# Posts tagged as “array”

Given an array of numbers arr. A sequence of numbers is called an arithmetic progression if the difference between any two consecutive elements is the same.

Return true if the array can be rearranged to form an arithmetic progression, otherwise, return false.

Example 1:

Input: arr = [3,5,1]
Output: true
Explanation: We can reorder the elements as [1,3,5] or [5,3,1] with differences 2 and -2 respectively, between each consecutive elements.


Example 2:

Input: arr = [1,2,4]
Output: false
Explanation: There is no way to reorder the elements to obtain an arithmetic progression.


Constraints:

• 2 <= arr.length <= 1000
• -10^6 <= arr[i] <= 10^6

## Solution 1: Sort and check.

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

## Solution 2: Rearrange the array

Find min / max / diff and put each element into its correct position by swapping elements in place.

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

## C++

Given a binary array nums, you should delete one element from it.

Return the size of the longest non-empty subarray containing only 1’s in the resulting array.

Return 0 if there is no such subarray.

Example 1:

Input: nums = [1,1,0,1]
Output: 3
Explanation: After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1's.

Example 2:

Input: nums = [0,1,1,1,0,1,1,0,1]
Output: 5
Explanation: After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with value of 1's is [1,1,1,1,1].

Example 3:

Input: nums = [1,1,1]
Output: 2
Explanation: You must delete one element.

Example 4:

Input: nums = [1,1,0,0,1,1,1,0,1]
Output: 4


Example 5:

Input: nums = [0,0,0]
Output: 0


Constraints:

• 1 <= nums.length <= 10^5
• nums[i] is either 0 or 1.

## Solution 1: DP

Preprocess:
l[i] := longest 1s from left side ends with nums[i], l[i] = nums[i] + nums[i] * l[i – 1]
r[i] := longest 1s from right side ends with nums[i], r[i] = nums[i] + nums[i] * r[i + 1]

Use each node as a bridge (ignored), the total number of consecutive 1s = l[i – 1] + r[i + 1].

ans = max{l[i-1] + r[i +1]}
Time complexity: O(n)
Space complexity: O(n)

## Solution 2: DP

dp[i][0] := longest subarray ends with nums[i] has no ones.
dp[i][0] := longest subarray ends with nums[i] has 1 one.
if nums[i] == 1:
dp[i][0] = dp[i – 1][0] + 1
dp[i][1] = dp[i – 1][1] + 1
if nums[i] == 0:
dp[i][0] = 0
dp[i][1] = dp[i – 1][0] + 1
Time complexity: O(n)
Space complexity: O(n) -> O(1)

## Solution 3: Sliding Window

Maintain a sliding window l ~ r s.t sum(num[l~r]) >= r – l. There can be at most one 0 in the window.
ans = max{r – l} for all valid windows.

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

## C++

Given an array of unique integers salary where salary[i] is the salary of the employee i.

Return the average salary of employees excluding the minimum and maximum salary.

Example 1:

Input: salary = [4000,3000,1000,2000]
Output: 2500.00000
Explanation: Minimum salary and maximum salary are 1000 and 4000 respectively.
Average salary excluding minimum and maximum salary is (2000+3000)/2= 2500


Example 2:

Input: salary = [1000,2000,3000]
Output: 2000.00000
Explanation: Minimum salary and maximum salary are 1000 and 3000 respectively.
Average salary excluding minimum and maximum salary is (2000)/1= 2000


Example 3:

Input: salary = [6000,5000,4000,3000,2000,1000]
Output: 3500.00000


Example 4:

Input: salary = [8000,9000,2000,3000,6000,1000]
Output: 4750.00000


Constraints:

• 3 <= salary.length <= 100
• 10^3 <= salary[i] <= 10^6
• salary[i] is unique.
• Answers within 10^-5 of the actual value will be accepted as correct.

## Solution: Brute Force

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

## C++

Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).

Return the running sum of nums.

Example 1:

Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].

Example 2:

Input: nums = [1,1,1,1,1]
Output: [1,2,3,4,5]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].

Example 3:

Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]


Constraints:

• 1 <= nums.length <= 1000
• -10^6 <= nums[i] <= 10^6

## Solution

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

## C++

Given the array prices where prices[i] is the price of the ith item in a shop. There is a special discount for items in the shop, if you buy the ith item, then you will receive a discount equivalent to prices[j] where j is the minimum index such that j > i and prices[j] <= prices[i], otherwise, you will not receive any discount at all.

Return an array where the ith element is the final price you will pay for the ith item of the shop considering the special discount.

Example 1:

Input: prices = [8,4,6,2,3]
Output: [4,2,4,2,3]
Explanation:
For item 0 with price[0]=8 you will receive a discount equivalent to prices[1]=4, therefore, the final price you will pay is 8 - 4 = 4.
For item 1 with price[1]=4 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 4 - 2 = 2.
For item 2 with price[2]=6 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 6 - 2 = 4.
For items 3 and 4 you will not receive any discount at all.


Example 2:

Input: prices = [1,2,3,4,5]
Output: [1,2,3,4,5]
Explanation: In this case, for all items, you will not receive any discount at all.


Example 3:

Input: prices = [10,1,1,6]
Output: [9,0,1,6]


Constraints:

• 1 <= prices.length <= 500
• 1 <= prices[i] <= 10^3

## Solution 1: Simulation

Time complexity: O(n^2)
Space complexity: O(1)

## Solution 2: Monotonic Stack

Use a stack to store monotonically increasing items, when the current item is cheaper than the top of the stack, we get the discount and pop that item. Repeat until the current item is no longer cheaper or the stack becomes empty.

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

index version

## C++

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