Word Break

The problem 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. Note that the same word in the dictionary may be reused multiple times in the segmentation. Examples Input: s = "leetcode", wordDict = ["leet","code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code". Input: s = "applepenapple", wordDict = ["apple","pen"] Output: true Explanation: Return true because "applepenapple" can be segmented as "apple pen apple". Note that you are allowed to reuse a dictionary word. Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"] Output: false Constraints 1 <= s.length <= 300 1 <= wordDict.length <= 1000 1 <= wordDict[i].length <= 20 s and wordDict[i] consist of only lowercase English letters. All the strings of wordDict are unique. Recursion solution func wordBreak(_ s: String, _ wordDict: [String]) -> Bool { let sCount = s.count let sArray = Array(s) func dfs(_ i: Int) -> Bool { if i == sCount { return true } for w in wordDict { let wCount = w.count if ((i + wCount) <= sCount && String(sArray[i ..< i + wCount]) == w ) { if dfs(i + wCount) { return true } } } return false } return dfs(0) } Explanation There are multiple ways to solve this problem. For example, we can check every character in input s and compare it with characters in wordDict, but this approach is inefficient from a time complexity perspective. Instead: ...