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Posts tagged as “medium”

花花酱 LeetCode 792. Number of Matching Subsequences

题目大意:给你一些单词,问有多少单词出现在字符串S的子序列中。

https://leetcode.com/problems/number-of-matching-subsequences/description/

Given string S and a dictionary of words words, find the number of words[i] that is a subsequence of S.

Note:

  • All words in words and S will only consists of lowercase letters.
  • The length of S will be in the range of [1, 50000].
  • The length of words will be in the range of [1, 5000].
  • The length of words[i] will be in the range of [1, 50].

Solution 1: Brute Force

Time complexity: O((S + L) * W)

C++ w/o cache TLE

Space complexity: O(1)

C++ w/ cache 155 ms

Space complexity: O(W * L)

Solution 2: Indexing+ Binary Search

Time complexity: O(S + W * L * log(S))

Space complexity: O(S)

S: length of S

W: number of words

L: length of a word

C++

Java

 

Python 3:

w/o cache

w/ cache

 

花花酱 LeetCode 795. Number of Subarrays with Bounded Maximum

题目大意:问一个数组中有多少个子数组的最大元素值在[L, R]的范围里。

We are given an array A of positive integers, and two positive integers L and R (L <= R).

Return the number of (contiguous, non-empty) subarrays such that the value of the maximum array element in that subarray is at least L and at most R.

Solution 1:

C++

Solution 2: One pass

C++

 

花花酱 LeetCode 794. Valid Tic-Tac-Toe State

题目大意:判断一个井字棋的棋盘是否有效。

A Tic-Tac-Toe board is given as a string array board. Return True if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.

The board is a 3 x 3 array, and consists of characters " ""X", and "O".  The ” ” character represents an empty square.

Here are the rules of Tic-Tac-Toe:

  • Players take turns placing characters into empty squares (” “).
  • The first player always places “X” characters, while the second player always places “O” characters.
  • “X” and “O” characters are always placed into empty squares, never filled ones.
  • The game ends when there are 3 of the same (non-empty) character filling any row, column, or diagonal.
  • The game also ends if all squares are non-empty.
  • No more moves can be played if the game is over.

Note:

  • board is a length-3 array of strings, where each string board[i] has length 3.
  • Each board[i][j] is a character in the set {" ", "X", "O"}.

Idea: Verify all rules

C++

 

 

花花酱 LeetCode 468. Validate IP Address

题目大意:给你一个字符串,让你判断是否是合法的ipv4或者ipv6地址,都不合法返回Neighter.

Write a function to check whether an input string is a valid IPv4 address or IPv6 address or neither.

IPv4 addresses are canonically represented in dot-decimal notation, which consists of four decimal numbers, each ranging from 0 to 255, separated by dots (“.”), e.g.,172.16.254.1;

Besides, leading zeros in the IPv4 is invalid. For example, the address 172.16.254.01 is invalid.

IPv6 addresses are represented as eight groups of four hexadecimal digits, each group representing 16 bits. The groups are separated by colons (“:”). For example, the address 2001:0db8:85a3:0000:0000:8a2e:0370:7334 is a valid one. Also, we could omit some leading zeros among four hexadecimal digits and some low-case characters in the address to upper-case ones, so 2001:db8:85a3:0:0:8A2E:0370:7334 is also a valid IPv6 address(Omit leading zeros and using upper cases).

However, we don’t replace a consecutive group of zero value with a single empty group using two consecutive colons (::) to pursue simplicity. For example, 2001:0db8:85a3::8A2E:0370:7334 is an invalid IPv6 address.

Besides, extra leading zeros in the IPv6 is also invalid. For example, the address 02001:0db8:85a3:0000:0000:8a2e:0370:7334 is invalid.

Note: You may assume there is no extra space or special characters in the input string.

Example 1:

Example 2:

Example 3:

Solution 1: String / Brute force 

 

花花酱 LeetCode 790. Domino and Tromino Tiling

题目大意:有两种不同形状的骨牌(1×2长条形,L型)无限多块。给你一个2xN的板子,问一共有多少不同的方式可以完全覆盖。

We have two types of tiles: a 2×1 domino shape, and an “L” tromino shape. These shapes may be rotated.

Given N, how many ways are there to tile a 2 x N board? Return your answer modulo 10^9 + 7.

(In a tiling, every square must be covered by a tile. Two tilings are different if and only if there are two 4-directionally adjacent cells on the board such that exactly one of the tilings has both squares occupied by a tile.)

 

Idea: DP

dp[i][0]: ways to cover i cols, both rows of i-th col are covered
dp[i][1]:  ways to cover i cols, only top row of i-th col is covered
dp[i][2]:  ways to cover i cols, only bottom row of i-th col is covered

Solution 1: DP

Time complexity: O(N)

Space complexity: O(N)

C++

C++ V2

Solution 2: DP

Another way to think about this problem

define: dp[i] ways to completely covert the i*2 board.

C++