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Posts tagged as “subsequence”

花花酱 LeetCode 334. Increasing Triplet Subsequence

Problem

题目大意:判断一个数组中是否存在长度为3的单调递增子序列。

Formally the function should:

Return true if there exists i, j, k 
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.

Your algorithm should run in O(n) time complexity and O(1) space complexity.

Examples:
Given [1, 2, 3, 4, 5],
return true.

Given [5, 4, 3, 2, 1],
return false.

Credits:
Special thanks to @DjangoUnchained for adding this problem and creating all test cases.

Solution: Greedy

Time complexity: O(n)

Space complexity: O(1)

C++

 

花花酱 LeetCode 516. Longest Palindromic Subsequence

Problem

题目大意:找出最长回文子序列的长度。

https://leetcode.com/problems/longest-palindromic-subsequence/description/

Given a string s, find the longest palindromic subsequence’s length in s. You may assume that the maximum length of s is 1000.

Example 1:
Input:

"bbbab"

Output:

4

One possible longest palindromic subsequence is “bbbb”.

Example 2:
Input:

"cbbd"

Output:

2

One possible longest palindromic subsequence is “bb”.

Solution: DP

Time complexity: O(n^2)

Space complexity: O(n^2)

C++

Time complexity: O(n^2)

Space complexity: O(n)

C++

Python3

C#

 

花花酱 LeetCode 594. Longest Harmonious Subsequence

Problem

https://leetcode.com/problems/longest-harmonious-subsequence/description/

题目大意:找一个最长子序列,要求子序列中最大值和最小值的差是1。

We define a harmonious array is an array where the difference between its maximum value and its minimum value is exactly 1.

Now, given an integer array, you need to find the length of its longest harmonious subsequence among all its possible subsequences.

Example 1:

Input: [1,3,2,2,5,2,3,7]
Output: 5
Explanation: The longest harmonious subsequence is [3,2,2,2,3].

Note: The length of the input array will not exceed 20,000.

Solution1: HashTable

Time complexity: O(n)
Space complexity: O(n)

C++

 

花花酱 LeetCode 392. Is Subsequence

题目大意:问s是不是t的子序列。

Problem:

https://leetcode.com/problems/is-subsequence/description/

Given a string s and a string t, check if s is subsequence of t.

You may assume that there is only lower case English letters in both s and tt is potentially a very long (length ~= 500,000) string, and sis a short string (<=100).

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not).

Example 1:
s = "abc"t = "ahbgdc"

Return true.

Example 2:
s = "axc"t = "ahbgdc"

Return false.

Follow up:
If there are lots of incoming S, say S1, S2, … , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?

Solution: Brute force

Time Complexity: O(|s| + |t|)

Space Complexity: O(1)

C++

 

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