Construct Target Array With Multiple Sums

Similar Problems:

Given an array of integers target. From a starting array, A consisting of all 1’s, you may perform the following procedure :

- let x be the sum of all elements currently in your array.
- choose index i, such that 0 <= i < target.size and set the value of A at index i to x.
- You may repeat this procedure as many times as needed.

Return True if it is possible to construct the target array from A otherwise return False.

Example 1:

Input: target = [9,3,5] Output: true Explanation: Start with [1, 1, 1] [1, 1, 1], sum = 3 choose index 1 [1, 3, 1], sum = 5 choose index 2 [1, 3, 5], sum = 9 choose index 0 [9, 3, 5] Done

Example 2:

Input: target = [1,1,1,2] Output: false Explanation: Impossible to create target array from [1,1,1,1].

Example 3:

Input: target = [8,5] Output: true

Constraints:

- N == target.length
- 1 <= target.length <= 5 * 10^4
- 1 <= target[i] <= 10^9

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

- Solution:

## https://code.dennyzhang.com/construct-target-array-with-multiple-sums ## Basic Ideas: heap ## ## Order doesn't matter ## Backtrack from the biggest value ## Sum become: sum-biggest ## The biggest value become: (2*biggest-sum) ## ## Complexity: Time ?, Space ? class Solution: def isPossible(self, target: List[int]) -> bool: s = sum(target) heapq._heapify_max(target) while len(target)>0 and target[0] != 1: b = heapq.heappop(target) v = 2*b-s s = s-b+v if v<=0: return False heapq.heappush(target, v) heapq._heapify_max(target) return target[0] == 1