# Posts tagged as “array”

You are given an integer array nums. The unique elements of an array are the elements that appear exactly once in the array.

Return the sum of all the unique elements of nums.

Example 1:

Input: nums = [1,2,3,2]
Output: 4
Explanation: The unique elements are [1,3], and the sum is 4.


Example 2:

Input: nums = [1,1,1,1,1]
Output: 0
Explanation: There are no unique elements, and the sum is 0.


Example 3:

Input: nums = [1,2,3,4,5]
Output: 15
Explanation: The unique elements are [1,2,3,4,5], and the sum is 15.


Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100

## Solution: Hashtable

Time complexity: O(n)
Space complexity: O(100)

## C++

There is an integer array nums that consists of n unique elements, but you have forgotten it. However, you do remember every pair of adjacent elements in nums.

You are given a 2D integer array adjacentPairs of size n - 1 where each adjacentPairs[i] = [ui, vi] indicates that the elements ui and vi are adjacent in nums.

It is guaranteed that every adjacent pair of elements nums[i] and nums[i+1] will exist in adjacentPairs, either as [nums[i], nums[i+1]] or [nums[i+1], nums[i]]. The pairs can appear in any order.

Return the original array nums. If there are multiple solutions, return any of them.

Example 1:

Input: adjacentPairs = [[2,1],[3,4],[3,2]]
Output: [1,2,3,4]
Explanation: This array has all its adjacent pairs in adjacentPairs.
Notice that adjacentPairs[i] may not be in left-to-right order.


Example 2:

Input: adjacentPairs = [[4,-2],[1,4],[-3,1]]
Output: [-2,4,1,-3]
Explanation: There can be negative numbers.
Another solution is [-3,1,4,-2], which would also be accepted.


Example 3:

Input: adjacentPairs = [[100000,-100000]]
Output: [100000,-100000]


Constraints:

• nums.length == n
• adjacentPairs.length == n - 1
• adjacentPairs[i].length == 2
• 2 <= n <= 105
• -105 <= nums[i], ui, vi <= 105
• There exists some nums that has adjacentPairs as its pairs.

## Solution: Hashtable

Reverse thinking! For a given input array, e.g.
[1, 2, 3, 4, 5]
it’s adjacent pairs are [1,2] , [2,3], [3,4], [4,5]
all numbers appeared exactly twice except 1 and 5, since they are on the boundary.
We just need to find the head or tail of the input array, and construct the rest of the array in order.

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

## C++

There is a biker going on a road trip. The road trip consists of n + 1 points at different altitudes. The biker starts his trip on point 0 with altitude equal 0.

You are given an integer array gain of length n where gain[i] is the net gain in altitude between points i​​​​​​ and i + 1 for all (0 <= i < n). Return the highest altitude of a point.

Example 1:

Input: gain = [-5,1,5,0,-7]
Output: 1
Explanation: The altitudes are [0,-5,-4,1,1,-6]. The highest is 1.


Example 2:

Input: gain = [-4,-3,-2,-1,4,3,2]
Output: 0
Explanation: The altitudes are [0,-4,-7,-9,-10,-6,-3,-1]. The highest is 0.


Constraints:

• n == gain.length
• 1 <= n <= 100
• -100 <= gain[i] <= 100

## Solution: Running Max

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

## C++

You are given an array nums of n positive integers.

You can perform two types of operations on any element of the array any number of times:

• If the element is evendivide it by 2.
• For example, if the array is [1,2,3,4], then you can do this operation on the last element, and the array will be [1,2,3,2].
• If the element is oddmultiply it by 2.
• For example, if the array is [1,2,3,4], then you can do this operation on the first element, and the array will be [2,2,3,4].

The deviation of the array is the maximum difference between any two elements in the array.

Return the minimum deviation the array can have after performing some number of operations.

Example 1:

Input: nums = [1,2,3,4]
Output: 1
Explanation: You can transform the array to [1,2,3,2], then to [2,2,3,2], then the deviation will be 3 - 2 = 1.


Example 2:

Input: nums = [4,1,5,20,3]
Output: 3
Explanation: You can transform the array after two operations to [4,2,5,5,3], then the deviation will be 5 - 2 = 3.


Example 3:

Input: nums = [2,10,8]
Output: 3


Constraints:

• n == nums.length
• 2 <= n <= 105
• 1 <= nums[i] <= 109

## Solution: Priority Queue

If we double an odd number it becomes an even number, then we can only divide it by two which gives us back the original number. So we can pre-double all the odd numbers and only do division in the following process.

We push all numbers including pre-doubled odd ones onto a priority queue, and track the difference between the largest and smallest number.

Each time, we pop the largest number out and divide it by two then put it back to the priority queue, until the largest number becomes odd. We can not discard it and divide any other smaller numbers by two will only increase the max difference, so we can stop here.

ex1: [3, 5, 8] => [6, 8, 10] (pre-double) => [5, 6, 8] => [4, 5, 6] => [3, 4, 5] max diff is 5 – 3 = 2
ex2: [4,1,5,20,3] => [2, 4, 6, 10, 20] (pre-double) => [2, 4, 6, 10] => [2, 4, 5, 6] => [2,3,4,5] max diff = 5-2 = 3

Time complexity: O(n*logm*logn)
Space complexity: O(n)

## C++/PQ

You are given an m x n integer grid accounts where accounts[i][j] is the amount of money the i​​​​​​​​​​​th​​​​ customer has in the j​​​​​​​​​​​th​​​​ bank. Return the wealth that the richest customer has.

A customer’s wealth is the amount of money they have in all their bank accounts. The richest customer is the customer that has the maximum wealth.

Example 1:

Input: accounts = [[1,2,3],[3,2,1]]
Output: 6
Explanation:
1st customer has wealth = 1 + 2 + 3 = 6
2nd customer has wealth = 3 + 2 + 1 = 6
Both customers are considered the richest with a wealth of 6 each, so return 6.


Example 2:

Input: accounts = [[1,5],[7,3],[3,5]]
Output: 10
Explanation:
1st customer has wealth = 6
2nd customer has wealth = 10
3rd customer has wealth = 8
The 2nd customer is the richest with a wealth of 10.

Example 3:

Input: accounts = [[2,8,7],[7,1,3],[1,9,5]]
Output: 17


Constraints:

• m == accounts.length
• n == accounts[i].length
• 1 <= m, n <= 50
• 1 <= accounts[i][j] <= 100

## Solution: Sum each row up

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