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Algorithm : Find the number of occurrence of each character presents in a string ?

+4 votes
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Algorithm : Find the number of occurrence of each character presents in a string ?
posted Jun 28, 2015 by Alok

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2 Answers

+1 vote

Below is my solution :

you only have a need of one array of int .

int count[256];
char  string[]="jhgdkwjhfoeirjbvndbjhegriuegbv";

for(int i=0;i<strlen(string); i++)
{
   count[(int)string[i]]++;
}

for(int i=0;i<256;i++)
{
  if(count[i])
     cout<<(char)(i)<<" "<<count[i];
}
answer Oct 31, 2015 by Bhagwat Singh
0 votes

This C Program counts the number of occurrence of each character ignoring the case and prints them.

/*
 * C Program to Count the Number of Occurrence of
 * each Character Ignoring the Case of Alphabets
 * & Display them
 */
#include <stdio.h>
#include <string.h>
#include <ctype.h>

struct detail
{
    char c;
    int freq;
};

int main()
{
    struct detail s[26];
    char string[100], c;
    int i = 0, index;

    for (i = 0; i < 26; i++)
    {
       s[i].c = i + 'a';
       s[i].freq = 0;
    }
    printf("Enter string: ");
    i = 0;
    do
    {
        fflush(stdin);
        c = getchar();
        string[i++] = c;
        if (c == '\n')
        {
            break;
        }
        c = tolower(c);
        index = c - 'a';
        s[index].freq++;
    } while (1);
    string[i - 1] = '\0';
    printf("The string entered is: %s\n", string);

    printf("*************************\nCharacter\tFrequency\n*************************\n");
    for (i = 0; i < 26; i++)
    {
        if (s[i].freq)
        {
            printf("     %c\t\t   %d\n", s[i].c, s[i].freq);
        }
    }

    return 0;
}

Enter string: A quIck brOwn fox JumpEd over a lazy dOg

The string entered is: A quIck brOwn fox JumpEd over a lazy dOg

Character Frequency

 a         3
 b         1
 c         1
 d         2
 e         2
 f         1
 g         1
 i         1
 j         1
 k         1
 l         1
 m          1
 n         1
 o         4
 p         1
 q         1
 r         2
 u         2
 v         1
 w           1
 x         1
 y         1
 z         1
answer Jul 14, 2015 by Mohammed Hussain
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