Segment Words

mediumcodingAlgorithmsDynamic ProgrammingStrings~30 min

Asked at Google, Meta, Stripe

Overview

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

The same word in the dictionary may be reused multiple times.

Constraints

1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
s and wordDict[i] consist of lowercase English letters.
All strings in wordDict are unique.

Examples

wordBreak("leetcode", ["leet", "code"]);
// => true  ("leet" + "code")

wordBreak("applepenapple", ["apple", "pen"]);
// => true  ("apple" + "pen" + "apple")

wordBreak("catsandog", ["cats", "dog", "sand", "and", "cat"]);
// => false

Solution

Reveal solution

function wordBreak(s, wordDict) {
  const wordSet = new Set(wordDict);
  const dp = new Array(s.length + 1).fill(false);
  dp[0] = true;

  for (let i = 1; i <= s.length; i++) {
    for (let j = 0; j < i; j++) {
      if (dp[j] && wordSet.has(s.substring(j, i))) {
        dp[i] = true;
        break;
      }
    }
  }

  return dp[s.length];
}

Bottom-up DP. dp[i] is true if s[0..i) can be segmented. For each position, check all possible last words.

Resources

segment-words.js

Segment Words

mediumcodingAlgorithmsDynamic Programming

Overview

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

The same word in the dictionary may be reused multiple times.

Constraints

1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
s and wordDict[i] consist of lowercase English letters.
All strings in wordDict are unique.

Examples

wordBreak("leetcode", ["leet", "code"]);
// => true  ("leet" + "code")

wordBreak("applepenapple", ["apple", "pen"]);
// => true  ("apple" + "pen" + "apple")

wordBreak("catsandog", ["cats", "dog", "sand", "and", "cat"]);
// => false

Solution

Reveal solution

function wordBreak(s, wordDict) {
  const wordSet = new Set(wordDict);
  const dp = new Array(s.length + 1).fill(false);
  dp[0] = true;

  for (let i = 1; i <= s.length; i++) {
    for (let j = 0; j < i; j++) {
      if (dp[j] && wordSet.has(s.substring(j, i))) {
        dp[i] = true;
        break;
      }
    }
  }

  return dp[s.length];
}

Bottom-up DP. dp[i] is true if s[0..i) can be segmented. For each position, check all possible last words.

Resources

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