What is the time complexity of Count Submatrices with Top-Left Element and Sum Less Than k?

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

What data structures are used to solve Count Submatrices with Top-Left Element and Sum Less Than k?

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

What are the key insights for solving Count Submatrices with Top-Left Element and Sum Less Than k?

Insight 1: The problem can be efficiently solved by utilizing prefix sums to quickly calculate the sum of elements within any submatrix originating from the top-left corner. Insight 2: The condition that the submatrix must contain the top-left element simplifies the problem because it reduces the search space significantly. We only need to consider submatrices with top-left at (0,0).

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Count Submatrices with Top-Left Element and Sum Less Than k - LeetCode 3070 Solution

Q: Which companies ask LeetCode 3070. Count Submatrices with Top-Left Element and Sum Less Than k in interviews?

This problem is frequently asked at Barclays.

Count Submatrices with Top-Left Element and Sum Less Than k - Complete Solution Guide

Count Submatrices with Top-Left Element and Sum Less Than k is LeetCode problem 3070, 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 matrix grid and an integer k . Return the number of submatrices that contain the top-left element of the grid , and have a sum less than or equal to k . Example 1: Input: grid = [[7,6,3],[6,6,1]], k = 18 Output: 4 Explanation: There are only 4 submatrices, shown in the image above, that contain the top-left element of grid, and have a sum less than or equal to 18. Example 2: Input: grid = [[7,2,9],[1,5,0],[2,6,6]], k = 20 Output: 6 Explanation: There are only 6

Detailed Explanation

The problem asks us to count the number of submatrices within a given matrix `grid` that satisfy two conditions: (1) the submatrix must include the top-left element of the `grid`, and (2) the sum of all elements within the submatrix must be less than or equal to a given integer `k`. The input is the `grid` (a 2D array of integers) and the integer `k`. The output is the total count of submatrices meeting both conditions. The matrix dimensions are limited to 1000x1000, and element values are between 0 and 1000, with `k` being between 1 and 10^9.

Solution Approach

The solution uses dynamic programming with prefix sums to efficiently calculate the sum of submatrices. First, it calculates the row-wise prefix sum. Then, it calculates the column-wise prefix sum on the row-wise prefix sum. For each possible submatrix defined by its bottom-right corner (r, c), the sum of its elements can be obtained from the prefix sum matrix. The code iterates through all possible bottom-right corners of submatrices that include the top-left element, and for each submatrix, it checks if the sum of its elements is less than or equal to `k`. If it is, the counter is incremented. A key optimization is that if, for a particular column, the sum exceeds `k`, we can break the inner loop because any submatrix extending further down in that column will also have a sum greater than `k`.

Step-by-Step Algorithm

Step 1: Iterate through each row of the grid (from row 0 to m-1).
Step 2: For each row, iterate through each column (from column 1 to n-1).
Step 3: Calculate the row-wise prefix sum: `grid[r][c] += grid[r][c-1]`. This transforms each element to represent the sum of elements from the beginning of that row until that column.
Step 4: Initialize a counter variable `count` to 0. This will store the number of submatrices meeting the criteria.
Step 5: Iterate through each column of the grid (from column 0 to n-1).
Step 6: For each column, iterate through each row (from row 0 to m-1).
Step 7: Calculate column-wise prefix sum on top of the row-wise prefix sum. if (r > 0) { grid[r][c] += grid[r-1][c]; }. This ensures grid[r][c] represents the sum of the submatrix from (0,0) to (r,c).
Step 8: Check if the sum of the submatrix (represented by `grid[r][c]`) is less than or equal to `k`.
Step 9: If the sum is less than or equal to `k`, increment the `count`.
Step 10: If the sum is greater than `k`, break the inner loop (row iteration) because all submatrices extending further down in that column will also have a sum greater than `k`.
Step 11: After iterating through all possible submatrices, return the `count`.

Key Insights

Insight 1: The problem can be efficiently solved by utilizing prefix sums to quickly calculate the sum of elements within any submatrix originating from the top-left corner.
Insight 2: The condition that the submatrix must contain the top-left element simplifies the problem because it reduces the search space significantly. We only need to consider submatrices with top-left at (0,0).
Insight 3: By calculating prefix sums in-place, we can avoid using extra space, optimizing the space complexity of the solution.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(1)

Topics

This problem involves: Array, Matrix, Prefix Sum.

Companies

Asked at: Barclays.

Count Submatrices with Top-Left Element and Sum Less Than k | LeetCode 3070

Medium

Array Matrix Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int countSubmatrices(int[][] grid, int k) {
        int m = grid.length;
        int n = grid[0].length;

        for (int r = 0; r < m; r++) {
            for (int c = 1; c < n; c++) {
                grid[r][c] += grid[r][c-1];
            }
        }

        int count = 0;
        for (int c = 0; c < n; c++) {
            for (int r = 0; r < m; r++) {
                if (r > 0) {
                    grid[r][c] += grid[r-1][c];
                }

                if (grid[r][c] <= k) {
                    count++;
                } else {
                    break;
                }
            }
        }

        return count;
    }
}

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Problem Info

Description

Problem 3070: Count Submatrices with Top-Left Element and Sum Less Than k - Detailed explanation and solution approaches.

Topics

Array Matrix Prefix Sum

Companies

Barclays

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3070

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

11 steps

1
Step 1: Iterate through each row of the grid (from row 0 to m-1).
2
Step 2: For each row, iterate through each column (from column 1 to n-1).
3
Step 3: Calculate the row-wise prefix sum: `grid[r][c] += grid[r][c-1]`. This transforms each element to represent the sum of elements from the beginning of that row until that column.
4
Step 4: Initialize a counter variable `count` to 0. This will store the number of submatrices meeting the criteria.
5
Step 5: Iterate through each column of the grid (from column 0 to n-1).
6
Step 6: For each column, iterate through each row (from row 0 to m-1).
7
Step 7: Calculate column-wise prefix sum on top of the row-wise prefix sum. if (r > 0) { grid[r][c] += grid[r-1][c]; }. This ensures grid[r][c] represents the sum of the submatrix from (0,0) to (r,c).
8
Step 8: Check if the sum of the submatrix (represented by `grid[r][c]`) is less than or equal to `k`.
9
Step 9: If the sum is less than or equal to `k`, increment the `count`.
10
Step 10: If the sum is greater than `k`, break the inner loop (row iteration) because all submatrices extending further down in that column will also have a sum greater than `k`.
11
Step 11: After iterating through all possible submatrices, return the `count`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the two nested loops, each iterating through the matrix. The space complexity is O(1) as the grid is modified in-place and no additional data structures dependent on the size of input are created.

📖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

Iterate through all possible submatrices starting from the top-left corner and calculate their sum directly for each submatrix. This involves calculating the sum of elements within each submatrix from scratch.

Time: O(m^2 * n^2), where m is the number of rows and n is the number of columns. This is because calculating sum of submatrix takes O(m*n) and we have O(m*n) such submatrices.Space: O(1), because it doesn't use any extra space proportional to the input size.

Trade-offs: This approach is significantly less efficient than the prefix sum approach due to its higher time complexity. It's simpler to understand but not practical for larger input sizes. Only good when matrix size is very small.

✨

Key Insights

1Insight 1: The problem can be efficiently solved by utilizing prefix sums to quickly calculate the sum of elements within any submatrix originating from the top-left corner.
2Insight 2: The condition that the submatrix must contain the top-left element simplifies the problem because it reduces the search space significantly. We only need to consider submatrices with top-left at (0,0).
3Insight 3: By calculating prefix sums in-place, we can avoid using extra space, optimizing the space complexity of the solution.

📚

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: Empty grid. The code implicitly handles the case where either m or n is 0 by returning 0 since loops will not execute. However, an explicit check for `if not grid or not grid[0]` at the beginning can improve readability.
Edge case 2: k is very small (e.g., 1). The code correctly handles this case because the count increments only when the submatrix sum is <= k. Many submatrices will exceed k quickly.
Edge case 3: k is very large (e.g., larger than the sum of all elements in the grid). The code handles this correctly because the count variable will increase for every submatrix it iterates over. The early breaking in the column only kicks in if submatrix sum exceeds k.
Edge case 4: All elements of grid are zero. In this case, every submatrix will have a sum of zero and provided that k >=0 all submatrices will be counted till the last element.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not correctly calculating prefix sums, leading to incorrect submatrix sums. It's important to build prefix sums correctly row-wise first, then column-wise or vice versa.

✗Common mistake 2: Not considering the constraint that the submatrix must contain the top-left element and trying to consider all possible submatrices in the grid, which would lead to unnecessary computations and a much less efficient solution.

✗Common mistake 3: Failing to break the inner loop when the submatrix sum exceeds `k`, which can lead to unnecessary iterations and slower performance.

✅

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, where you might need to calculate the sum of pixel values within certain rectangular regions of an image. This could be used for feature extraction or image filtering.

🏦Real-world scenario 2: Data analysis, where you might need to find the sum of values in certain ranges of a dataset, for example, to calculate rolling sums or moving averages of time-series 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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: June 11, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

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

@AlgoMasterJul 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.

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 "Count Submatrices with Top-Left Element and Sum Less Than k"?

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

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