C program to find the frequency of characters in a string
Given a string S containing lowercase English characters, the task is to find the frequency of all the characters in the string.
Examples:
Input: S=”geeksforgeeks”
Output:
e – 4
f – 1
g – 2
k – 2
o – 1
r – 1
s – 2Input: S=”gfg”
Output:
f – 1
g – 2
Approach: Follow the steps to solve the problem:
- Initialize an array freq[] to store the frequency of each alphabet in the given string. The 0th index stores the frequency of the character ‘a’, 1st/sup> index stores the frequency of the character ‘b’ and so on.
- Iterate over the given string S and increment the frequency of each character encountered by 1, by performing freq[S[i] – ‘a’] += 1. If S[i] = ‘a’, then S[i] – ‘a’ is equal to 0, therefore the frequency of ‘a’ is incremented in the array.=
- After complete traversal of the string, print the frequency of all the characters in the string by traversing the array freq[].
Below is the implementation of the above approach:
C
// C program for the above approach#include <stdio.h>#include <string.h> // Function to print the frequencies// of each character of the stringvoid printFrequency(int freq[]){ for (int i = 0; i < 26; i++) { // If frequency of the // alphabet is non-zero if (freq[i] != 0) { // Print the character and // its respective frequency printf("%c - %d\n", i + 'a', freq[i]); } }} // Function to calculate the frequencies// of each character of the stringvoid findFrequncy(char S[]){ int i = 0; // Stores the frequencies // of each character int freq[26] = { 0 }; // Traverse over the string while (S[i] != '\0') { // Increment the count of // each character by 1 freq[S[i] - 'a']++; // Increment the index i++; } // Function call to print // the frequencies printFrequency(freq);} // Driver Codeint main(){ char S[100] = "geeksforgeeks"; findFrequncy(S);} |
Output:
e - 4 f - 1 g - 2 k - 2 o - 1 r - 1 s - 2
Time Complexity: O(N)
Auxiliary Space: O(26)
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