LeetCode #1074 — HARD

Number of Submatrices That Sum to Target

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Given a matrix and a target, return the number of non-empty submatrices that sum to target.

A submatrix x1, y1, x2, y2 is the set of all cells matrix[x][y] with x1 <= x <= x2 and y1 <= y <= y2.

Two submatrices (x1, y1, x2, y2) and (x1', y1', x2', y2') are different if they have some coordinate that is different: for example, if x1 != x1'.

Example 1:

Input: matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
Output: 4
Explanation: The four 1x1 submatrices that only contain 0.

Example 2:

Input: matrix = [[1,-1],[-1,1]], target = 0
Output: 5
Explanation: The two 1x2 submatrices, plus the two 2x1 submatrices, plus the 2x2 submatrix.

Example 3:

Input: matrix = [[904]], target = 0
Output: 0

Constraints:

1 <= matrix.length <= 100
1 <= matrix[0].length <= 100
-1000 <= matrix[i][j] <= 1000
-10^8 <= target <= 10^8

Step 01

Brute Force Baseline

Problem summary: Given a matrix and a target, return the number of non-empty submatrices that sum to target. A submatrix x1, y1, x2, y2 is the set of all cells matrix[x][y] with x1 <= x <= x2 and y1 <= y <= y2. Two submatrices (x1, y1, x2, y2) and (x1', y1', x2', y2') are different if they have some coordinate that is different: for example, if x1 != x1'.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

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

Example 2

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

Example 3

[[904]]
0

Core Insight

What unlocks the optimal approach

Using a 2D prefix sum, we can query the sum of any submatrix in O(1) time. Now for each (r1, r2), we can find the largest sum of a submatrix that uses every row in [r1, r2] in linear time using a sliding window.

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

Largest constraint values

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 #1074: Number of Submatrices That Sum to Target
class Solution {
    public int numSubmatrixSumTarget(int[][] matrix, int target) {
        int m = matrix.length, n = matrix[0].length;
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            int[] col = new int[n];
            for (int j = i; j < m; ++j) {
                for (int k = 0; k < n; ++k) {
                    col[k] += matrix[j][k];
                }
                ans += f(col, target);
            }
        }
        return ans;
    }

    private int f(int[] nums, int target) {
        Map<Integer, Integer> d = new HashMap<>();
        d.put(0, 1);
        int s = 0, cnt = 0;
        for (int x : nums) {
            s += x;
            cnt += d.getOrDefault(s - target, 0);
            d.merge(s, 1, Integer::sum);
        }
        return cnt;
    }
}

// Accepted solution for LeetCode #1074: Number of Submatrices That Sum to Target
func numSubmatrixSumTarget(matrix [][]int, target int) (ans int) {
	m, n := len(matrix), len(matrix[0])
	for i := 0; i < m; i++ {
		col := make([]int, n)
		for j := i; j < m; j++ {
			for k := 0; k < n; k++ {
				col[k] += matrix[j][k]
			}
			ans += f(col, target)
		}
	}
	return
}

func f(nums []int, target int) (cnt int) {
	d := map[int]int{0: 1}
	s := 0
	for _, x := range nums {
		s += x
		if v, ok := d[s-target]; ok {
			cnt += v
		}
		d[s]++
	}
	return
}

# Accepted solution for LeetCode #1074: Number of Submatrices That Sum to Target
class Solution:
    def numSubmatrixSumTarget(self, matrix: List[List[int]], target: int) -> int:
        def f(nums: List[int]) -> int:
            d = defaultdict(int)
            d[0] = 1
            cnt = s = 0
            for x in nums:
                s += x
                cnt += d[s - target]
                d[s] += 1
            return cnt

        m, n = len(matrix), len(matrix[0])
        ans = 0
        for i in range(m):
            col = [0] * n
            for j in range(i, m):
                for k in range(n):
                    col[k] += matrix[j][k]
                ans += f(col)
        return ans

// Accepted solution for LeetCode #1074: Number of Submatrices That Sum to Target
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn num_submatrix_sum_target(matrix: Vec<Vec<i32>>, target: i32) -> i32 {
        let n = matrix.len();
        let m = matrix[0].len();
        let mut prefix = vec![vec![]; n];
        for i in 0..n {
            let mut prev = 0;
            for j in 0..m {
                prev += matrix[i][j];
                prefix[i].push(prev);
            }
        }
        let mut res = 0;
        for j1 in 0..m {
            for j2 in j1..m {
                let mut hm: HashMap<i32, usize> = HashMap::new();
                hm.insert(0, 1);
                let mut sum = 0;
                for i in 0..n {
                    let cur = if j1 == 0 {
                        prefix[i][j2]
                    } else {
                        prefix[i][j2] - prefix[i][j1 - 1]
                    };
                    sum += cur;
                    res += *hm.entry(sum - target).or_default();
                    *hm.entry(sum).or_default() += 1;
                }
            }
        }
        res as i32
    }
}

#[test]
fn test() {
    let matrix = vec_vec_i32![[0, 1, 0], [1, 1, 1], [0, 1, 0]];
    let target = 0;
    let res = 4;
    assert_eq!(Solution::num_submatrix_sum_target(matrix, target), res);
    let matrix = vec_vec_i32![[1, -1], [-1, 1]];
    let target = 0;
    let res = 5;
    assert_eq!(Solution::num_submatrix_sum_target(matrix, target), res);

    let matrix = vec_vec_i32![
        [0, 1, 1, 1, 0, 1],
        [0, 0, 0, 0, 0, 1],
        [0, 0, 1, 0, 0, 1],
        [1, 1, 0, 1, 1, 0],
        [1, 0, 0, 1, 0, 0]
    ];
    let target = 0;

    let res = 43;
    assert_eq!(Solution::num_submatrix_sum_target(matrix, target), res);
}

// Accepted solution for LeetCode #1074: Number of Submatrices That Sum to Target
function numSubmatrixSumTarget(matrix: number[][], target: number): number {
    const m = matrix.length;
    const n = matrix[0].length;
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        const col: number[] = new Array(n).fill(0);
        for (let j = i; j < m; ++j) {
            for (let k = 0; k < n; ++k) {
                col[k] += matrix[j][k];
            }
            ans += f(col, target);
        }
    }
    return ans;
}

function f(nums: number[], target: number): number {
    const d: Map<number, number> = new Map();
    d.set(0, 1);
    let cnt = 0;
    let s = 0;
    for (const x of nums) {
        s += x;
        if (d.has(s - target)) {
            cnt += d.get(s - target)!;
        }
        d.set(s, (d.get(s) || 0) + 1);
    }
    return cnt;
}

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(n)

Space

O(1)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.