[CodeForces] Perform the Combo

An interesting problem using suffix sum.

For each wrong tries that is correct until after index p[i], we add 1 to cnt[p[i]]. This represents that all characters before and including s[p[i]] have 1 more appearance. After processing all wrong tries, we sum up each cnt[i] backward, from right to left. The reason is that any wrong try always start from the first character. So cnt[j] <= cnt[i], if j > i. In other word, for a given cnt[i] that represents the count of s[i] for all wrong tries, it is equal to the number of wrong tries that goes correct at least as far as s[i], which is essentially the sum for all cnt[j], j >= i. Hence the name suffix sum.

    private static void solve(int q, FastScanner in, PrintWriter out) {
        for (int qq = 0; qq < q; qq++) {
            int n = in.nextInt();
            int m = in.nextInt();
            String s = in.next();
            int[] p = new int[m];
            for(int i = 0; i < m; i++) {
                p[i] = in.nextInt() - 1;
            }

            long[] cnt = new long[n];
            for(int i = 0; i < m; i++) {
                cnt[p[i]]++;
            }
            for(int i = n - 2; i >= 0; i--) {
                cnt[i] += cnt[i + 1];
            }
            long[] ans = new long[26];
            for(int i = 0; i < n; i++) {
                ans[s.charAt(i) - 'a'] += (cnt[i] + 1);
            }
            for(int j = 0; j < 26; j++) {
                out.print(ans[j] + " ");
            }
            out.println();
        }
        out.close();
    }