Why this code is giving SIGSEGV error?

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The seg fault issue is due to incorrect operation with strings and improper separation between local and global variables.

First, all char a[20] have to be changed to char a[21]. Same for char b[100][20] - it should be char b[100][21]. Per problem statement, the maximum size of the words is 20 characters. That means, however, that in order to store it and operate on it as a string object, we need to allocate 21 characters to reserve an extra space for null-terminated character.

As a follow-up, when making a copy of a string, you have to copy strlen(a)+1 characters in the strncpy call.

Second, when you accumulate the typed words, you keep the count in the global variable o but the accumulation happens in the local array b[100][20]. Therefore on every time call the b[100][20] array is brand new, and it is not in sync with the o counter.

Once all these issues are fixed, you also need to adjust the output since per problem statement an answer for each test case should be on a new line - you have to add \n to printf.

Here’s the adjusted code that gets AC:

#include <stdio.h>
#include <string.h>
int N,o=0,p;
char b[100][21]; // should be global otherwise modifications are
                 // inconsistent with 'o' counter changes
int time(char a[21])
{
    int wt=2;
    for(int i=1;i<strlen(a);i++)
    {
        if(a[i-1]==a[i])
            wt=wt+4;
        else if((a[i-1]=='d'&&a[i]=='f')||(a[i-1]=='f'&&a[i]=='d'))
            wt=wt+4;
        else if((a[i-1]=='j'&&a[i]=='k')||(a[i-1]=='k'&&a[i]=='j'))
            wt=wt+4;
        else
            wt=wt+2;
    }
    for(int j=0;j<o;j++)
    {
        p=strncmp(b[j],a,strlen(a));
        if((p==0) && (strlen(b[j]) == strlen(a)))
        {
            wt=wt/2;
            break;
        }
    }
    strncpy(b[o],a,strlen(a)+1); // +1 is for copying the null-terminating character
    o++;
    return wt;
}
int main()
{
    int j,t=0,sum=0;
    scanf("%d",&t);
    for(int r=0;r<t;r++)
    {
        o = 0;
        sum=0;
        scanf("%d",&N);
        char a[21];
        for(int i=0;i<N;i++)
        {
            scanf("%s",&a);
            sum=sum+time(a);
        }
        printf("%d\n",sum); // the end-of-line was missing in the output
    }
    return 0;
}

However I would solve the problem a bit differently. As you have seen, operating with strings on character level is quite error-prone. IMHO, it is easier and more robust to operate with strings directly, and also use some of the STL data structures like std::vector and std::set. Here’s how I see a better organized solution:

#include <bits/stdc++.h>
using namespace std;

int solve_word(const string& s) {
    vector<int> hand;
    for(int i=0; i<s.size(); ++i) {
        hand.push_back((s[i]=='d' || s[i]=='f') ? 0 : 1);
    }
    int res = 2;
    for(int i=1; i<s.size(); ++i) {
        res += (hand[i-1] == hand[i]) ? 4 : 2;
    }
    return res;
}

int solve_test_case(const vector<string>& w) {
    const int N = w.size();
    set<string> typed;
    int res = 0;
    for(int i=0; i<N; ++i) {
        int v = solve_word(w[i]);
        if (typed.count(w[i])) {
            v /= 2;
        }
        res += v;
        typed.insert(w[i]);
    }
    return res;
}

int main() {
    int T, N;
    cin >> T;
    for(int t=0; t<T; ++t) {
        cin >> N;
        vector<string> w(N);
        for(int i=0; i<N; ++i) {
            cin >> w[i];
        }
        int res = solve_test_case(w);
        cout << res << endl;
    }
    return 0;
}
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Thank you for solution.