3081. Replace Question Marks in String to Minimize Its Value

Description

You are given a string s. s[i] is either a lowercase English letter or '?'.

For a string t having length m containing only lowercase English letters, we define the function cost(i) for an index i as the number of characters equal to t[i] that appeared before it, i.e. in the range [0, i - 1].

The value of t is the sum of cost(i) for all indices i.

For example, for the string t = "aab":

• cost(0) = 0
• cost(1) = 1
• cost(2) = 0
• Hence, the value of "aab" is 0 + 1 + 0 = 1.

Your task is to replace all occurrences of '?' in s with any lowercase English letter so that the value of s is minimized.

Return a string denoting the modified string with replaced occurrences of '?'. If there are multiple strings resulting in the minimum value, return the lexicographically smallest one.

 

Example 1:

Input: s = "???"

Output: "abc"

Explanation: In this example, we can replace the occurrences of '?' to make s equal to "abc".

For "abc", cost(0) = 0, cost(1) = 0, and cost(2) = 0.

The value of "abc" is 0.

Some other modifications of s that have a value of 0 are "cba", "abz", and, "hey".

Among all of them, we choose the lexicographically smallest.

Example 2:

Input: s = "a?a?"

Output: "abac"

Explanation: In this example, the occurrences of '?' can be replaced to make s equal to "abac".

For "abac", cost(0) = 0, cost(1) = 0, cost(2) = 1, and cost(3) = 0.

The value of "abac" is 1.

 

Constraints:

• 1 <= s.length <= 105
• s[i] is either a lowercase English letter or '?'.

Solutions

Solution 1: Greedy + Priority Queue

According to the problem, we can find that if a letter $c$ appears $v$ times, then the score it contributes to the answer is $1 + 2 + \cdots + (v - 1) = \frac{v \times (v - 1)}{2}$. To make the answer as small as possible, we should replace the question marks with those letters that appear less frequently.

Therefore, we can use a priority queue to maintain the occurrence times of each letter, take out the letter with the least occurrence times each time, record it in the array $t$, then increase its occurrence times by one, and put it back into the priority queue. Finally, we sort the array $t$, and then traverse the string $s$, replacing each question mark with the letters in the array $t$ in turn.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string $s$.

Python3JavaC++Go

class Solution:
    def minimizeStringValue(self, s: str) -> str:
        cnt = Counter(s)
        pq = [(cnt[c], c) for c in ascii_lowercase]
        heapify(pq)
        t = []
        for _ in range(s.count("?")):
            v, c = pq[0]
            t.append(c)
            heapreplace(pq, (v + 1, c))
        t.sort()
        cs = list(s)
        j = 0
        for i, c in enumerate(s):
            if c == "?":
                cs[i] = t[j]
                j += 1
        return "".join(cs)

class Solution {
    public String minimizeStringValue(String s) {
        int[] cnt = new int[26];
        int n = s.length();
        int k = 0;
        char[] cs = s.toCharArray();
        for (char c : cs) {
            if (c == '?') {
                ++k;
            } else {
                ++cnt[c - 'a'];
            }
        }
        PriorityQueue<int[]> pq
            = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
        for (int i = 0; i < 26; ++i) {
            pq.offer(new int[] {cnt[i], i});
        }
        int[] t = new int[k];
        for (int j = 0; j < k; ++j) {
            int[] p = pq.poll();
            t[j] = p[1];
            pq.offer(new int[] {p[0] + 1, p[1]});
        }
        Arrays.sort(t);

        for (int i = 0, j = 0; i < n; ++i) {
            if (cs[i] == '?') {
                cs[i] = (char) (t[j++] + 'a');
            }
        }
        return new String(cs);
    }
}

class Solution {
public:
    string minimizeStringValue(string s) {
        int cnt[26]{};
        int k = 0;
        for (char& c : s) {
            if (c == '?') {
                ++k;
            } else {
                ++cnt[c - 'a'];
            }
        }
        priority_queue<pair<int, int>, vector<pair<