What is the time complexity of Image Smoother?

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

What data structures are used to solve Image Smoother?

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

What are the key insights for solving Image Smoother?

Insight 1: The core logic involves iterating through each cell of the image and calculating a local average based on its neighbors. Boundary checks are vital to avoid accessing out-of-bounds elements. Insight 2: The use of nested loops is essential to visit each cell in the 2D matrix and then another set of (potentially bounded) nested loops to iterate through its neighbors.

Which companies ask LeetCode 661. Image Smoother in interviews?

This problem is frequently asked at Roblox, Toptal, Verkada, Visa.

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Image Smoother - LeetCode 661 Solution

Q: What is the space complexity of Image Smoother?

The space complexity is O(m*n).

Image Smoother - Complete Solution Guide

Image Smoother is LeetCode problem 661, 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

An image smoother is a filter of the size 3 x 3 that can be applied to each cell of an image by rounding down the average of the cell and the eight surrounding cells (i.e., the average of the nine cells in the blue smoother). If one or more of the surrounding cells of a cell is not present, we do not consider it in the average (i.e., the average of the four cells in the red smoother). Given an m x n integer matrix img representing the grayscale of an image, return the image after applying the sm

Detailed Explanation

The "Image Smoother" problem asks us to apply a smoothing filter to an image represented as a 2D integer matrix (grayscale). For each cell in the image, we need to calculate the average of the cell and its eight surrounding neighbors. The crucial point is that if a neighbor is outside the bounds of the image, it's not included in the average. The result should be an integer matrix where each cell contains the floor of the calculated average for the corresponding cell in the original image. The input is an m x n matrix of integers, and the output is a new m x n matrix with the smoothed values.

Solution Approach

The provided solutions iterate through each cell of the input image. For each cell, they iterate through its neighbors (including itself) within a 3x3 window. They sum the values of the valid neighbors and count the number of valid neighbors. Finally, they divide the sum by the count (integer division, which implicitly applies the 'floor' operation) to get the smoothed value for that cell, and store the smoothed value in a new output matrix.

Step-by-Step Algorithm

Step 1: Get the dimensions (m and n) of the input image matrix.
Step 2: Create a new matrix, `smoothed`, of the same dimensions as the input matrix to store the smoothed image.
Step 3: Iterate through each cell (i, j) of the input matrix using nested loops.
Step 4: For each cell (i, j), initialize `sum` and `count` to 0.
Step 5: Iterate through the neighbors of the current cell using nested loops. The neighbor coordinates range from (i-1, j-1) to (i+1, j+1).
Step 6: Within the inner loops, check if the neighbor coordinates (x, y) are within the bounds of the image matrix. The range should be 0 <= x < m and 0 <= y < n.
Step 7: If the neighbor coordinates are valid, add the value of `img[x][y]` to `sum` and increment `count`.
Step 8: After iterating through all the neighbors, calculate the smoothed value by dividing `sum` by `count` (integer division).
Step 9: Store the smoothed value in `smoothed[i][j]`.
Step 10: After iterating through all cells in the input matrix, return the `smoothed` matrix.

Key Insights

Insight 1: The core logic involves iterating through each cell of the image and calculating a local average based on its neighbors. Boundary checks are vital to avoid accessing out-of-bounds elements.
Insight 2: The use of nested loops is essential to visit each cell in the 2D matrix and then another set of (potentially bounded) nested loops to iterate through its neighbors.
Insight 3: The min/max functions (or equivalent conditional statements in C) provide a concise way to handle edge cases where neighbor coordinates might go out of bounds, simplifying boundary checks.
Insight 4: Using integer division ensures the floor of the average is calculated as required by the problem statement.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Matrix.

Companies

Asked at: Roblox, Toptal, Verkada, Visa.

Image Smoother | LeetCode 661

Easy

Array Matrix

Time: O(m*n)Space: O(m*n)

Solution Code

Available in:

class Solution {
    public int[][] imageSmoother(int[][] img) {
        int m = img.length;
        int n = img[0].length;
        int[][] result = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int sum = 0;
                int count = 0;
                for (int x = Math.max(0, i - 1); x <= Math.min(m - 1, i + 1); x++) {
                    for (int y = Math.max(0, j - 1); y <= Math.min(n - 1, j + 1); y++) {
                        sum += img[x][y];
                        count++;
                    }
                }
                result[i][j] = sum / count;
            }
        }
        return result;
    }
}

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Problem Info

Description

Problem 661: Image Smoother - Detailed explanation and solution approaches.

Topics

Array Matrix

Companies

RobloxToptalVerkadaVisa

Stats

Difficulty:Easy

Languages:4

Companies:4

Problem #:661

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

10 steps

1
Step 1: Get the dimensions (m and n) of the input image matrix.
2
Step 2: Create a new matrix, `smoothed`, of the same dimensions as the input matrix to store the smoothed image.
3
Step 3: Iterate through each cell (i, j) of the input matrix using nested loops.
4
Step 4: For each cell (i, j), initialize `sum` and `count` to 0.
5
Step 5: Iterate through the neighbors of the current cell using nested loops. The neighbor coordinates range from (i-1, j-1) to (i+1, j+1).
6
Step 6: Within the inner loops, check if the neighbor coordinates (x, y) are within the bounds of the image matrix. The range should be 0 <= x < m and 0 <= y < n.
7
Step 7: If the neighbor coordinates are valid, add the value of `img[x][y]` to `sum` and increment `count`.
8
Step 8: After iterating through all the neighbors, calculate the smoothed value by dividing `sum` by `count` (integer division).
9
Step 9: Store the smoothed value in `smoothed[i][j]`.
10
Step 10: After iterating through all cells in the input matrix, return the `smoothed` matrix.

💡

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*n)

💾

Space Complexity

O(m*n)

📝

Why?

The algorithm iterates through each cell of the m x n input matrix. For each cell, it iterates through at most 9 neighbors (a constant). Therefore, the time complexity is O(m*n). A new matrix of size m x n is created to store the smoothed image. Thus, the space complexity is O(m*n).

📖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

In-place Smoothing (Potentially Tricky)

Attempt to modify the original image directly instead of creating a new one. This requires careful consideration of how modifications affect subsequent calculations. This might be possible with bit manipulation tricks if the range of img[i][j] is limited.

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

Trade-offs: Saves space but can be more complex to implement and understand, and could be less readable. Generally not recommended due to the added complexity without substantial performance gain.

✨

Key Insights

1Insight 1: The core logic involves iterating through each cell of the image and calculating a local average based on its neighbors. Boundary checks are vital to avoid accessing out-of-bounds elements.
2Insight 2: The use of nested loops is essential to visit each cell in the 2D matrix and then another set of (potentially bounded) nested loops to iterate through its neighbors.
3Insight 3: The min/max functions (or equivalent conditional statements in C) provide a concise way to handle edge cases where neighbor coordinates might go out of bounds, simplifying boundary checks.
4Insight 4: Using integer division ensures the floor of the average is calculated as required by the problem statement.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: Cells at the boundaries of the image have fewer than 9 neighbors. The algorithm correctly handles this by checking the bounds of the neighbor coordinates and only including valid neighbors in the average.
Edge case 2: A single-pixel image. The algorithm will still work correctly, summing only the single pixel and dividing by 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check the bounds of the neighbor coordinates, leading to out-of-bounds array accesses and runtime errors.

✗Common mistake 2: Incorrectly calculating the average by not counting the correct number of neighbors or summing the values correctly.

✗Common mistake 3: Not using integer division (or floor function) to get the correct integer result, as required by the problem description.

✗Common mistake 4: Modifying the original image while calculating the smoothed values, which can affect the average calculations for subsequent cells.

✅

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: Image processing: Smoothing is a fundamental operation in image processing used to reduce noise, blur images, and prepare images for other algorithms like edge detection.

🏦Real-world scenario 2: Signal processing: Smoothing techniques are also used in signal processing to remove high-frequency components and reveal underlying trends in 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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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 30, 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 13

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

@InterviewProJul 6

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

@CodeNinjaJun 29

❓

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 "Image Smoother"?

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

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