# C program to find the frequency of characters in a string

Last Updated : 01 Jun, 2023

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 - 2

Input: S="gfg"
Output:
f - 1
g - 2```

### Approach

Follow the steps to solve the problem:

1. 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’, the 1st index stores the frequency of the character ‘b’, and so on.
2. 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.
3. 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 to find the frequency ` `// of characters in a string` `#include ` `#include `   `// Function to print the frequencies` `// of each character of the string` `void` `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 string` `void` `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 Code` `int` `main()` `{` `    ``char` `S[100] = ``"geeksforgeeks"``;` `    ``findFrequncy(S);` `}`

Output:

```e - 4
f - 1
g - 2
k - 2
o - 1
r - 1
s - 2```

#### Complexity of the above method

Time Complexity: O(N)

Auxiliary Space: O(26)

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