Given a set of words (without duplicates), find all word squares you can build from them. A sequence of words forms a valid word square if the kth row and column read the exact same string, where 0 ≤ k < max(numRows, numColumns). For example, the wor…
Write a bash script to calculate the frequency of each word in a text file words.txt. For simplicity sake, you may assume: words.txt contains only lowercase characters and space ' ' characters. Each word must consist of lowercase characters only. Wor…
Given an array of n distinct non-empty strings, you need to generate minimal possible abbreviations for every word following rules below. Begin with the first character and then the number of characters abbreviated, which followed by the last charact…
Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be us…
class Solution: def lengthOfLastWord(self, s): """ :type s: str :rtype: int """ l = [] for i in s[::-1].lstrip(): if i != ' ': l.append(i) else: break return len(l) w = Solution() print(w.lengthOfLastWord("hello world&qu…
Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary For example, Given:start…
121. 买卖股票的最佳时机(简单) [分类]:模拟.思维 [题解]:可以用O(n)的复杂度完成,只需要在遍历的时候记录到当前位置为止买入股票的最小价格minn,再维护一个当前卖出股票价(a-minn)的最大值即可. [代码]: C++: class Solution { public: int maxProfit(vector<int>& prices) { int minn = 0x3f3f3f3f; int ans = 0; for(int i = 0; i < price…
