1354. Construct Target Array With Multiple Sums (H)

Construct Target Array With Multiple Sums (H)

题目

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

---

题意

给定一个全为1的数组，每次可以进行如下操作：计算数组之和为x；任选数组中一个元素将它替换为x。问能否经过若干次操作后，使得初始全为1的数组变为目标数组。

思路

倒推，数组中最大的数一定是通过上述流程得到的。每次选择最大的数top，将其减去剩余数之和remain得到top执行一次流程前的值，重复操作直到所有数被还原为1。注意注意特殊情况的处理。

---

代码实现

Java

class Solution {
    public boolean isPossible(int[] target) {
        if (target.length == 1 && target[0] != 1) return false;
            
        long sum = 0;			// 和可能越界
        Queue<Integer> pq = new PriorityQueue<>(Comparator.reverseOrder());

        for (int num : target) {
            sum += num;
            pq.offer(num);
        }

        while (pq.peek() > 1) {
            int top = pq.poll();
            sum -= top;
            if (sum == 1) return true;			// 如果剩余之和为1，那么一定能将top还原为1，如：[4,1]
            if (top <= sum || top % sum == 0) return false;		// 特殊情况：[2,2]
            top %= sum;			// 还原top
            sum += top;
            pq.offer((int) top);
        }

        return true;
    }
}