LeetCode #1504 — MEDIUM

Count Submatrices With All Ones

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

The Problem

Problem Statement

Given an m x n binary matrix mat, return the number of submatrices that have all ones.

Example 1:

Input: mat = [[1,0,1],[1,1,0],[1,1,0]]
Output: 13
Explanation: 
There are 6 rectangles of side 1x1.
There are 2 rectangles of side 1x2.
There are 3 rectangles of side 2x1.
There is 1 rectangle of side 2x2. 
There is 1 rectangle of side 3x1.
Total number of rectangles = 6 + 2 + 3 + 1 + 1 = 13.

Example 2:

Input: mat = [[0,1,1,0],[0,1,1,1],[1,1,1,0]]
Output: 24
Explanation: 
There are 8 rectangles of side 1x1.
There are 5 rectangles of side 1x2.
There are 2 rectangles of side 1x3. 
There are 4 rectangles of side 2x1.
There are 2 rectangles of side 2x2. 
There are 2 rectangles of side 3x1. 
There is 1 rectangle of side 3x2. 
Total number of rectangles = 8 + 5 + 2 + 4 + 2 + 2 + 1 = 24.

Constraints:

1 <= m, n <= 150
mat[i][j] is either 0 or 1.

Step 01

Brute Force Baseline

Problem summary: Given an m x n binary matrix mat, return the number of submatrices that have all ones.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Stack

Example 1

[[1,0,1],[1,1,0],[1,1,0]]

Example 2

[[0,1,1,0],[0,1,1,1],[1,1,1,0]]

Core Insight

What unlocks the optimal approach

For each row i, create an array nums where: if mat[i][j] == 0 then nums[j] = 0 else nums[j] = nums[j-1] +1.
In the row i, number of rectangles between column j and k(inclusive) and ends in row i, is equal to SUM(min(nums[j, .. idx])) where idx go from j to k. Expected solution is O(n^3).

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 #1504: Count Submatrices With All Ones
class Solution {
    public int numSubmat(int[][] mat) {
        int m = mat.length, n = mat[0].length;
        int[][] g = new int[m][n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (mat[i][j] == 1) {
                    g[i][j] = j == 0 ? 1 : 1 + g[i][j - 1];
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int col = 1 << 30;
                for (int k = i; k >= 0 && col > 0; --k) {
                    col = Math.min(col, g[k][j]);
                    ans += col;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1504: Count Submatrices With All Ones
func numSubmat(mat [][]int) (ans int) {
	m, n := len(mat), len(mat[0])
	g := make([][]int, m)
	for i := range g {
		g[i] = make([]int, n)
		for j := range g[i] {
			if mat[i][j] == 1 {
				if j == 0 {
					g[i][j] = 1
				} else {
					g[i][j] = 1 + g[i][j-1]
				}
			}
		}
	}
	for i := range g {
		for j := range g[i] {
			col := 1 << 30
			for k := i; k >= 0 && col > 0; k-- {
				col = min(col, g[k][j])
				ans += col
			}
		}
	}
	return
}

# Accepted solution for LeetCode #1504: Count Submatrices With All Ones
class Solution:
    def numSubmat(self, mat: List[List[int]]) -> int:
        m, n = len(mat), len(mat[0])
        g = [[0] * n for _ in range(m)]
        for i in range(m):
            for j in range(n):
                if mat[i][j]:
                    g[i][j] = 1 if j == 0 else 1 + g[i][j - 1]
        ans = 0
        for i in range(m):
            for j in range(n):
                col = inf
                for k in range(i, -1, -1):
                    col = min(col, g[k][j])
                    ans += col
        return ans

// Accepted solution for LeetCode #1504: Count Submatrices With All Ones
impl Solution {
    pub fn num_submat(mat: Vec<Vec<i32>>) -> i32 {
        let m = mat.len();
        let n = mat[0].len();
        let mut g = vec![vec![0; n]; m];

        for i in 0..m {
            for j in 0..n {
                if mat[i][j] == 1 {
                    if j == 0 {
                        g[i][j] = 1;
                    } else {
                        g[i][j] = 1 + g[i][j - 1];
                    }
                }
            }
        }

        let mut ans = 0;
        for i in 0..m {
            for j in 0..n {
                let mut col = i32::MAX;
                let mut k = i as i32;
                while k >= 0 && col > 0 {
                    col = col.min(g[k as usize][j]);
                    ans += col;
                    k -= 1;
                }
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #1504: Count Submatrices With All Ones
function numSubmat(mat: number[][]): number {
    const m = mat.length;
    const n = mat[0].length;
    const g: number[][] = Array.from({ length: m }, () => Array(n).fill(0));

    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            if (mat[i][j]) {
                g[i][j] = j === 0 ? 1 : 1 + g[i][j - 1];
            }
        }
    }

    let ans = 0;
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            let col = Infinity;
            for (let k = i; k >= 0; k--) {
                col = Math.min(col, g[k][j]);
                ans += col;
            }
        }
    }

    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^2 × n)

Space

O(m × n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.