Given an array of strings
words representing an English Dictionary, return the longest word in
words that can be built one character at a time by other words in
If there is more than one possible answer, return the longest word with the smallest lexicographical order. If there is no answer, return the empty string.
Input: words = ["w","wo","wor","worl","world"]
Explanation: The word "world" can be built one character at a time by "w", "wo", "wor", and "worl".
I knew that the best solution is “Tire”. But, I used Hash first and that is method of brute force. I will used “Tire” later.
Sort first. check the word with hashset one by one. If those words are include in hashset, I will compare length of result and the word, and replace result to longer word.
Given a string
s and a string array
dictionary, return the longest string in the dictionary that can be formed by deleting some of the given string characters. If there is more than one possible result, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.
Input: s = "abpcplea", dictionary = ["ale","apple","monkey","plea"]
Because return the longest word with the smallest lexicographical order. So, I need sort dictionary first. Each of two arrays have a pointer. Base on dictionary array and check s string one by one to find the same as char of dictionary. Record the string of dictionary, if satisfaction condition. I compare and get the longest word of dictionary, then return it.
Given a string
true if the
s can be palindrome after deleting at most one character from it.
Input: s = "aba"
Right point start from 0 and shift to left. Left point start from length-1 and shift to right. Until find different character of right and left point. Check remain string is it a palindrome string, when right point shift to left 1 step or left point shift to right 1 step. I need check two conditions. If both true, return true. Otherwise, return false.
Given a non-negative integer
c, decide whether there're two integers
b such that
a2 + b2 = c.
Input: c = 5
Explanation: 1 * 1 + 2 * 2 = 5
I suppose first this question is extension of Two Sum. Then I used HashSet to solve this question. But… The RunTime is too bad. So, I change my thinking to Two Points. small point is start from 0 and shift to right, large point is start from squrt(c) and shift to left. If sum of square of small point and square of large point lesser equal c, return true. If lesser than c, small point shift to right. If larger than c, large point shift to left.
You are given two integer arrays
nums2, sorted in non-decreasing order, and two integers
n, representing the number of elements in
nums2 into a single array sorted in non-decreasing order.
The final sorted array should not be returned by…
Given an array of integers
numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific
Return the indices of the two numbers (1-indexed) as an integer array
answer of size
1 <= answer < answer <= numbers.length.
Given an array
n integers, your task is to check if it could become non-decreasing by modifying at most one element.
We define an array is non-decreasing if
nums[i] <= nums[i + 1] holds for every
i (0-based) such that (
0 <= i <= n - 2).
You are given an array of people,
people, which are the attributes of some people in a queue (not necessarily in order). …
Given two arrays of integers
index. Your task is to create target array under the following rules:
nums[i]in target array.
Return the target array.
It is guaranteed that the insertion operations will be valid.
Input: nums = [0,1,2,3,4], index = [0,1,2,2,1]
You are given a string
s. We want to partition the string into as many parts as possible so that each letter appears in at most one part.
Return a list of integers representing the size of these parts.
Input: s = "ababcbacadefegdehijhklij"
The partition is "ababcbaca", "defegde", "hijhklij".
This is a partition so that each letter appears in at most one part.
A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits s into less parts.
Array fast than HashMap.
Recording the last position of every letter and saving in Array or HashMap.
Sweeping string of input again. setting the end point is the maximum. when the index position is equal to the end point, it means that an interval ends. calculate the difference with start point and end point.