LeetCode in TypeScript
30. Substring with Concatenation of All Words
Hard
You are given a string s and an array of strings words of the same length. Return all starting indices of substring(s) in s that is a concatenation of each word in words exactly once, in any order, and without any intervening characters.
You can return the answer in any order.
Example 1:
Input: s = “barfoothefoobarman”, words = [“foo”,”bar”]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are “barfoo” and “foobar” respectively. The output order does not matter, returning [9,0] is fine too.
Example 2:
Input: s = “wordgoodgoodgoodbestword”, words = [“word”,”good”,”best”,”word”]
Output: []
Example 3:
Input: s = “barfoofoobarthefoobarman”, words = [“bar”,”foo”,”the”]
Output: [6,9,12]
Constraints:
1 <= s.length <= 104sconsists of lower-case English letters.1 <= words.length <= 50001 <= words[i].length <= 30words[i]consists of lower-case English letters.
Solution
function findSubstring(s: string, words: string[]): number[] {
let ans: number[] = []
let n1 = words[0].length
let n2 = s.length
let map1 = new Map<string, number>()
for (let ch of words) {
map1.set(ch, (map1.get(ch) ?? 0) + 1)
}
for (let i = 0; i < n1; i++) {
let left = i
let j = i
let c = 0
let map2 = new Map<string, number>()
while (j + n1 <= n2) {
let word1 = s.substring(j, j + n1)
j += n1
if (map1.has(word1)) {
map2.set(word1, (map2.get(word1) ?? 0) + 1)
c++
while ((map2.get(word1) ?? 0) > (map1.get(word1) ?? 0)) {
let word2 = s.substring(left, left + n1)
map2.set(word2, (map2.get(word2) ?? 0) - 1)
left += n1
c--
}
if (c === words.length) {
ans.push(left)
}
} else {
map2.clear()
c = 0
left = j
}
}
}
return ans
}
export { findSubstring }