LeetCode #2906 — MEDIUM

Construct Product Matrix

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

The Problem

Problem Statement

Given a 0-indexed 2D integer matrix grid of size n * m, we define a 0-indexed 2D matrix p of size n * m as the product matrix of grid if the following condition is met:

Each element p[i][j] is calculated as the product of all elements in grid except for the element grid[i][j]. This product is then taken modulo 12345.

Return the product matrix of grid.

Example 1:

Input: grid = [[1,2],[3,4]]
Output: [[24,12],[8,6]]
Explanation: p[0][0] = grid[0][1] * grid[1][0] * grid[1][1] = 2 * 3 * 4 = 24
p[0][1] = grid[0][0] * grid[1][0] * grid[1][1] = 1 * 3 * 4 = 12
p[1][0] = grid[0][0] * grid[0][1] * grid[1][1] = 1 * 2 * 4 = 8
p[1][1] = grid[0][0] * grid[0][1] * grid[1][0] = 1 * 2 * 3 = 6
So the answer is [[24,12],[8,6]].

Example 2:

Input: grid = [[12345],[2],[1]]
Output: [[2],[0],[0]]
Explanation: p[0][0] = grid[0][1] * grid[0][2] = 2 * 1 = 2.
p[0][1] = grid[0][0] * grid[0][2] = 12345 * 1 = 12345. 12345 % 12345 = 0. So p[0][1] = 0.
p[0][2] = grid[0][0] * grid[0][1] = 12345 * 2 = 24690. 24690 % 12345 = 0. So p[0][2] = 0.
So the answer is [[2],[0],[0]].

Constraints:

1 <= n == grid.length <= 10⁵
1 <= m == grid[i].length <= 10⁵
2 <= n * m <= 10⁵
1 <= grid[i][j] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: Given a 0-indexed 2D integer matrix grid of size n * m, we define a 0-indexed 2D matrix p of size n * m as the product matrix of grid if the following condition is met: Each element p[i][j] is calculated as the product of all elements in grid except for the element grid[i][j]. This product is then taken modulo 12345. Return the product matrix of grid.

Baseline thinking

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

Pattern signal: Array

Example 1

[[1,2],[3,4]]

Example 2

[[12345],[2],[1]]

Core Insight

What unlocks the optimal approach

Try to solve this without using the <code>'/'</code> (division operation).
Create two 2D arrays for <b>suffix</b> and <b>prefix</b> product, and use them to find the product for each position.

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 #2906: Construct Product Matrix
class Solution {
    public int[][] constructProductMatrix(int[][] grid) {
        final int mod = 12345;
        int n = grid.length, m = grid[0].length;
        int[][] p = new int[n][m];
        long suf = 1;
        for (int i = n - 1; i >= 0; --i) {
            for (int j = m - 1; j >= 0; --j) {
                p[i][j] = (int) suf;
                suf = suf * grid[i][j] % mod;
            }
        }
        long pre = 1;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                p[i][j] = (int) (p[i][j] * pre % mod);
                pre = pre * grid[i][j] % mod;
            }
        }
        return p;
    }
}

// Accepted solution for LeetCode #2906: Construct Product Matrix
func constructProductMatrix(grid [][]int) [][]int {
	const mod int = 12345
	n, m := len(grid), len(grid[0])
	p := make([][]int, n)
	for i := range p {
		p[i] = make([]int, m)
	}
	suf := 1
	for i := n - 1; i >= 0; i-- {
		for j := m - 1; j >= 0; j-- {
			p[i][j] = suf
			suf = suf * grid[i][j] % mod
		}
	}
	pre := 1
	for i := 0; i < n; i++ {
		for j := 0; j < m; j++ {
			p[i][j] = p[i][j] * pre % mod
			pre = pre * grid[i][j] % mod
		}
	}
	return p
}

# Accepted solution for LeetCode #2906: Construct Product Matrix
class Solution:
    def constructProductMatrix(self, grid: List[List[int]]) -> List[List[int]]:
        n, m = len(grid), len(grid[0])
        p = [[0] * m for _ in range(n)]
        mod = 12345
        suf = 1
        for i in range(n - 1, -1, -1):
            for j in range(m - 1, -1, -1):
                p[i][j] = suf
                suf = suf * grid[i][j] % mod
        pre = 1
        for i in range(n):
            for j in range(m):
                p[i][j] = p[i][j] * pre % mod
                pre = pre * grid[i][j] % mod
        return p

// Accepted solution for LeetCode #2906: Construct Product Matrix
impl Solution {
    pub fn construct_product_matrix(grid: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        let modulo: i32 = 12345;
        let n = grid.len();
        let m = grid[0].len();
        let mut p: Vec<Vec<i32>> = vec![vec![0; m]; n];
        let mut suf = 1;

        for i in (0..n).rev() {
            for j in (0..m).rev() {
                p[i][j] = suf;
                suf = (((suf as i64) * (grid[i][j] as i64)) % (modulo as i64)) as i32;
            }
        }

        let mut pre = 1;

        for i in 0..n {
            for j in 0..m {
                p[i][j] = (((p[i][j] as i64) * (pre as i64)) % (modulo as i64)) as i32;
                pre = (((pre as i64) * (grid[i][j] as i64)) % (modulo as i64)) as i32;
            }
        }

        p
    }
}

// Accepted solution for LeetCode #2906: Construct Product Matrix
function constructProductMatrix(grid: number[][]): number[][] {
    const mod = 12345;
    const [n, m] = [grid.length, grid[0].length];
    const p: number[][] = Array.from({ length: n }, () => Array.from({ length: m }, () => 0));
    let suf = 1;
    for (let i = n - 1; ~i; --i) {
        for (let j = m - 1; ~j; --j) {
            p[i][j] = suf;
            suf = (suf * grid[i][j]) % mod;
        }
    }
    let pre = 1;
    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < m; ++j) {
            p[i][j] = (p[i][j] * pre) % mod;
            pre = (pre * grid[i][j]) % mod;
        }
    }
    return p;
}

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.