LeetCode #2789 — MEDIUM

Largest Element in an Array after Merge Operations

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed array nums consisting of positive integers.

You can do the following operation on the array any number of times:

Choose an index i such that 0 <= i < nums.length - 1 and nums[i] <= nums[i + 1]. Replace the element nums[i + 1] with nums[i] + nums[i + 1] and delete the element nums[i] from the array.

Return the value of the largest element that you can possibly obtain in the final array.

Example 1:

Input: nums = [2,3,7,9,3]
Output: 21
Explanation: We can apply the following operations on the array:
- Choose i = 0. The resulting array will be nums = [5,7,9,3].
- Choose i = 1. The resulting array will be nums = [5,16,3].
- Choose i = 0. The resulting array will be nums = [21,3].
The largest element in the final array is 21. It can be shown that we cannot obtain a larger element.

Example 2:

Input: nums = [5,3,3]
Output: 11
Explanation: We can do the following operations on the array:
- Choose i = 1. The resulting array will be nums = [5,6].
- Choose i = 0. The resulting array will be nums = [11].
There is only one element in the final array, which is 11.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array nums consisting of positive integers. You can do the following operation on the array any number of times: Choose an index i such that 0 <= i < nums.length - 1 and nums[i] <= nums[i + 1]. Replace the element nums[i + 1] with nums[i] + nums[i + 1] and delete the element nums[i] from the array. Return the value of the largest element that you can possibly obtain in the final array.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[2,3,7,9,3]

Example 2

[5,3,3]

Core Insight

What unlocks the optimal approach

Start from the end of the array and keep merging elements together until it is no longer possible.
The answer will be the resulting element from the last merge operation.

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

Upper-end input sizes

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 #2789: Largest Element in an Array after Merge Operations
class Solution {
    public long maxArrayValue(int[] nums) {
        int n = nums.length;
        long ans = nums[n - 1], t = nums[n - 1];
        for (int i = n - 2; i >= 0; --i) {
            if (nums[i] <= t) {
                t += nums[i];
            } else {
                t = nums[i];
            }
            ans = Math.max(ans, t);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2789: Largest Element in an Array after Merge Operations
func maxArrayValue(nums []int) int64 {
	n := len(nums)
	ans, t := nums[n-1], nums[n-1]
	for i := n - 2; i >= 0; i-- {
		if nums[i] <= t {
			t += nums[i]
		} else {
			t = nums[i]
		}
		ans = max(ans, t)
	}
	return int64(ans)
}

# Accepted solution for LeetCode #2789: Largest Element in an Array after Merge Operations
class Solution:
    def maxArrayValue(self, nums: List[int]) -> int:
        for i in range(len(nums) - 2, -1, -1):
            if nums[i] <= nums[i + 1]:
                nums[i] += nums[i + 1]
        return max(nums)

// Accepted solution for LeetCode #2789: Largest Element in an Array after Merge Operations
/**
 * [2789] Largest Element in an Array after Merge Operations
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn max_array_value(nums: Vec<i32>) -> i64 {
        let mut nums: Vec<i64> = nums.iter().map(|x| *x as i64).collect();

        let mut result = nums[0];
        for i in (0..(nums.len() - 1)).rev() {
            if nums[i] <= nums[i + 1] {
                nums[i] = nums[i] + nums[i + 1];

                result = result.max(nums[i]);
            }
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2789() {
        assert_eq!(21, Solution::max_array_value(vec![2, 3, 7, 9, 3]));
        assert_eq!(11, Solution::max_array_value(vec![5, 3, 3]));
    }
}

// Accepted solution for LeetCode #2789: Largest Element in an Array after Merge Operations
function maxArrayValue(nums: number[]): number {
    for (let i = nums.length - 2; i >= 0; --i) {
        if (nums[i] <= nums[i + 1]) {
            nums[i] += nums[i + 1];
        }
    }
    return Math.max(...nums);
}

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(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.