1354. Construct Target Array With Multiple Sums

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

 

Solutions

Solution: Priority Queue

• Time complexity: O(nlogM*logn)
  • M is Max(target[i])
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} target
 * @return {boolean}
 */
const isPossible = function (target) {
  const queue = new MaxPriorityQueue();
  let sum = target.reduce((result, num) => result + num);

  for (const num of target) {
    queue.enqueue(num);
  }

  while (queue.front().element > 1) {
    const num = queue.dequeue().element;
    const restSum = sum - num;

    if (restSum === 0) return false;
    if (restSum === 1) return true;
    const nextNum = num % restSum;

    if (nextNum < 1 || nextNum === num) return false;
    queue.enqueue(nextNum);
    sum = sum - num + nextNum;
  }
  return true;
};