1354-construct-target-array-with-multiple-sums

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Description

You are given an array target of n integers. From a starting array arr consisting of n 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 < n and set the value of arr 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 arr, otherwise, return false.

 

Example 1:

Input: target = [9,3,5]
Output: true
Explanation: Start with arr = [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 <= n <= 5 * 104
• 1 <= target[i] <= 109

Solution(Python)

class Solution:
    def isPossible(self, target: List[int]) -> bool:
        if len(target) == 1:
            return target[0] == 1

        total_sum = sum(target)
        # Use a min-heap by pushing negative values to simulate a max-heap
        max_heap = [-num for num in target]
        heapq.heapify(max_heap)

        while -max_heap[0] > 1:
            max_val = -heapq.heappop(max_heap)
            rest_sum = total_sum - max_val

            # Edge cases and impossibility checks
            if rest_sum == 1:
                return True
            if rest_sum == 0 or max_val <= rest_sum:
                return False

            # Calculate the previous value using modulo
            prev_val = max_val% rest_sum
            
            # If prev_val is 0, it indicates an impossible state
            if prev_val == 0:
                return False

            total_sum = rest_sum + prev_val
            heapq.heappush(max_heap, -prev_val)

        return True