Question 1299 on leetcode

Date : August 30, 2024
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Question 1299 on LeetCode: A Deep Dive into the Solution

LeetCode is a fantastic platform for enhancing your coding skills through a variety of programming challenges. One such problem is Question 1299, which tasks you with replacing every element in an array with the greatest element among the elements to its right. The last element should be replaced with -1 since there are no elements to its right. In this blog post, we’ll explore a solution to this problem, analyze its efficiency, and discuss potential improvements.

Problem Statement

Given an array of integers, your goal is to create a new array where each element at index i is replaced by the greatest element to the right of i in the original array. If there are no elements to the right, replace it with -1.

Example

For example, given the input array:

arr = [17, 18, 5, 4, 6, 1]

The output array should be:

[18, 6, 6, 6, 1, -1]

The Solution

Here’s the solution I came up with for this problem:

javascript /**

@param {number[]} arr
@return {number[]} */ var replaceElements = function(arr) { let max; let array = new Array(); for(let i = arr.length - 1; i >= 0; i–) { if(i == arr.length - 1) { array[i] = -1; max = arr[i]; } else { array[i] = max; if(arr[i] > max) { max = arr[i]; } } } return array; };

Explanation of the Code

Initialization: We declare a variable max to keep track of the maximum value we have encountered so far while iterating through the array from the end. We also create an empty array array to store the results.
Iterating Backwards: We loop through the array in reverse order, starting from the last index. If we’re at the last element, we set array[i] to -1 since there are no elements to the right. We then assign that last element to max.
Updating the Result Array: For each other element, we simply assign the current max to array[i]. If the current element arr[i] is greater than max, we update max to this new value.
Return the Result: Finally, we return the newly constructed array.

Time and Space Complexity

Time Complexity: The algorithm runs in O(n) time, where n is the number of elements in the input array. This is efficient as we only make a single pass through the array.
Space Complexity: The space complexity is O(n) as well, due to the additional array used to store results. If we want to optimize further, we could modify the input array in place.

Suggested Improvements

While the solution works correctly, here are a few improvements we could consider:

In-place Modification: Instead of creating a new array, we could modify the input array directly. This would reduce the space complexity to O(1) (excluding the input and output arrays).
Use of Descriptive Variable Names: While max is understandable, renaming it to something like currentMax could clarify its purpose.
Edge Case Handling: We should ensure to handle edge cases, such as when the input array is empty. Currently, the function will return an empty array, which is acceptable, but it’s always good practice to document these scenarios in your function.
Commenting: Adding more comments throughout the code can help others (and your future self) understand the logic at a glance.

Conclusion

The solution provided for LeetCode Question 1299 effectively solves the problem with a time-efficient approach. By iterating through the array backwards and keeping track of the maximum element, we can construct the desired output efficiently. While the basic solution is solid, there are opportunities for optimization and improvement in terms of code clarity and space efficiency.

I would love to hear your thoughts on this solution! Do you have any alternative approaches? Feel free to share in the comments below. Happy coding!