# Posts tagged as “array”

Given an array of integers A sorted in non-decreasing order, return an array of the squares of each number, also in sorted non-decreasing order.

Example 1:

Input: [-4,-1,0,3,10]
Output: [0,1,9,16,100]


Example 2:

Input: [-7,-3,2,3,11]
Output: [4,9,9,49,121]


Note:

1. 1 <= A.length <= 10000
2. -10000 <= A[i] <= 10000
3. A is sorted in non-decreasing order.

## Solution: Two pointers + Merge two sorted arrays

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

## c++

Given an array A of positive lengths, return the largest perimeter of a triangle with non-zero area, formed from 3 of these lengths.

If it is impossible to form any triangle of non-zero area, return 0.

Example 1:

Input: [2,1,2]
Output: 5


Example 2:

Input: [1,2,1]
Output: 0


Example 3:

Input: [3,2,3,4]
Output: 10


Example 4:

Input: [3,6,2,3]
Output: 8


Note:

1. 3 <= A.length <= 10000
2. 1 <= A[i] <= 10^6

## Solution: Greedy

Answer must be 3 consecutive numbers in the sorted array
if A[i] >= A[i+1] + A[i+2], then A[i] >= A[i+j] + A[i+k], 1 < j < k
if A[i] < A[i+1] + A[i+2], then A[i] + A[i+1] + A[i+2] is the answer

# Problem

Given an array A of integers, return true if and only if it is a valid mountain array.

Recall that A is a mountain array if and only if:

• A.length >= 3
• There exists some i with 0 < i < A.length - 1 such that:
• A < A < ... A[i-1] < A[i]
• A[i] > A[i+1] > ... > A[B.length - 1]

Example 1:

Input: [2,1]
Output: false


Example 2:

Input: [3,5,5]
Output: false


Example 3:

Input: [0,3,2,1]
Output: true

Note:

1. 0 <= A.length <= 10000
2. 0 <= A[i] <= 10000

# Solution

Use has_up and has_down to track whether we have A[i] > A[i – 1] and A[i] < A[i – 1] receptively.

return false if any of the following happened:

1. size(A) < 3
2. has_down happened before has_up
3. not has_down or not has_up
4. A[i – 1] < A[i] after has_down
5. A[i – 1] > A[i] before has_up

Time complexity: O(n)

Space complexity: O(n)

# Problem

Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively. It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}

# Solution:

Time complexity: O(n)

Space complexity: O(1)

## C++

An array is monotonic if it is either monotone increasing or monotone decreasing.

An array A is monotone increasing if for all i <= jA[i] <= A[j].  An array A is monotone decreasing if for all i <= jA[i] >= A[j].

Return true if and only if the given array A is monotonic.

# Solution:

## Python

Mission News Theme by Compete Themes.