What is the time complexity of Find Peak Element?

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

What data structures are used to solve Find Peak Element?

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

What are the key insights for solving Find Peak Element?

Insight 1: The constraint of O(log n) strongly suggests using Binary Search. Insight 2: Even though the problem isn't explicitly sorted, the property that elements are not equal to their neighbors allows us to apply a modified binary search. The comparison between nums[mid] and nums[mid+1] is crucial for deciding which half of the array contains a peak.

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Find Peak Element - LeetCode 162 Solution

Q: Which companies ask LeetCode 162. Find Peak Element in interviews?

This problem is frequently asked at Accenture, Commvault, Databricks, Flipkart, Goldman Sachs and 13 more companies.

Find Peak Element - Complete Solution Guide

Find Peak Element is LeetCode problem 162, 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

A peak element is an element that is strictly greater than its neighbors. Given a 0-indexed integer array nums , find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks . You may imagine that nums[-1] = nums[n] = -∞ . In other words, an element is always considered to be strictly greater than a neighbor that is outside the array. You must write an algorithm that runs in O(log n) time. Example 1: Input: nums = [1,2,3,1] Output: 2

Detailed Explanation

The problem asks us to find a 'peak element' within a given integer array. A peak element is defined as an element that is strictly greater than its adjacent neighbors. The array is 0-indexed. The crucial constraint is to find *any* peak element's index in O(log n) time. We can imagine that the array extends to negative infinity on both ends, meaning nums[-1] = nums[n] = -infinity.

Solution Approach

The solution utilizes a modified binary search algorithm to efficiently find a peak element. The core idea is that if `nums[mid] < nums[mid+1]`, then there must be a peak element in the right half of the array (including `mid+1`). Conversely, if `nums[mid] > nums[mid+1]`, then there must be a peak element in the left half of the array (including `mid`). By repeatedly halving the search space based on this comparison, we can quickly converge to a peak element.

Step-by-Step Algorithm

Step 1: Initialize `low` to 0 and `high` to `len(nums) - 1`. These represent the bounds of our search space.
Step 2: Enter a `while` loop that continues as long as `low < high`. This ensures that we have a search space of at least two elements.
Step 3: Calculate the middle index `mid = low + (high - low) // 2`. This prevents potential integer overflow issues.
Step 4: Compare `nums[mid]` with `nums[mid + 1]`. If `nums[mid] < nums[mid + 1]`, it implies that the slope is increasing, meaning a peak exists to the right of `mid`. Update `low = mid + 1`.
Step 5: Otherwise, if `nums[mid] >= nums[mid + 1]`, it implies that the slope is decreasing or flat, meaning a peak exists at or to the left of `mid`. Update `high = mid`.
Step 6: Repeat steps 3-5 until `low` and `high` converge. At this point, `low` (or `high`) will be the index of a peak element.
Step 7: Return `low` (or `high`).

Key Insights

Insight 1: The constraint of O(log n) strongly suggests using Binary Search.
Insight 2: Even though the problem isn't explicitly sorted, the property that elements are not equal to their neighbors allows us to apply a modified binary search. The comparison between nums[mid] and nums[mid+1] is crucial for deciding which half of the array contains a peak.
Insight 3: The condition nums[i] != nums[i+1] guarantees that there is always at least one peak element. This ensures that the binary search will always converge on a valid peak.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: Accenture, Commvault, Databricks, Flipkart, Goldman Sachs, IXL, Nvidia, Oracle, PayPal, ServiceNow, Snap, Tekion, TikTok, Visa, Waymo, Wix, Zepto, eBay.

Find Peak Element | LeetCode 162

Medium

Array Binary Search

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

Solution Code

Available in:

class Solution {
    public int findPeakElement(int[] nums) {
        int low = 0;
        int high = nums.length - 1;

        while (low < high) {
            int mid = low + (high - low) / 2;
            if (nums[mid] < nums[mid + 1]) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }

        return low;
    }
}

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Problem Info

Description

Problem 162: Find Peak Element - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

AccentureCommvaultDatabricksFlipkartGoldman SachsIXLNvidiaOracle

Stats

Difficulty:Medium

Languages:4

Companies:18

Problem #:162

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

7 steps

1
Step 1: Initialize `low` to 0 and `high` to `len(nums) - 1`. These represent the bounds of our search space.
2
Step 2: Enter a `while` loop that continues as long as `low < high`. This ensures that we have a search space of at least two elements.
3
Step 3: Calculate the middle index `mid = low + (high - low) // 2`. This prevents potential integer overflow issues.
4
Step 4: Compare `nums[mid]` with `nums[mid + 1]`. If `nums[mid] < nums[mid + 1]`, it implies that the slope is increasing, meaning a peak exists to the right of `mid`. Update `low = mid + 1`.
5
Step 5: Otherwise, if `nums[mid] >= nums[mid + 1]`, it implies that the slope is decreasing or flat, meaning a peak exists at or to the left of `mid`. Update `high = mid`.
6
Step 6: Repeat steps 3-5 until `low` and `high` converge. At this point, `low` (or `high`) will be the index of a peak element.
7
Step 7: Return `low` (or `high`).

💡

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(log n)

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(log n) because we are using binary search to reduce the search space by half in each iteration. The space complexity is O(1) because we are only using a few constant extra variables (`low`, `high`, `mid`) regardless of the input array 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

Linear Scan

Iterate through the array and check for each element if it is greater than its neighbors. This approach directly implements the definition of a peak element.

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

Trade-offs: The linear scan is simple to understand, but it's less efficient than the binary search approach, especially for large arrays. Use it only if the array is very small or if O(log n) solution isn't required.

✨

Key Insights

1Insight 1: The constraint of O(log n) strongly suggests using Binary Search.
2Insight 2: Even though the problem isn't explicitly sorted, the property that elements are not equal to their neighbors allows us to apply a modified binary search. The comparison between nums[mid] and nums[mid+1] is crucial for deciding which half of the array contains a peak.
3Insight 3: The condition nums[i] != nums[i+1] guarantees that there is always at least one peak element. This ensures that the binary search will always converge on a valid peak.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: Single element array. If the array has only one element, that element is considered a peak. The loop condition `low < high` will not be entered, and the initial value of `low` (which is 0) will be returned correctly.
Edge case 2: Array is monotonically increasing. The loop will iterate, always moving `low` to `mid + 1`. Eventually, `low` and `high` will converge on the last element, which is indeed a peak in this case.
Edge case 3: Array is monotonically decreasing. `high` will continuously become `mid`. Finally, `low` and `high` will converge on index `0` which is indeed the peak.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow when calculating `mid`. Using `mid = low + (high - low) // 2` avoids potential overflow issues that can occur with `mid = (low + high) // 2` when `low` and `high` are large.

✗Common mistake 2: Incorrectly updating `low` and `high`. It's crucial to include `mid` in the possible search space when updating `high`, because `mid` itself might be a peak. This is why we update `high = mid` instead of `high = mid - 1`.

✗Common mistake 3: Not considering the edge case of a single-element array. The solution should handle this gracefully without causing an error.

✅

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: Finding the maximum value in unimodal functions. Unimodal functions are functions that increase up to a certain point and then decrease. The peak element algorithm can be used to find the maximum value of such functions efficiently.

🏦Real-world scenario 2: Resource allocation. Imagine allocating resources to a project, and the return on investment increases initially, then decreases. Finding the optimal resource allocation corresponds to finding the 'peak' of the return on investment curve.

💡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: November 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

❓

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 avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

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 "Find Peak Element"?

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 binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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