লিটকোড ১৩৯ | LEARN PROBLEM SOLVING with PYTHON | PART 72 : Leetcode 139 : Word Break

[@itvaya]

Algorithm: Word Break Using DFS with Memoization Definitions: - `s`: Input string to check if it can be segmented into words from the dictionary. - `wordDict`: List of valid words for segmentation. - `word_set`: A set of words from `wordDict` for fast lookups. - `cache`: A dictionary to store results of subproblems for memoization. Class: Solution Method: wordBreak(s, wordDict) Input: - s (string): Input string to segment. - wordDict (List[string]): List of valid words. Output: Boolean indicating whether `s` can be segmented into words from `wordDict`. 1. Convert `wordDict` into a set `word_set` for efficient lookups. 2. Initialize an empty dictionary `cache` for memoization. 3. Define a helper function `dfs(s)` for recursive depth-first search: a. If `s` is in `cache`, return `cache[s]` (result for the subproblem is already computed). b. If `s` is in `word_set`, return `True` (the whole string `s` is a valid word). c. For `i` from 1 to `len(s) - 1`: i. Set `pre` to `s[:i]` (prefix of the string). ii. If `pre` is in `word_set` and `dfs(s[i:])` (the suffix can be segmented), do the following: - Set `cache[s] = True`. - Return `True`. d. If no valid segmentation is found, set `cache[s] = False` and return `False`. 4. Call `dfs(s)` to determine if the string `s` can be segmented and return its result. Complexity Analysis: - Time Complexity: O(n²), where `n` is the length of the string. The recursive function evaluates each substring, and slicing takes O(n) time for each split. - Space Complexity: O(n), for the recursion stack and memoization dictionary.

[@HasipSheikh-r3x]

Amazing.

[@wahidurrahman77]

vai Java language theke ashe Python dekhle mone hoi ulonggo

[@itvaya]

🤣🤣🤣