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Posts published in “Divide and conquer”

花花酱 LeetCode 148. Sort List

Problem

Sort a linked list in O(n log n) time using constant space complexity.

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

Solution: Merge Sort

Top-down (recursion)

Time complexity: O(nlogn)

Space complexity: O(logn)

C++

Java

Python3

 

bottom up

Time complexity: O(nlogn)

Space complexity: O(1)

花花酱 LeetCode 775. Global and Local Inversions

Problem

We have some permutation A of [0, 1, ..., N - 1], where N is the length of A.

The number of (global) inversions is the number of i < j with 0 <= i < j < N and A[i] > A[j].

The number of local inversions is the number of i with 0 <= i < N and A[i] > A[i+1].

Return true if and only if the number of global inversions is equal to the number of local inversions.

Example 1:

Input: A = [1,0,2]
Output: true
Explanation: There is 1 global inversion, and 1 local inversion.

Example 2:

Input: A = [1,2,0]
Output: false
Explanation: There are 2 global inversions, and 1 local inversion.

Note:

  • A will be a permutation of [0, 1, ..., A.length - 1].
  • A will have length in range [1, 5000].
  • The time limit for this problem has been reduced.

Solution1: Brute Force (TLE)

Time complexity: O(n^2)

Space complexity: O(1)

C++

 

Solution2: MergeSort

Time complexity: O(nlogn)

Space complexity: O(n)

C#

 

Solution3: Input Property

Input is a permutation of [0, 1, …, N – 1]

Time Complexity: O(n)

Space Complexity: O(1)

 

 

花花酱 LeetCode 169. Majority Element

题目大意:给你一个数组,其中一个数出现超过n/2次,问你出现次数最多的那个数是什么?

Problem:

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

Ideas:


Solution 1:

Hash table O(n) / O(n)

 

Solution 2:

BST O(nlogk) / O(n)

 

Solution 3:

Randomization O(n) / O(1)

 

Solution 4:

Bit voting O(n) / O(1)

 

Solution 5:

Moore Voting O(n) / O(1)

 

Solution 6:

Full sorting O(nlogn) / O(1)

 

Solution 7:

Partial sorting O(n) / O(1)

 

Solution 8:

Divide and conquer O(nlogn) / O(logn)

Divide and conquer O(nlogn) / O(logn)

 

花花酱 LeetCode 719. Find K-th Smallest Pair Distance

题目大意:给你一个数组,返回所有数对中,绝对值差第k小的值。

Problem:

Given an integer array, return the k-th smallest distance among all the pairs. The distance of a pair (A, B) is defined as the absolute difference between A and B.

Example 1:

Note:

  1. 2 <= len(nums) <= 10000.
  2. 0 <= nums[i] < 1000000.
  3. 1 <= k <= len(nums) * (len(nums) - 1) / 2.

Idea

Bucket sort

Binary search / dp

Solution

C++ / binary search

 

C++ / bucket sort w/ vector O(n^2)

Related Problems:

花花酱 LeetCode 154. Find Minimum in Rotated Sorted Array II

Problem:

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

The array may contain duplicates.

Idea: 

Divide and conquer

Time complexity:

Average: O(logn)

Worst: O(n)

Solution:

 

Related Problems: