char in string is not in order wanted - string

Question

I am a beginner and I need some help here. This program prints out the frequency of char in the string, e.g. if user enters zzaaa it prints out a3z2 and what I need to print is z2a3 since z is entered first before a. But I am having a hard time switching the order around. Thanks in advance!
int main
{
int ib, i=0, j=0, k=0;
int count[26] = {0};
char chh[3][10];
for (ib = 0; ib < 3; ib++) // get 3 input
gets(chh[ib]);
for (i = 0; i < 3; i++)
{
for (j = 0; j < 10; j++)
{
if (chh[i][j] >= 'a' && chh[i][j] <= 'z')
{
count[chh[i][j] - 'a']++;
}
}
for (k = 0; k < 26; k++)
{
if (count[k] != 0) // if array location is not equals to 0
printf("%c%d", k + 'a', count[k]);
}
memset(count, 0, sizeof(count)); //reset integer array
printf("\n");
}

Answer 1

It prints a before z because you arranged count from a to z by alphabetic priority not entering priority:
count[chh[i][j] - 'a']
if you want to print them by entering priority you should change it. there are several ways to do this. like this:
#include <stdio.h>
#include <string.h>
int main()
{
int ib, i=0, j=0,k=0, kk=0,c=0,found=0;
int count[26][2];
char chh[3][10];
for (ib = 0; ib < 3; ib++) // get 3 input
gets(chh[ib]);
printf("output is:\n");
for (i=0;i<26;i++)
{
count[i][0]=0;
count[i][1]=0;
}
for (i = 0; i < 3; i++)
{
for (j = 0; j < 10; j++)
{
if (chh[i][j] >= 'a' && chh[i][j] <= 'z')
{
found=0;
for (c=0;c<kk;c++)
if (count[c][0]==chh[i][j])
{
count[c][1]++;
found=1;
break;
}
if (!found)
{
count[c][0]=chh[i][j];
count[c][1]++;
kk++;
}
}
}
for (k = 0; k < 26; k++)
{
if (count[k][1] != 0) // if array location is not equals to 0
printf("%c%d", count[k][0], count[k][1]);
}
memset(count, 0, sizeof(count)); //reset integer array
printf("\n");
}
}