Construct Target Array With Multiple Sums

Problem Statement

In this task, you are provided with an array referred to as target which contains n integers. To address the challenge, you start with another array called arr initialized with n elements, all of which are 1s. Through a sequence of operations, you aim to transform arr into exactly matching the target array. Each allowed operation involves calculating x, the sum of all current elements in arr, and then replacing the content of arr at a particular index i (where 0 <= i < n) with this sum x. The process can be iterated as many times as necessary. You need to determine if it is possible to use these operations to turn arr into target and return true if you can, or false if you cannot.

Examples

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

Approach and Intuition

The operations on the arr can be visualized as iteratively building the target by accumulating the current sum of arr into its individual elements until arr equals to target. Let's go step by step to understand and solve this:

1. Visual Verification Through Backward Construction:

   • Instead of trying to transform arr into target directly, the intuition is to think of how target could have been formed in reverse. This involves working backwards to see if a valid x can be identified for each position in target.

2. Simulate the Operations:

   • From the examples given, clearly, the operations accumulate values in the arr. Start from target and repeatedly work backwards. Identify prev (the previous state of arr), which would be derived from subtracting from the largest item in current arr and setting it into the next biggest index until the initial state [1, 1, ..., 1] is reached or consistency falls apart.

3. Validating the Process and Integer Constraints:

   • With each iteration, validate if the transformation (or backward step) adheres to the constraints and actually edges closer to the initial state without breaking any rule (like turning any number negative or zero, which isn't feasible operationally according to problem constraints).

4. Optimal Checks:

   • The possibilities can be quickly pruned or decided by comparing the sums of arr and target. If at any step the sum of values in the arr derived via backward reconstruction doesn’t match the summed integers series leading to target, it’s an indication that forming target is infeasible under the given operations.

5. Continue Until Completion

   • Continue this simulated process with control mechanisms until a conclusive decision point (true or false) is reached based on feasible backtracking or resolvable series matching.

The intuition here relies significantly on these transformations and validation checkpoints, making sure the starting arr can truly evolve into the target with the given procedures.

Solutions

class Solution {
    public boolean canTransform(int[] target) {
        
        // Early return for single element arrays
        if (target.length == 1) {
            return target[0] == 1;
        }
        
        int sumElements = Arrays.stream(target).sum();
        
        PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(Collections.reverseOrder());
        for (int element : target) {
            maxHeap.add(element);
        }
        
        while (maxHeap.peek() > 1) {
            int maxElement = maxHeap.poll();
            int remainingSum = sumElements - maxElement;
            
            // Directly return for two elements scenario
            if (remainingSum == 1) {
                return true;
            }
            int remainder = maxElement % remainingSum;
            
            // Check for invalid or non-changing cases
            if (remainder == 0 || remainder == maxElement) {
                return false;
            }
            maxHeap.add(remainder);
            sumElements = sumElements - maxElement + remainder;
        }
        
        return true; 
    }
}