HACKER RANK Algorithm Strings Challenges - Gemstones Solution.'C'

Algorithm Strings Challenges - Gemstones Solution

Problem Statement

John has discovered various rocks. Each rock is composed of various elements, and each element is represented by a lower-case Latin letter from 'a' to 'z'. An element can be present multiple times in a rock. An element is called a gem-element if it occurs at least once in each of the rocks.

Given the list of N rocks with their compositions, display the number of gem-elements that exist in those rocks.

Input Format

The first line consists of an integer, N, the number of rocks.
Each of the next N lines contains a rock's composition. Each composition consists of lower-case letters of English alphabet.

Constraints
1≤N≤100
Each composition consists of only lower-case Latin letters ('a'-'z').
1≤ length of each composition ≤100

Output Format

Print the number of gem-elements that are common in these rocks. If there are none, print 0.

Sample Input

3
abcdde
baccd
eeabg

Sample Output

2

Explanation

Only "a" and "b" are the two kinds of gem-elements, since these are the only characters that occur in every rock's composition.

Solution:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
    int i,j,n,k,cnt=0,c=0;
    scanf("%d",&n);
    char str[n][100];
    int h1[n][26];
    for(i=0;i<n;i++)
     for(j=0;j<26;j++)
          h1[i][j]=0;
    for(i=0;i<n;i++)
        {
        fflush(stdin);
        scanf("%s",str[i]);
        for(j=0;j<strlen(str[i]);j++)
            {
           h1[i][(str[i][j])-97]=1;       
        }
       }
        i=0;
        for(j=0;j<26;j++)
        {
            cnt=0;
            for(k=1;k<n;k++)
            {
              if(h1[i][j]!=0)
              {
               if((h1[i][j]==h1[k][j]))
                    cnt++;
              }
            }
           if(cnt==n-1)
           {
                c++;
           }
    }
    printf("%d",c);
    return 0;
}

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