LeetCode #3070 — MEDIUM

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

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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 submatrices, shown in the image above, that contain the top-left element of grid, and have a sum less than or equal to 20.

Constraints:

m == grid.length
n == grid[i].length
1 <= n, m <= 1000
0 <= grid[i][j] <= 1000
1 <= k <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[7,6,3],[6,6,1]]
18

Example 2

[[7,2,9],[1,5,0],[2,6,6]]
20

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3070: Count Submatrices with Top-Left Element and Sum Less Than k
class Solution {
    public int countSubmatrices(int[][] grid, int k) {
        int m = grid.length, n = grid[0].length;
        int[][] s = new int[m + 1][n + 1];
        int ans = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + grid[i - 1][j - 1];
                if (s[i][j] <= k) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3070: Count Submatrices with Top-Left Element and Sum Less Than k
func countSubmatrices(grid [][]int, k int) (ans int) {
	s := make([][]int, len(grid)+1)
	for i := range s {
		s[i] = make([]int, len(grid[0])+1)
	}
	for i, row := range grid {
		for j, x := range row {
			s[i+1][j+1] = s[i+1][j] + s[i][j+1] - s[i][j] + x
			if s[i+1][j+1] <= k {
				ans++
			}
		}
	}
	return
}

# Accepted solution for LeetCode #3070: Count Submatrices with Top-Left Element and Sum Less Than k
class Solution:
    def countSubmatrices(self, grid: List[List[int]], k: int) -> int:
        s = [[0] * (len(grid[0]) + 1) for _ in range(len(grid) + 1)]
        ans = 0
        for i, row in enumerate(grid, 1):
            for j, x in enumerate(row, 1):
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + x
                ans += s[i][j] <= k
        return ans

// Accepted solution for LeetCode #3070: Count Submatrices with Top-Left Element and Sum Less Than k
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3070: Count Submatrices with Top-Left Element and Sum Less Than k
// class Solution {
//     public int countSubmatrices(int[][] grid, int k) {
//         int m = grid.length, n = grid[0].length;
//         int[][] s = new int[m + 1][n + 1];
//         int ans = 0;
//         for (int i = 1; i <= m; ++i) {
//             for (int j = 1; j <= n; ++j) {
//                 s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + grid[i - 1][j - 1];
//                 if (s[i][j] <= k) {
//                     ++ans;
//                 }
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3070: Count Submatrices with Top-Left Element and Sum Less Than k
function countSubmatrices(grid: number[][], k: number): number {
    const m = grid.length;
    const n = grid[0].length;
    const s: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
    let ans: number = 0;
    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + grid[i - 1][j - 1];
            if (s[i][j] <= k) {
                ++ans;
            }
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(m × n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.