[LC] 3297 Count Substrings That Can Be Rearranged to Contain a String I

Java

/**
 * <a href="https://leetcode.cn/problems/count-substrings-that-can-be-rearranged-to-contain-a-string-i/">Count Substrings That Can Be Rearranged to Contain a String I</a>
 * 哈希表；字符串；滑动窗口
 */
class Solution {
    public long validSubstringCount(String word1, String word2) {
        int[] cnt = new int[26];
        int diff = 0;    // word2中剩余的字符种类数
        for (char c : word2.toCharArray()) {
            int index = c - 'a';
            if (cnt[index] == 0) {
                diff++;
            }
            cnt[index]++;
        }

        long res = 0;
        char[] arr = word1.toCharArray();
        for (int left = 0, right = 0; right < arr.length; right++) {
            int push = arr[right] - 'a';
            cnt[push]--;
            if (cnt[push] == 0) {
                diff--;
            }
            while (diff == 0) {
                int pop = arr[left++] - 'a';
                if (cnt[pop] == 0) {
                    diff++;
                }
                cnt[pop]++;
            }
            // 每当diff为0时，说明当前窗口内的子串可以重新排列成word2
            // 此时窗口左边界left之前的所有字符都可以作为窗口的起始位置，因此可以形成left个满足条件的子串
            res += left;    // 将left加到res中，累加满足条件的子串数量
        }

        return res;
    }
}

Go

func validSubstringCount(word1 string, word2 string) int64 {
    cnt := make([]int, 26)
    diff := 0
    for _, c := range word2 {
        index := c - 'a'
        if cnt[index] == 0 {
            diff++
        }
        cnt[index]++
    }

    var res int64 = 0
    for left, right := 0, 0; right < len(word1); right++ {
        push := word1[right] - 'a'
        cnt[push]--
        if cnt[push] == 0 {
            diff--
        }
        for diff == 0 {
            pop := word1[left] - 'a'
            left++
            if cnt[pop] == 0 {
                diff++
            }
            cnt[pop]++
        }
        res += int64(left)
    }

    return res
}