Medium
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) = 0cost(1) = 1cost(2) = 0"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 <= 105s[i] is either a lowercase English letter or '?'.public class Solution {
public String minimizeStringValue(String s) {
int n = s.length();
int time = 0;
int[] count = new int[26];
int[] res = new int[26];
char[] str = s.toCharArray();
for (char c : str) {
if (c != '?') {
count[c - 'a']++;
} else {
time++;
}
}
int minTime = Integer.MAX_VALUE;
for (int i = 0; i < 26; i++) {
minTime = Math.min(minTime, count[i]);
}
while (time > 0) {
for (int j = 0; j < 26; j++) {
if (count[j] == minTime) {
res[j]++;
count[j]++;
time--;
}
if (time == 0) {
break;
}
}
minTime++;
}
int start = 0;
for (int i = 0; i < n; i++) {
if (str[i] == '?') {
while (res[start] == 0) {
start++;
}
str[i] = (char) ('a' + start);
res[start]--;
}
}
return new String(str);
}
}