What is the time complexity of Replace Elements with Greatest Element on Right Side?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Replace Elements with Greatest Element on Right Side?

This problem uses Array. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Replace Elements with Greatest Element on Right Side?

**Right-to-Left Traversal is Crucial**: Processing the array from `len(arr) - 1` down to `0` allows us to compute the 'greatest element to the right' on the fly. Each element `arr[i]` can be replaced because `max_right` already holds the maximum of all elements `arr[j]` where `j > i`. **`max_right` Accumulates the Future Maximum**: The `max_right` variable acts as a running maximum of all elements *already seen* during the right-to-left traversal. When we are at `arr[i]`, `max_right` accurately represents the greatest element among `arr[i+1]` through `arr[len(arr)-1]`, directly fulfilling the problem's requirement for `arr[i]`'s replacement.

Which companies ask LeetCode 1299. Replace Elements with Greatest Element on Right Side in interviews?

This problem is frequently asked at Zoho.

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Replace Elements with Greatest Element on Right Side - LeetCode 1299 Solution

Replace Elements with Greatest Element on Right Side - Complete Solution Guide

Replace Elements with Greatest Element on Right Side is LeetCode problem 1299, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an array arr , replace every element in that array with the greatest element among the elements to its right, and replace the last element with -1 . After doing so, return the array. Example 1: Input: arr = [17,18,5,4,6,1] Output: [18,6,6,6,1,-1] Explanation: - index 0 --> the greatest element to the right of index 0 is index 1 (18). - index 1 --> the greatest element to the right of index 1 is index 4 (6). - index 2 --> the greatest element to the right of index 2 is index 4 (6). - index

Detailed Explanation

This problem asks us to transform an array by replacing each of its elements with the largest value found strictly among the elements to its right. It's crucial to understand that we're looking for a *localized* maximum for each position, not the overall maximum of the entire array. For instance, if you're at index `i`, you need to scan `arr[i+1]`, `arr[i+2]`, ..., up to `arr[len(arr)-1]` to find the greatest among them. The final element in the array presents a special case: since there are no elements to its right, its replacement is fixed to -1. Consider the example `arr = [17,18,5,4,6,1]`. For the element `17` at index 0, the greatest element to its right is `18` (at index 1). For `18` at index 1, the greatest to its right is `6` (at index 4). This pattern continues, leading to the output `[18,6,6,6,1,-1]`. The primary challenge is to perform this operation efficiently across the entire array without repeatedly scanning large sub-arrays, which could quickly lead to poor performance.

Solution Approach

The elegant solution provided tackles this problem by iterating through the array from right to left, rather than the intuitive left-to-right approach. The key insight is that when you're at `arr[i]` and moving leftwards, the 'greatest element to its right' has already been computed or considered. We maintain a single variable, `max_right`, which always stores the maximum value encountered so far from the rightmost end of the array up to the element just past our current position. Here's the step-by-step logic: We initialize `max_right` to -1. This serves two critical purposes: it correctly handles the replacement for the very last element (as its initial value when we process the last element), and it ensures that any positive number encountered will immediately become the new `max_right`. As we iterate from `len(arr) - 1` down to `0`: we first store `arr[i]` in a temporary variable. Then, we replace `arr[i]` with the *current* `max_right` value. This works because the current `max_right` correctly reflects the greatest element among `arr[i+1]` to `arr[len(arr)-1]`. Finally, we update `max_right` by comparing it with the `temp` (original `arr[i]`). This update ensures that for the *next* element we process (which is `arr[i-1]`), `max_right` will correctly include `arr[i]` in its consideration of 'elements to its right.' This single reverse pass, with its clever `max_right` accumulation, makes the solution highly efficient.

Step-by-Step Algorithm

Initialize `max_right` to -1. This handles the case where there are no elements to the right of the last element.
Iterate through the array from the second to last element (index n-2) down to the first element (index 0).
For each element at index `i`:
Store the current element's value in a temporary variable `temp`.
Replace the element at index `i` with the current value of `max_right`.
Update `max_right` to be the maximum of `max_right` and `temp`.

Key Insights

**Right-to-Left Traversal is Crucial**: Processing the array from `len(arr) - 1` down to `0` allows us to compute the 'greatest element to the right' on the fly. Each element `arr[i]` can be replaced because `max_right` already holds the maximum of all elements `arr[j]` where `j > i`.
**`max_right` Accumulates the Future Maximum**: The `max_right` variable acts as a running maximum of all elements *already seen* during the right-to-left traversal. When we are at `arr[i]`, `max_right` accurately represents the greatest element among `arr[i+1]` through `arr[len(arr)-1]`, directly fulfilling the problem's requirement for `arr[i]`'s replacement.
**Efficient In-Place Modification with Sentinel Value**: The solution modifies the array in-place, eliminating the need for extra memory. The initial `max_right = -1` serves as a sentinel, cleanly handling the last element's replacement and ensuring that any valid positive array element will correctly become the maximum as the traversal proceeds.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: Zoho.

Replace Elements with Greatest Element on Right Side | LeetCode 1299

Easy

Array

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int[] replaceElements(int[] arr) {
        int n = arr.length;
        if (n == 0) return arr;
        int maxRight = -1;
        for (int i = n - 1; i >= 0; i--) {
            int temp = arr[i];
            arr[i] = maxRight;
            maxRight = Math.max(maxRight, temp);
        }
        return arr;
    }
}

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Problem Info

Description

Problem 1299: Replace Elements with Greatest Element on Right Side - Detailed explanation and solution approaches.

Topics

Array

Companies

Zoho

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1299

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Initialize `max_right` to -1. This handles the case where there are no elements to the right of the last element.
2
Iterate through the array from the second to last element (index n-2) down to the first element (index 0).
3
For each element at index `i`:
4
Store the current element's value in a temporary variable `temp`.
5
Replace the element at index `i` with the current value of `max_right`.
6
Update `max_right` to be the maximum of `max_right` and `temp`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through the array once from right to left. No additional data structures proportional to the input size are used; only a constant number of variables (`max_right`, `temp`, and loop counter) are needed. Therefore, both time and space complexities are linear in the size of the input array and constant respectively.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a separate result array

Create a new array of the same size as the input array. Iterate through the input array from right to left, calculating the maximum element to the right for each element and storing it in the new array. Then, return the new array.

Time: O(n)Space: O(n)

Trade-offs: This approach is less efficient in terms of space complexity because it requires creating a new array. It's simpler to understand for beginners but less optimal than the in-place approach.

✨

Key Insights

1**Right-to-Left Traversal is Crucial**: Processing the array from `len(arr) - 1` down to `0` allows us to compute the 'greatest element to the right' on the fly. Each element `arr[i]` can be replaced because `max_right` already holds the maximum of all elements `arr[j]` where `j > i`.
2**`max_right` Accumulates the Future Maximum**: The `max_right` variable acts as a running maximum of all elements *already seen* during the right-to-left traversal. When we are at `arr[i]`, `max_right` accurately represents the greatest element among `arr[i+1]` through `arr[len(arr)-1]`, directly fulfilling the problem's requirement for `arr[i]`'s replacement.
3**Efficient In-Place Modification with Sentinel Value**: The solution modifies the array in-place, eliminating the need for extra memory. The initial `max_right = -1` serves as a sentinel, cleanly handling the last element's replacement and ensuring that any valid positive array element will correctly become the maximum as the traversal proceeds.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Empty array: The solutions correctly handle this case by either returning the empty array or handling it specifically in a conditional statement.
Array with one element: The solutions correctly return an array containing only -1.

⚠️

Common Mistakes to Avoid

✗Iterating from left to right: This approach would require maintaining a data structure (e.g., a max-heap) to track the maximum element to the right efficiently, leading to a higher time and space complexity.

✗Incorrect handling of edge cases: Forgetting to handle empty arrays or arrays with only one element can lead to incorrect results or runtime errors.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Data preprocessing: This type of transformation might be useful in preparing data for certain algorithms or machine learning models where the relationship between an element and its subsequent elements is important.

🏦Signal processing: Finding the maximum value to the right of a signal could be used to detect peaks or trends in the data.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 10, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Replace Elements with Greatest Element on Right Side"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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