LeetCode-in-Java

1354. Construct Target Array With Multiple Sums

Hard

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

public class Solution {
    public boolean isPossible(int[] target) {
        long sum = target[0];
        int maxIndex = 0;
        for (int i = 1; i < target.length; i++) {
            sum += target[i];
            if (target[i] > target[maxIndex]) {
                maxIndex = i;
            }
        }
        long remainingSum = sum - target[maxIndex];
        if (target[maxIndex] == 1 || remainingSum == 1) {
            return true;
        }
        if (remainingSum >= target[maxIndex]
                || remainingSum == 0
                || target[maxIndex] % remainingSum == 0) {
            return false;
        }
        target[maxIndex] %= (int) remainingSum;
        return isPossible(target);
    }
}

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