# Solving Leetcode Interviews in Seconds with AI: Largest Element in an Array after Merge Operations


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2789" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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 integer 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  

	# Explanation
	*   **Greedy Approach:** Iterate through the array from left to right. If `nums[i] <= nums[i+1]`, merge them by adding `nums[i]` to `nums[i+1]` and removing `nums[i]`. This is greedy because we want to accumulate the largest possible value towards the end.
*   **In-place Modification:** Modify the input array in place to avoid extra space. Since we remove elements during the process, we maintain an index representing the valid portion of the array.
*   **Finding the Maximum:** After performing all possible merge operations, the largest element in the resulting array is returned.

*   **Runtime Complexity:** O(n), **Storage Complexity:** O(1)

	
	# Code
	```python
	def solve():
    n = int(input())
    nums = list(map(int, input().split()))

    i = 0
    while i < len(nums) - 1:
        if nums[i] <= nums[i + 1]:
            nums[i + 1] += nums[i]
            nums.pop(i)
        else:
            i += 1

    return max(nums)

def maximum_final_value(nums):
    """
    Finds the maximum possible value in the final array after applying the given operations.

    Args:
        nums: A list of positive integers.

    Returns:
        The maximum possible value in the final array.
    """
    i = 0
    while i < len(nums) - 1:
        if nums[i] <= nums[i + 1]:
            nums[i + 1] += nums[i]
            nums.pop(i)
            i = 0 # Reset index to 0 to ensure all possible merges are done. Optimizes for cases where removing element i unblocks potential merges earlier in the array
        else:
            i += 1

    return max(nums)
	```