3412. Find Mirror Score of a String

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3412. Find Mirror Score of a String

Description

You are given a string s.

We define the mirror of a letter in the English alphabet as its corresponding letter when the alphabet is reversed. For example, the mirror of 'a' is 'z', and the mirror of 'y' is 'b'.

Initially, all characters in the string s are unmarked .

You start with a score of 0, and you perform the following process on the string s:

  • Iterate through the string from left to right.
  • At each index i, find the closest unmarked index j such that j < i and s[j] is the mirror of s[i]. Then, mark both indices i and j, and add the value i - j to the total score.
  • If no such index j exists for the index i, move on to the next index without making any changes.

Return the total score at the end of the process.

Example 1:

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Input: s = "aczzx"

Output: 5

Explanation:

  • i = 0. There is no index j that satisfies the conditions, so we skip.
  • i = 1. There is no index j that satisfies the conditions, so we skip.
  • i = 2. The closest index j that satisfies the conditions is j = 0, so we mark both indices 0 and 2, and then add 2 - 0 = 2 to the score.
  • i = 3. There is no index j that satisfies the conditions, so we skip.
  • i = 4. The closest index j that satisfies the conditions is j = 1, so we mark both indices 1 and 4, and then add 4 - 1 = 3 to the score.

Example 2:

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Input: s = "abcdef"

Output: 0

Explanation:

For each index i, there is no index j that satisfies the conditions.

Constraints:

  • 1 <= s.length <= 10^5
  • s consists only of lowercase English letters.

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    long long calculateScore(string s) {
        vector<stack<int>> nums(26);
        long long res = 0;
        int n = s.size();
        for (int i = 0; i < n; i++) {
            int idx = s[i] - 'a';
            int reverseIdx = 25 - idx;
            if (!nums[reverseIdx].empty()) {
                int prev = nums[reverseIdx].top();
                nums[reverseIdx].pop();
                res += i - prev;
            } else {
                nums[idx].push(i);
            }
        }
        return res;
    }
};