How do you solve Construct Target Array With Multiple Sums on LeetCode?

Construct Target Array With Multiple Sums (LeetCode #1354) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: The largest element must be formed last, as it is the sum of the rest.

What is the brute force solution for Construct Target Array With Multiple Sums?

The brute force approach for Construct Target Array With Multiple Sums is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try to construct the target array by repeatedly setting elements in a starting array of 1s to the sum of all elements. This approach checks every possible combination, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Construct Target Array With Multiple Sums?

Construct Target Array With Multiple Sums uses the following concepts: Array, Heap (Priority Queue). The recommended patterns to study are: Heap (Priority Queue), Array.

What are the interview tips for Construct Target Array With Multiple Sums?

Think about the problem in reverse — it can simplify your approach. Always check edge cases, such as when all elements are the same. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Construct Target Array With Multiple Sums?

Failing to realize that the largest element must be handled carefully. Not considering the case where the sum of the rest can never equal the largest element.

#1354

Construct Target Array With Multiple Sums

Hard

ArrayHeap (Priority Queue)Heap (Priority Queue)Array

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

Instead of building the target array from the starting array, we can reverse the process. By continuously subtracting the largest element from the sum of the rest, we can determine if we can reach the starting state of all 1s.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total sum of the target array.
2Step 2: While the largest element is greater than 1, subtract the largest element by the sum of the rest of the elements.
3Step 3: Update the largest element and repeat until all elements are 1 or the largest element is less than or equal to the sum of the rest.

solution.py14 lines

1# Full working Python code
2class Solution:
3    def isPossible(self, target):
4        total = sum(target)
5        while True:
6            max_val = max(target)
7            if max_val == 1:
8                return True
9            total -= max_val
10            if total <= 0 or total >= max_val:
11                return False
12            target[target.index(max_val)] = max_val % total
13            total += target[target.index(max_val)]
14

ℹ

Complexity note: The time complexity is O(n log n) because we need to find the maximum element in the array repeatedly, which takes O(n) time, and this is done in a loop until we reach the base case.

1The largest element must be formed last, as it is the sum of the rest.
2The process can be reversed to check if we can reach the starting state.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.