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

Solution: Backwards Simulation

Start with the largest number in the array, since it should be the sum of all previous numbers.

[9,3,5] => [9 – (3+5), 3, 5] => [1, 3, 5] => [1, 3, 5 – (1+3)] => [1, 3, 1] => [1, 3 – (1+1), 1] => [1,1,1] done

Time complexity: O(n + log(n*t)*logn)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  bool isPossible(vector<int>& target) {
    priority_queue<int> q(begin(target), end(target));
    long sum = accumulate(begin(target), end(target), 0LL);    
    while (true) {      
      int t = q.top(); q.pop();
      sum -= t;
      if (t == 1 || sum == 1) return true;      
      if (t < sum || sum == 0 || t % sum == 0) return false;
      t %= sum;
      sum += t;
      q.push(t);
    }
    return true;
  }
};

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