What is the time complexity of Maximize the Topmost Element After K Moves?

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

What data structures are used to solve Maximize the Topmost Element After K Moves?

This problem uses Array, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximize the Topmost Element After K Moves?

Insight 1: The key is to consider two main scenarios: either the top element after 'k' moves is an element that was temporarily removed and then added back, or it's an element that was reached after removing 'k' elements from the original pile. Insight 2: Parity plays a crucial role. If we remove an element at index `i`, we need `k-i` moves to bring it back to the top. Thus, `k-i` must be an even number for this to be possible.

Which companies ask LeetCode 2202. Maximize the Topmost Element After K Moves in interviews?

This problem is frequently asked at American Express.

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Maximize the Topmost Element After K Moves - LeetCode 2202 Solution

Maximize the Topmost Element After K Moves - Complete Solution Guide

Maximize the Topmost Element After K Moves is LeetCode problem 2202, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 0-indexed integer array nums representing the contents of a pile , where nums[0] is the topmost element of the pile. In one move, you can perform either of the following: If the pile is not empty, remove the topmost element of the pile. If there are one or more removed elements, add any one of them back onto the pile. This element becomes the new topmost element. You are also given an integer k , which denotes the total number of moves to be made. Return the maximum value of the

Detailed Explanation

The problem presents a pile of integers represented by an array `nums`, where `nums[0]` is the top element. We're allowed two types of moves: removing the top element (if the pile is not empty) or adding a previously removed element back to the top. Given a fixed number of moves `k`, the task is to find the maximum possible value of the topmost element after exactly `k` moves. If a non-empty pile cannot be obtained after `k` moves, return -1.

Solution Approach

The solution iterates through the possible elements that could be on top after `k` moves. It considers two main cases. The first case considers elements from index 0 up to `min(n, k-1)` that can be removed and re-added. The second considers element at index k if k < n. For elements in first case, we need k-i to be even i.e., the number of moves remaining to bring them to the top after removing them should be even. Among all of the considered values, we update the maximum value if the current value is greater than max_val. Special cases when array length is 1, and when k = 0 are also handled.

Step-by-Step Algorithm

Step 1: Handle base cases: If k is 0, return nums[0]. If the array has only one element, return -1 if k is odd and return nums[0] if k is even.
Step 2: Initialize max_val to -1. This variable stores the maximum value of the topmost element possible after k moves.
Step 3: Iterate through the first min(n, k-1) elements of the array (representing the elements that can potentially be removed and added back). We check for the possibility of the parity by using (k - i) % 2 == 0, and update max_val accordingly.
Step 4: If k is less than the size of the array, then there will be one element left in the array after k deletions. Check if `nums[k]` is larger than `max_val`, and update it.
Step 5: Return the value of max_val.

Key Insights

Insight 1: The key is to consider two main scenarios: either the top element after 'k' moves is an element that was temporarily removed and then added back, or it's an element that was reached after removing 'k' elements from the original pile.
Insight 2: Parity plays a crucial role. If we remove an element at index `i`, we need `k-i` moves to bring it back to the top. Thus, `k-i` must be an even number for this to be possible.
Insight 3: Special cases need to be handled: When k=0, we return nums[0]. When the array has only one element, we can return a non-empty pile only if `k` is even.

Complexity Analysis

Time Complexity: O(min(n, k))

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy.

Companies

Asked at: American Express.

Maximize the Topmost Element After K Moves | LeetCode 2202

Medium

Array Greedy

Time: O(min(n, k))Space: O(1)

Solution Code

Available in:

class Solution {
    public int maximumTop(int[] nums, int k) {
        int n = nums.length;

        if (k == 0) {
            return nums[0];
        }

        if (n == 1) {
            if (k % 2 == 1) {
                return -1;
            } else {
                return nums[0];
            }
        }

        int max_val = -1;

        // Case 1: The top element is one of the first k-1 elements,
        // which is removed and then added back.
        // This requires i <= k-2 and k and i have the same parity.
        // The elements we can potentially do this with are nums[0...min(n-1, k-2)]
        int limit_idx = k - 2;
        if (limit_idx >= 0) {
            for (int i = 0; i < Math.min(n, limit_idx + 1); i++) {
                if ((k - i) % 2 == 0) {
                    if (nums[i] > max_val) {
                        max_val = nums[i];
                    }
                }
            }
        }

        // Case 2: The top element is nums[k], reached by k removals.
        if (k < n) {
            if (nums[k] > max_val) {
                max_val = nums[k];
            }
        }

        return max_val;
    }
}

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Problem Info

Description

Problem 2202: Maximize the Topmost Element After K Moves - Detailed explanation and solution approaches.

Topics

Array Greedy

Companies

American Express

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2202

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

5 steps

1
Step 1: Handle base cases: If k is 0, return nums[0]. If the array has only one element, return -1 if k is odd and return nums[0] if k is even.
2
Step 2: Initialize max_val to -1. This variable stores the maximum value of the topmost element possible after k moves.
3
Step 3: Iterate through the first min(n, k-1) elements of the array (representing the elements that can potentially be removed and added back). We check for the possibility of the parity by using (k - i) % 2 == 0, and update max_val accordingly.
4
Step 4: If k is less than the size of the array, then there will be one element left in the array after k deletions. Check if `nums[k]` is larger than `max_val`, and update it.
5
Step 5: Return the value of max_val.

💡

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(min(n, k))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the for loop. In worst case scenario, the for loop iterates for `min(n, k)` times. Rest of the code executes in constant time. The space complexity is determined by the variables used in the program. No auxiliary space is used which scales with input size.

📖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

Brute Force

Generate all possible sequences of moves and keep track of the maximum top element. This is impractical and will lead to TLE.

Time: ExponentialSpace: Exponential

Trade-offs: Avoid this approach. It is much slower. It may be useful for very small arrays and values of k, however.

✨

Key Insights

1Insight 1: The key is to consider two main scenarios: either the top element after 'k' moves is an element that was temporarily removed and then added back, or it's an element that was reached after removing 'k' elements from the original pile.
2Insight 2: Parity plays a crucial role. If we remove an element at index `i`, we need `k-i` moves to bring it back to the top. Thus, `k-i` must be an even number for this to be possible.
3Insight 3: Special cases need to be handled: When k=0, we return nums[0]. When the array has only one element, we can return a non-empty pile only if `k` is even.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: nums.length == 1. If k is odd, it's impossible to have a non-empty pile, so return -1. If k is even, the element remains the same.
Edge case 2: k == 0. The topmost element remains nums[0].
Edge case 3: k >= n. We remove all elements. If k > n, we can add any removed value back. If nums length is 1, we are stuck. If nums length is > 1, we can take the maximum of the nums[0:n-1]
Edge case 4: nums = [5, 2, 2, 4, 0, 6], k = 4

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge cases where nums.length is 1.

✗Common mistake 2: Incorrectly computing the parity of moves. You must remove all elements before index 'i' and then make sure you have an even number of moves remaining to put the element back on top.

✗Common mistake 3: Not considering k >= n as a possible edge case and not correctly thinking about if k == n or if k > n.

✅

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?

💼Real-world scenario 1: Stack manipulation scenarios in memory management or data processing pipelines.

🏦Real-world scenario 2: Resource allocation problems where resources are taken and returned and the goal is to maximize a specific resource characteristic at the end of a fixed number of steps.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 13, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximize the Topmost Element After K Moves"?

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

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