What is the time complexity of Find a Peak Element II?

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

What data structures are used to solve Find a Peak Element II?

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

What are the key insights for solving Find a Peak Element II?

Insight 1: The constraint that no two adjacent cells are equal ensures that a peak always exists. This makes the problem solvable without explicitly checking for its existence. Insight 2: Binary search can be applied on either rows or columns to reduce the search space in each iteration.

Which companies ask LeetCode 1901. Find a Peak Element II in interviews?

This problem is frequently asked at Zeta.

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Find a Peak Element II - LeetCode 1901 Solution

Find a Peak Element II - Complete Solution Guide

Find a Peak Element II is LeetCode problem 1901, 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 in a 2D grid is an element that is strictly greater than all of its adjacent neighbors to the left, right, top, and bottom. Given a 0-indexed m x n matrix mat where no two adjacent cells are equal , find any peak element mat[i][j] and return the length 2 array [i,j] . You may assume that the entire matrix is surrounded by an outer perimeter with the value -1 in each cell. You must write an algorithm that runs in O(m log(n)) or O(n log(m)) time. Example 1: Input: mat = [[1,4],[3,2]

Detailed Explanation

The problem asks us to find a peak element in a 2D matrix. A peak element is defined as an element that is strictly greater than all its adjacent neighbors (left, right, top, and bottom). The matrix has the property that no two adjacent cells are equal, and the entire matrix is surrounded by an outer perimeter of -1. We need to return the row and column indices of any peak element, and the algorithm must run in O(m log n) or O(n log m) time.

Solution Approach

The provided solution uses a binary search approach on the columns of the matrix. For each column selected by the binary search, it finds the maximum element in that column. Then, it checks if this maximum element is greater than its left and right neighbors. If it is, we have found a peak and we return its coordinates. If the left neighbor is greater, it means there must be a peak to the left, so we narrow the search to the left half of the matrix. If the right neighbor is greater, it means there must be a peak to the right, so we narrow the search to the right half of the matrix.

Step-by-Step Algorithm

Step 1: Initialize `low_col` to 0 and `high_col` to n - 1, representing the boundaries of the column indices.
Step 2: While `low_col` is less than or equal to `high_col`, perform a binary search:
Step 3: Calculate the middle column index `mid_col` as the average of `low_col` and `high_col`.
Step 4: Find the maximum element in the `mid_col` column and its row index `max_row`.
Step 5: Get the value of the current element `current_val` at `mat[max_row][mid_col]`, the left neighbor `left_val` (or -1 if there's no left neighbor), and the right neighbor `right_val` (or -1 if there's no right neighbor).
Step 6: If `current_val` is greater than both `left_val` and `right_val`, then `mat[max_row][mid_col]` is a peak element. Return `[max_row, mid_col]`.
Step 7: If `left_val` is greater than `current_val`, there must be a peak to the left. Set `high_col = mid_col - 1`.
Step 8: Otherwise, `right_val` must be greater than `current_val`, indicating there must be a peak to the right. Set `low_col = mid_col + 1`.

Key Insights

Insight 1: The constraint that no two adjacent cells are equal ensures that a peak always exists. This makes the problem solvable without explicitly checking for its existence.
Insight 2: Binary search can be applied on either rows or columns to reduce the search space in each iteration.
Insight 3: Instead of checking all neighbors explicitly, focus on finding the maximum element in each column and comparing it with its left and right neighbors in the same row to determine a potential peak.

Complexity Analysis

Time Complexity: O(m log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Matrix.

Companies

Asked at: Zeta.

Find a Peak Element II | LeetCode 1901

Medium

Array Binary Search Matrix

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

Solution Code

Available in:

class Solution {
    public int[] findPeakGrid(int[][] mat) {
        int m = mat.length;
        int n = mat[0].length;

        int low_col = 0;
        int high_col = n - 1;

        while (low_col <= high_col) {
            int mid_col = low_col + (high_col - low_col) / 2;
            
            int max_row = 0;
            for (int i = 0; i < m; i++) {
                if (mat[i][mid_col] > mat[max_row][mid_col]) {
                    max_row = i;
                }
            }
            
            int current_val = mat[max_row][mid_col];
            int left_val = (mid_col > 0) ? mat[max_row][mid_col - 1] : -1;
            int right_val = (mid_col < n - 1) ? mat[max_row][mid_col + 1] : -1;
            
            if (current_val > left_val && current_val > right_val) {
                return new int[]{max_row, mid_col};
            } else if (left_val > current_val) {
                high_col = mid_col - 1;
            } else {
                low_col = mid_col + 1;
            }
        }
        return new int[]{-1, -1}; // Should not reach here, but added for completeness
    }
}

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Problem Info

Description

Problem 1901: Find a Peak Element II - Detailed explanation and solution approaches.

Topics

Array Binary Search Matrix

Companies

Zeta

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1901

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

8 steps

1
Step 1: Initialize `low_col` to 0 and `high_col` to n - 1, representing the boundaries of the column indices.
2
Step 2: While `low_col` is less than or equal to `high_col`, perform a binary search:
3
Step 3: Calculate the middle column index `mid_col` as the average of `low_col` and `high_col`.
4
Step 4: Find the maximum element in the `mid_col` column and its row index `max_row`.
5
Step 5: Get the value of the current element `current_val` at `mat[max_row][mid_col]`, the left neighbor `left_val` (or -1 if there's no left neighbor), and the right neighbor `right_val` (or -1 if there's no right neighbor).
6
Step 6: If `current_val` is greater than both `left_val` and `right_val`, then `mat[max_row][mid_col]` is a peak element. Return `[max_row, mid_col]`.
7
Step 7: If `left_val` is greater than `current_val`, there must be a peak to the left. Set `high_col = mid_col - 1`.
8
Step 8: Otherwise, `right_val` must be greater than `current_val`, indicating there must be a peak to the right. Set `low_col = mid_col + 1`.

💡

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(m log n)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the binary search performed on the columns (log n), and the linear scan of each column (m) during each iteration. The space complexity is constant as we only use a few variables to store indices and temporary values. The C version requires allocation of memory for the result array, but its size is constant.

📖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

Binary Search on Rows

Instead of performing binary search on columns, we could perform binary search on rows. For each row selected by the binary search, find the maximum element. Then, check if this maximum element is greater than its top and bottom neighbors. If it is, we have found a peak. If the top neighbor is greater, it means there must be a peak to the top, so we narrow the search to the top half of the matrix. If the bottom neighbor is greater, it means there must be a peak to the bottom half of the matrix.

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

Trade-offs: This approach is symmetric to the original approach, and is preferred when the number of rows is significantly smaller than the number of columns.

✨

Key Insights

1Insight 1: The constraint that no two adjacent cells are equal ensures that a peak always exists. This makes the problem solvable without explicitly checking for its existence.
2Insight 2: Binary search can be applied on either rows or columns to reduce the search space in each iteration.
3Insight 3: Instead of checking all neighbors explicitly, focus on finding the maximum element in each column and comparing it with its left and right neighbors in the same row to determine a potential 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: Matrix with only one column (n=1). The binary search will quickly converge and find the maximum value in that column, which will be the peak.
Edge case 2: Matrix with only one row (m=1). The solution will still work correctly as it iterates through columns and compares the value with its immediate left and right neighbors. The algorithm correctly handles boundary conditions where `mid_col` is 0 or n-1 by considering edge cells have value -1.
Edge case 3: A peak at the edge of the matrix. The code handles edge cases by assigning -1 to the value of non-existing neighbours (left and right) thus the algorithm treats these edges as valid cases and returns correct result.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge cases where the peak is located at the boundary of the matrix. The provided solution handles this by treating out-of-bounds elements as -1.

✗Common mistake 2: Incorrectly implementing the binary search. Ensuring that the `low_col` and `high_col` are updated correctly is crucial for the binary search to converge.

✗Common mistake 3: Confusing row and column indices. It's important to keep track of which dimension you're operating on.

✅

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 local maxima in terrain elevation data. Representing terrain as a matrix and identifying peak elevations can be useful in geographical analysis or pathfinding for robots.

🏦Real-world scenario 2: Identifying hot spots in a heat map. In data analysis, this algorithm can be used to find regions with maximum intensity in a heat map or image, which could represent areas of high activity or interest.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: September 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 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.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Find a Peak Element II"?

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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