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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:

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:

Input: s = "abcdef"

Output: 0

Explanation:

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

 

Constraints:

  • 1 <= s.length <= 105
  • s consists only of lowercase English letters.

Solutions

Solution 1: Hash Table

We can use a hash table $\textit{d}$ to store the index list of each unmarked character, where the key is the character and the value is the list of indices.

We traverse the string $\textit{s}$, and for each character $\textit{x}$, we find its mirror character $\textit{y}$. If $\textit{d}$ contains $\textit{y}$, we take out the index list $\textit{ls}$ corresponding to $\textit{y}$, take out the last element $\textit{j}$ from $\textit{ls}$, and remove $\textit{j}$ from $\textit{ls}$. If $\textit{ls}$ becomes empty, we remove $\textit{y}$ from $\textit{d}$. At this point, we have found a pair of indices $(\textit{j}, \textit{i})$ that meet the condition, and we add $\textit{i} - \textit{j}$ to the answer. Otherwise, we add $\textit{x}$ to $\textit{d}$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $\textit{s}$.

  • class Solution { public long calculateScore(String s) { Map<Character, List<Integer>> d = new HashMap<>(26); int n = s.length(); long ans = 0; for (int i = 0; i < n; ++i) { char x = s.charAt(i); char y = (char) ('a' + 'z' - x); if (d.containsKey(y)) { var ls = d.get(y); int j = ls.remove(ls.size() - 1); if (ls.isEmpty()) { d.remove(y); } ans += i - j; } else { d.computeIfAbsent(x, k -> new ArrayList<>()).add(i); } } return ans; } } 
  • class Solution { public: long long calculateScore(string s) { unordered_map<char, vector<int>> d; int n = s.length(); long long ans = 0; for (int i = 0; i < n; ++i) { char x = s[i]; char y = 'a' + 'z' - x; if (d.contains(y)) { vector<int>& ls = d[y]; int j = ls.back(); ls.pop_back(); if (ls.empty()) { d.erase(y); } ans += i - j; } else { d[x].push_back(i); } } return ans; } }; 
  • class Solution: def calculateScore(self, s: str) -> int: d = defaultdict(list) ans = 0 for i, x in enumerate(s): y = chr(ord("a") + ord("z") - ord(x)) if d[y]: j = d[y].pop() ans += i - j else: d[x].append(i) return ans 
  • func calculateScore(s string) (ans int64) { d := make(map[rune][]int) for i, x := range s { y := 'a' + 'z' - x if ls, ok := d[y]; ok { j := ls[len(ls)-1] d[y] = ls[:len(ls)-1] if len(d[y]) == 0 { delete(d, y) } ans += int64(i - j) } else { d[x] = append(d[x], i) } } return } 
  • function calculateScore(s: string): number { const d: Map<string, number[]> = new Map(); const n = s.length; let ans = 0; for (let i = 0; i < n; i++) { const x = s[i]; const y = String.fromCharCode('a'.charCodeAt(0) + 'z'.charCodeAt(0) - x.charCodeAt(0)); if (d.has(y)) { const ls = d.get(y)!; const j = ls.pop()!; if (ls.length === 0) { d.delete(y); } ans += i - j; } else { if (!d.has(x)) { d.set(x, []); } d.get(x)!.push(i); } } return ans; }

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