JavaScript // Function to find the character with the maximum count in a string function findMaxCharacterCount(str) { let countMap = {}; // Count occurrences of each character for (let ch of str) { if (countMap[ch]) countMap[ch]++; else countMap[ch] = 1; } // Find the character with the maximum count let maxChar = ''; let maxCount = 0; for (let [char, count] of Object.entries(countMap)) { if (count > maxCount) { maxChar = char; maxCount = count; } } let result = {}; result[maxChar] = maxCount; return result; } // Main function function main() { let str = "geeksforgeeks"; // Call the function and print the result let result = findMaxCharacterCount(str); let maxEntry = Object.entries(result)[0]; console.log(`Character: ${maxEntry[0]}, Count: ${maxEntry[1]}`); } // Invoke main function main(); OutputCharacter: e, Count: 4 Time Complexity: O(N), Traversing the string of length N one time.Auxiliary Space: O(1) Return maximum occurring character in an input string Comment More infoAdvertise with us Next Article Asymptotic Notations for Analysis of Algorithms K kartik Follow Improve Article Tags : Strings Hash DSA Morgan Stanley Practice Tags : Morgan StanleyHashStrings Similar Reads Basics & PrerequisitesTime and Space ComplexityMany times there are more than one ways to solve a problem with different algorithms and we need a way to compare multiple ways. 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