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.
\n
Example 1:
\n\nInput: s = "???"
\n\nOutput: "abc"
\n\nExplanation: 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.
\nExample 2:
\n\nInput: s = "a?a?"
\n\nOutput: "abac"
\n\nExplanation: 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.
\n
Constraints:
\n\n1 <= s.length <= 105s[i] is either a lowercase English letter or '?'.给你一个字符串 s 。s[i] 要么是小写英文字母，要么是问号 '?' 。
对于长度为 m 且 只 含有小写英文字母的字符串 t ，我们定义函数 cost(i) 为下标 i 之前（也就是范围 [0, i - 1] 中）出现过与 t[i] 相同 字符出现的次数。
字符串 t 的 分数 为所有下标 i 的 cost(i) 之 和 。
比方说，字符串 t = \"aab\" ：
cost(0) = 0cost(1) = 1cost(2) = 0\"aab\" 的分数为 0 + 1 + 0 = 1 。你的任务是用小写英文字母 替换 s 中 所有 问号，使 s 的 分数最小 。
请你返回替换所有问号 '?' 之后且分数最小的字符串。如果有多个字符串的 分数最小 ，那么返回字典序最小的一个。
\n\n
示例 1：
\n\n输入：s = \"???\"
\n\n输出： \"abc\"
\n\n解释：这个例子中，我们将 s 中的问号 '?' 替换得到 \"abc\" 。
对于字符串 \"abc\" ，cost(0) = 0 ，cost(1) = 0 和 cost(2) = 0 。
\"abc\" 的分数为 0 。
其他修改 s 得到分数 0 的字符串为 \"cba\" ，\"abz\" 和 \"hey\" 。
这些字符串中，我们返回字典序最小的。
\n示例 2：
\n\n输入： s = \"a?a?\"
\n\n输出： \"abac\"
\n\n解释：这个例子中，我们将 s 中的问号 '?' 替换得到 \"abac\" 。
对于字符串 \"abac\" ，cost(0) = 0 ，cost(1) = 0 ，cost(2) = 1 和 cost(3) = 0 。
\"abac\" 的分数为 1 。
\n\n
提示：
\n\n1 <= s.length <= 105s[i] 要么是小写英文字母，要么是 '?' 。The cost does not depend on the order of characters. If a character c appears x times, the cost is exactly 0 + 1 + 2 + … + (x − 1) = x * (x − 1) / 2.
We know the total number of question marks; for each one, we should select the letter with the minimum frequency to replace it.
", "The letter selection can be achieved by a min-heap (or even by brute-forcing the 26 possibilities).
So, we know the extra letters we need to replace finally. However, we must put those letters in order from left to right so that the resulting string is the lexicographically smallest one.
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