## 442. Find All Duplicates in an Array

Description:

https://leetcode.com/problems/find-all-duplicates-in-an-array/description/

Code:

Time & Space:
O(n) & O(1)

## 448. Find All Numbers Disappeared in an Array

Description:

https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/description/

Code:

Code for fastest algorithm:

Time & Space:
O(n) & O(1)

## 495. Teemo Attacking

Description:

https://leetcode.com/problems/teemo-attacking/description/

Code:

Time & Space:
O(n) & O(1)

## 532. K-diff Pairs in an Array

Description:

https://leetcode.com/problems/k-diff-pairs-in-an-array/description

Code:

Time & Space:
O(nlog(n)) & O(n)

## 485. Max Consecutive Ones

Description:

https://leetcode.com/problems/max-consecutive-ones/description/

Code:

Time & Space:
O(n) & O(1)

## 628. Maximum Product of Three Numbers

Description:

https://leetcode.com/problems/maximum-product-of-three-numbers/description/

Code:

Time & Space:

O(n) & O(1)

Description:

Code:

Time & Space:
Record: O(1)
Sort: O(nlogn)
Calculate idle: O(1)

Space: O(1)

## 611. Valid Triangle Number

Description:

https://leetcode.com/problems/valid-triangle-number/description/

Algorithm1:

Use for loop for three times.

It works but slow (beats 23.60%)

Code1:

Timing & Space:

O(n^3) & O(1)

Algorithm2:

Code2:

Timing & Space:

Time:

Worst: O(n^2)

Best: O(1)

Space: O(1)

## 565. Array Nesting

Description:

https://leetcode.com/problems/array-nesting/description/

Algorithm:

Actually we are separating the array into several groups by determining S[k]. The numbers inside of S[k] will form a cycle.

For example:

Input: A = [5,4,0,3,1,6,2]

Output: 4

Explanation:

A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.

One of the longest S[K]:

S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}

S[5] = {A[5], A[6], A[2], A[0]} = {6, 2, 0, 5}

S[6] = {A[6], A[2], A[0], A[5]} = {2, 0, 5, 6}

S[2] = {A[2], A[0], A[5], A[6]} = {0, 5, 6, 2}

S[1] = {A[1], A[4]} = {4,1}

S[4] = {A[4], A[1]} = {1,4}

S[3] = {A[3]} = {3}

So just mark the used number in the finding process. If meets used number -> stop.

Code:

Timing & Space:

fastest

O(n) & O(1)