LeetCode #1605 — MEDIUM

Find Valid Matrix Given Row and Column Sums

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

The Problem

Problem Statement

You are given two arrays rowSum and colSum of non-negative integers where rowSum[i] is the sum of the elements in the i^th row and colSum[j] is the sum of the elements of the j^th column of a 2D matrix. In other words, you do not know the elements of the matrix, but you do know the sums of each row and column.

Find any matrix of non-negative integers of size rowSum.length x colSum.length that satisfies the rowSum and colSum requirements.

Return a 2D array representing any matrix that fulfills the requirements. It's guaranteed that at least one matrix that fulfills the requirements exists.

Example 1:

Input: rowSum = [3,8], colSum = [4,7]
Output: [[3,0],
         [1,7]]
Explanation: 
0^th row: 3 + 0 = 3 == rowSum[0]
1^st row: 1 + 7 = 8 == rowSum[1]
0^th column: 3 + 1 = 4 == colSum[0]
1^st column: 0 + 7 = 7 == colSum[1]
The row and column sums match, and all matrix elements are non-negative.
Another possible matrix is: [[1,2],
                             [3,5]]

Example 2:

Input: rowSum = [5,7,10], colSum = [8,6,8]
Output: [[0,5,0],
         [6,1,0],
         [2,0,8]]

Constraints:

1 <= rowSum.length, colSum.length <= 500
0 <= rowSum[i], colSum[i] <= 10⁸
sum(rowSum) == sum(colSum)

Step 01

Brute Force Baseline

Problem summary: You are given two arrays rowSum and colSum of non-negative integers where rowSum[i] is the sum of the elements in the ith row and colSum[j] is the sum of the elements of the jth column of a 2D matrix. In other words, you do not know the elements of the matrix, but you do know the sums of each row and column. Find any matrix of non-negative integers of size rowSum.length x colSum.length that satisfies the rowSum and colSum requirements. Return a 2D array representing any matrix that fulfills the requirements. It's guaranteed that at least one matrix that fulfills the requirements exists.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[3,8]
[4,7]

Example 2

[5,7,10]
[8,6,8]

Core Insight

What unlocks the optimal approach

Find the smallest rowSum or colSum, and let it be x. Place that number in the grid, and subtract x from rowSum and colSum. Continue until all the sums are satisfied.

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 #1605: Find Valid Matrix Given Row and Column Sums
class Solution {
    public int[][] restoreMatrix(int[] rowSum, int[] colSum) {
        int m = rowSum.length;
        int n = colSum.length;
        int[][] ans = new int[m][n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int x = Math.min(rowSum[i], colSum[j]);
                ans[i][j] = x;
                rowSum[i] -= x;
                colSum[j] -= x;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1605: Find Valid Matrix Given Row and Column Sums
func restoreMatrix(rowSum []int, colSum []int) [][]int {
	m, n := len(rowSum), len(colSum)
	ans := make([][]int, m)
	for i := range ans {
		ans[i] = make([]int, n)
	}
	for i := range rowSum {
		for j := range colSum {
			x := min(rowSum[i], colSum[j])
			ans[i][j] = x
			rowSum[i] -= x
			colSum[j] -= x
		}
	}
	return ans
}

# Accepted solution for LeetCode #1605: Find Valid Matrix Given Row and Column Sums
class Solution:
    def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
        m, n = len(rowSum), len(colSum)
        ans = [[0] * n for _ in range(m)]
        for i in range(m):
            for j in range(n):
                x = min(rowSum[i], colSum[j])
                ans[i][j] = x
                rowSum[i] -= x
                colSum[j] -= x
        return ans

// Accepted solution for LeetCode #1605: Find Valid Matrix Given Row and Column Sums
struct Solution;

impl Solution {
    fn restore_matrix(mut row_sum: Vec<i32>, mut col_sum: Vec<i32>) -> Vec<Vec<i32>> {
        let n = row_sum.len();
        let m = col_sum.len();
        let mut i = 0;
        let mut j = 0;
        let mut res = vec![vec![0; m]; n];
        while i < n && j < m {
            let x = row_sum[i].min(col_sum[j]);
            row_sum[i] -= x;
            col_sum[j] -= x;
            res[i][j] = x;
            if row_sum[i] == 0 {
                i += 1;
            } else {
                j += 1;
            }
        }
        res
    }
}

#[test]
fn test() {
    let row_sum = vec![3, 8];
    let col_sum = vec![4, 7];
    let res = vec_vec_i32![[3, 0], [1, 7]];
    assert_eq!(Solution::restore_matrix(row_sum, col_sum), res);
}

// Accepted solution for LeetCode #1605: Find Valid Matrix Given Row and Column Sums
function restoreMatrix(rowSum: number[], colSum: number[]): number[][] {
    const m = rowSum.length;
    const n = colSum.length;
    const ans = Array.from(new Array(m), () => new Array(n).fill(0));
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            const x = Math.min(rowSum[i], colSum[j]);
            ans[i][j] = x;
            rowSum[i] -= x;
            colSum[j] -= x;
        }
    }
    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

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.