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substring-with-concatenation-of-all-words

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	`1`	`+# Substring with Concatenation of All Words`
	`2`	`+`
	`3`	`+You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.`
	`4`	`+`
	`5`	`+## Example 1`
	`6`	`+`
	`7`	`+Input:`
	`8`	`+`
	`9`	`+ s = "barfoothefoobarman", words = ["foo","bar"]`
	`10`	`+`
	`11`	`+Output: [0,9]`
	`12`	`+`
	`13`	`+Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.`
	`14`	`+The output order does not matter, returning [9,0] is fine too.`
	`15`	`+`
	`16`	`+## Example 2`
	`17`	`+`
	`18`	`+Input:`
	`19`	`+`
	`20`	`+ s = "wordgoodgoodgoodbestword",`
	`21`	`+`
	`22`	`+ words = ["word","good","best","word"]`
	`23`	`+`
	`24`	`+Output: []`
	`25`	`+`
	`26`	`+## More info`
	`27`	`+`
	`28`	`+<https://leetcode.com/problems/substring-with-concatenation-of-all-words/>`

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	`1`	`+/**`
	`2`	`+ * @param {string} s`
	`3`	`+ * @param {string[]} words`
	`4`	`+ * @return {number[]}`
	`5`	`+ */`
	`6`	`+var findSubstring = function (s, words) {`
	`7`	`+ if (!words \|\| words.length === 0) return [];`
	`8`	`+`
	`9`	`+ let uniqueWordsMap = {};`
	`10`	`+ let n = words.length, m = words[0].length;`
	`11`	`+ let len = n * m;`
	`12`	`+ let resp = [];`
	`13`	`+ for (word of words) uniqueWordsMap[word] = ~~uniqueWordsMap[word] + 1;`
	`14`	`+ for (let i = 0; i < s.length - len + 1; i++) {`
	`15`	`+ let temp = Object.assign({}, uniqueWordsMap);`
	`16`	`+ for (let j = i; j < i + len; j += m) {`
	`17`	`+ const str = s.substr(j, m);`
	`18`	`+ if (!(str in temp)) break;`
	`19`	`+ if (--temp[str] === 0) delete temp[str];`
	`20`	`+ }`
	`21`	`+ if (Object.keys(temp).length === 0) resp.push(i);`
	`22`	`+ }`
	`23`	`+ return resp;`
	`24`	`+};`
	`25`	`+`
	`26`	`+console.log(findSubstring("", []))`

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