What is the time complexity of Range Addition II?

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

What data structures are used to solve Range Addition II?

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

What are the key insights for solving Range Addition II?

Insight 1: The maximum value in the matrix will be determined by the smallest `a_i` and `b_i` across all operations, since the cells within the rectangle defined by these minimum values will be incremented the most times. Insight 2: We don't need to actually update the matrix. We only need to find the minimum `a` and minimum `b` from the operations and then multiply them.

Which companies ask LeetCode 598. Range Addition II in interviews?

This problem is frequently asked at IXL.

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Range Addition II - LeetCode 598 Solution

Q: How to solve Range Addition II?

The solution efficiently finds the minimum `m` and `n` values after applying all operations. The number of maximum integers in the matrix is the product of the final `m` and `n` values. Iterating through the `ops` array, we update `m` to be the minimum of the current `m` and `ops[i][0]`, and similarly for `n` and `ops[i][1]`. The final `m * n` gives the count of the maximum elements.

Range Addition II - Complete Solution Guide

Range Addition II is LeetCode problem 598, 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

You are given an m x n matrix M initialized with all 0 's and an array of operations ops , where ops[i] = [a i , b i ] means M[x][y] should be incremented by one for all 0 <= x < a i and 0 <= y < b i . Count and return the number of maximum integers in the matrix after performing all the operations . Example 1: Input: m = 3, n = 3, ops = [[2,2],[3,3]] Output: 4 Explanation: The maximum integer in M is 2, and there are four of it in M. So return 4. Example 2: Input: m = 3, n = 3, ops = [[2,2],[3,

Detailed Explanation

The problem describes a matrix `M` of size `m x n` initialized with all zeros. We are given an array of operations `ops`, where each operation `ops[i] = [a_i, b_i]` instructs us to increment all elements `M[x][y]` by one for `0 <= x < a_i` and `0 <= y < b_i`. After applying all operations, we need to find the number of elements in `M` that hold the maximum value.

Solution Approach

The solution efficiently finds the minimum `m` and `n` values after applying all operations. The number of maximum integers in the matrix is the product of the final `m` and `n` values. Iterating through the `ops` array, we update `m` to be the minimum of the current `m` and `ops[i][0]`, and similarly for `n` and `ops[i][1]`. The final `m * n` gives the count of the maximum elements.

Step-by-Step Algorithm

Step 1: Initialize `m` and `n` with the given matrix dimensions.
Step 2: Iterate through the `ops` array.
Step 3: In each iteration, update `m` to be the minimum of the current `m` and `ops[i][0]`.
Step 4: Similarly, update `n` to be the minimum of the current `n` and `ops[i][1]`.
Step 5: After iterating through all the operations, return the product `m * n`.

Key Insights

Insight 1: The maximum value in the matrix will be determined by the smallest `a_i` and `b_i` across all operations, since the cells within the rectangle defined by these minimum values will be incremented the most times.
Insight 2: We don't need to actually update the matrix. We only need to find the minimum `a` and minimum `b` from the operations and then multiply them.
Insight 3: If the `ops` array is empty, the maximum value remains 0, and the number of maximum integers is simply m * n, as the matrix is filled with 0s initially.

Complexity Analysis

Time Complexity: O(k)

Space Complexity: O(1)

Topics

This problem involves: Array, Math.

Companies

Asked at: IXL.

Range Addition II | LeetCode 598

Easy

Array Math

Time: O(k)Space: O(1)

Solution Code

Available in:

class Solution {
    public int maxCount(int m, int n, int[][] ops) {
        for (int[] op : ops) {
            m = Math.min(m, op[0]);
            n = Math.min(n, op[1]);
        }
        return m * n;
    }
}

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Problem Info

Description

Problem 598: Range Addition II - Detailed explanation and solution approaches.

Topics

Array Math

Companies

IXL

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:598

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: Initialize `m` and `n` with the given matrix dimensions.
2
Step 2: Iterate through the `ops` array.
3
Step 3: In each iteration, update `m` to be the minimum of the current `m` and `ops[i][0]`.
4
Step 4: Similarly, update `n` to be the minimum of the current `n` and `ops[i][1]`.
5
Step 5: After iterating through all the operations, return the product `m * n`.

💡

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(k)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the loop that iterates through the `ops` array. The space complexity is constant because we are only using a few variables to store the minimum `m` and `n` values regardless of the 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

No alternative needed - Approach is already Optimal

Due to the problem constraints and objective (find the overlap rectangle among multiple range operations), the provided approach is already the most efficient and optimal. No other algorithms can significantly improve the O(k) time complexity and O(1) space complexity.

Time: O(k)Space: O(1)

Trade-offs: N/A

✨

Key Insights

1Insight 1: The maximum value in the matrix will be determined by the smallest `a_i` and `b_i` across all operations, since the cells within the rectangle defined by these minimum values will be incremented the most times.
2Insight 2: We don't need to actually update the matrix. We only need to find the minimum `a` and minimum `b` from the operations and then multiply them.
3Insight 3: If the `ops` array is empty, the maximum value remains 0, and the number of maximum integers is simply m * n, as the matrix is filled with 0s initially.

📚

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: `ops` is empty. In this case, no operations are performed, and the whole matrix remains filled with zeros. The correct result is `m * n`. The provided code handles this implicitly because if `ops` is empty, the loop doesn't execute, and the initial values of `m` and `n` are returned.
Edge case 2: `m` or `n` is 1. The code correctly handles this because the `min` function will still determine the smallest increment area.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Trying to simulate the actual matrix updates. This would result in a Time Limit Exceeded (TLE) error, as the matrix could be very large (up to 4 * 10^4 x 4 * 10^4).

✗Common mistake 2: Not handling the case where `ops` is empty. This would lead to an incorrect result if the initial matrix dimensions are not considered.

✅

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: Resource allocation. Imagine `m x n` represents available resources, and `ops` represent requests to allocate resources in a specific rectangular area. The algorithm can efficiently determine the area with the highest resource demand.

🏦Real-world scenario 2: Image processing. Suppose `m x n` is an image, and `ops` represents a series of filters applied to rectangular regions of the image. The algorithm can determine the area of the image where the most filters have been applied.

💡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: July 29, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 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 should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Range Addition II"?

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

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