LeetCode #1354 — HARD

Construct Target Array With Multiple Sums

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[9,3,5]

Example 2

[1,1,1,2]

Example 3

[8,5]

Core Insight

What unlocks the optimal approach

Given that the sum is strictly increasing, the largest element in the target must be formed in the last step by adding the total sum in the previous step. Thus, we can simulate the process in a reversed way.
Subtract the largest with the rest of the array, and put the new element into the array. Repeat until all elements become one

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1354: Construct Target Array With Multiple Sums
class Solution {
    public boolean isPossible(int[] target) {
        PriorityQueue<Long> pq = new PriorityQueue<>(Collections.reverseOrder());
        long s = 0;
        for (int x : target) {
            s += x;
            pq.offer((long) x);
        }
        while (pq.peek() > 1) {
            long mx = pq.poll();
            long t = s - mx;
            if (t == 0 || mx - t < 1) {
                return false;
            }
            long x = mx % t;
            if (x == 0) {
                x = t;
            }
            pq.offer(x);
            s = s - mx + x;
        }
        return true;
    }
}

// Accepted solution for LeetCode #1354: Construct Target Array With Multiple Sums
func isPossible(target []int) bool {
	pq := &hp{target}
	s := 0
	for _, x := range target {
		s += x
	}
	heap.Init(pq)
	for target[0] > 1 {
		mx := target[0]
		t := s - mx
		if t < 1 || mx-t < 1 {
			return false
		}
		x := mx % t
		if x == 0 {
			x = t
		}
		target[0] = x
		heap.Fix(pq, 0)
		s = s - mx + x
	}
	return true
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
func (hp) Pop() (_ any)         { return }
func (hp) Push(any)             {}

# Accepted solution for LeetCode #1354: Construct Target Array With Multiple Sums
class Solution:
    def isPossible(self, target: List[int]) -> bool:
        s = sum(target)
        pq = [-x for x in target]
        heapify(pq)
        while -pq[0] > 1:
            mx = -heappop(pq)
            t = s - mx
            if t == 0 or mx - t < 1:
                return False
            x = (mx % t) or t
            heappush(pq, -x)
            s = s - mx + x
        return True

// Accepted solution for LeetCode #1354: Construct Target Array With Multiple Sums
use std::collections::BinaryHeap;

impl Solution {
    pub fn is_possible(target: Vec<i32>) -> bool {
        let mut pq = BinaryHeap::from(target.clone());
        let mut s: i64 = target.iter().map(|&x| x as i64).sum();

        while let Some(&mx) = pq.peek() {
            if mx == 1 {
                break;
            }
            let mx = pq.pop().unwrap() as i64;
            let t = s - mx;
            if t < 1 || mx - t < 1 {
                return false;
            }
            let x = if mx % t == 0 { t } else { mx % t };
            pq.push(x as i32);
            s = s - mx + x;
        }
        true
    }
}

// Accepted solution for LeetCode #1354: Construct Target Array With Multiple Sums
function isPossible(target: number[]): boolean {
    const pq = new MaxPriorityQueue<number>();
    let s = 0;
    for (const x of target) {
        s += x;
        pq.enqueue(x);
    }
    while (pq.front() > 1) {
        const mx = pq.dequeue();
        const t = s - mx;
        if (t < 1 || mx - t < 1) {
            return false;
        }
        const x = mx % t || t;
        pq.enqueue(x);
        s = s - mx + x;
    }
    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.