LeetCode #1253 — MEDIUM

Reconstruct a 2-Row Binary Matrix

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

Solve on LeetCode

The Problem

Problem Statement

Given the following details of a matrix with n columns and 2 rows :

The matrix is a binary matrix, which means each element in the matrix can be 0 or 1.
The sum of elements of the 0-th(upper) row is given as upper.
The sum of elements of the 1-st(lower) row is given as lower.
The sum of elements in the i-th column(0-indexed) is colsum[i], where colsum is given as an integer array with length n.

Your task is to reconstruct the matrix with upper, lower and colsum.

Return it as a 2-D integer array.

If there are more than one valid solution, any of them will be accepted.

If no valid solution exists, return an empty 2-D array.

Example 1:

Input: upper = 2, lower = 1, colsum = [1,1,1]
Output: [[1,1,0],[0,0,1]]
Explanation: [[1,0,1],[0,1,0]], and [[0,1,1],[1,0,0]] are also correct answers.

Example 2:

Input: upper = 2, lower = 3, colsum = [2,2,1,1]
Output: []

Example 3:

Input: upper = 5, lower = 5, colsum = [2,1,2,0,1,0,1,2,0,1]
Output: [[1,1,1,0,1,0,0,1,0,0],[1,0,1,0,0,0,1,1,0,1]]

Constraints:

1 <= colsum.length <= 10^5
0 <= upper, lower <= colsum.length
0 <= colsum[i] <= 2

Step 01

Brute Force Baseline

Problem summary: Given the following details of a matrix with n columns and 2 rows : The matrix is a binary matrix, which means each element in the matrix can be 0 or 1. The sum of elements of the 0-th(upper) row is given as upper. The sum of elements of the 1-st(lower) row is given as lower. The sum of elements in the i-th column(0-indexed) is colsum[i], where colsum is given as an integer array with length n. Your task is to reconstruct the matrix with upper, lower and colsum. Return it as a 2-D integer array. If there are more than one valid solution, any of them will be accepted. If no valid solution exists, return an empty 2-D array.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

2
1
[1,1,1]

Example 2

2
3
[2,2,1,1]

Example 3

5
5
[2,1,2,0,1,0,1,2,0,1]

Core Insight

What unlocks the optimal approach

You cannot do anything about colsum[i] = 2 case or colsum[i] = 0 case. Then you put colsum[i] = 1 case to the upper row until upper has reached. Then put the rest into lower row.
Fill 0 and 2 first, then fill 1 in the upper row or lower row in turn but be careful about exhausting permitted 1s in each row.

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 #1253: Reconstruct a 2-Row Binary Matrix
class Solution {
    public List<List<Integer>> reconstructMatrix(int upper, int lower, int[] colsum) {
        int n = colsum.length;
        List<Integer> first = new ArrayList<>();
        List<Integer> second = new ArrayList<>();
        for (int j = 0; j < n; ++j) {
            int a = 0, b = 0;
            if (colsum[j] == 2) {
                a = b = 1;
                upper--;
                lower--;
            } else if (colsum[j] == 1) {
                if (upper > lower) {
                    upper--;
                    a = 1;
                } else {
                    lower--;
                    b = 1;
                }
            }
            if (upper < 0 || lower < 0) {
                break;
            }
            first.add(a);
            second.add(b);
        }
        return upper == 0 && lower == 0 ? List.of(first, second) : List.of();
    }
}

// Accepted solution for LeetCode #1253: Reconstruct a 2-Row Binary Matrix
func reconstructMatrix(upper int, lower int, colsum []int) [][]int {
	n := len(colsum)
	ans := make([][]int, 2)
	for i := range ans {
		ans[i] = make([]int, n)
	}
	for j, v := range colsum {
		if v == 2 {
			ans[0][j], ans[1][j] = 1, 1
			upper--
			lower--
		}
		if v == 1 {
			if upper > lower {
				upper--
				ans[0][j] = 1
			} else {
				lower--
				ans[1][j] = 1
			}
		}
		if upper < 0 || lower < 0 {
			break
		}
	}
	if upper != 0 || lower != 0 {
		return [][]int{}
	}
	return ans
}

# Accepted solution for LeetCode #1253: Reconstruct a 2-Row Binary Matrix
class Solution:
    def reconstructMatrix(
        self, upper: int, lower: int, colsum: List[int]
    ) -> List[List[int]]:
        n = len(colsum)
        ans = [[0] * n for _ in range(2)]
        for j, v in enumerate(colsum):
            if v == 2:
                ans[0][j] = ans[1][j] = 1
                upper, lower = upper - 1, lower - 1
            if v == 1:
                if upper > lower:
                    upper -= 1
                    ans[0][j] = 1
                else:
                    lower -= 1
                    ans[1][j] = 1
            if upper < 0 or lower < 0:
                return []
        return ans if lower == upper == 0 else []

// Accepted solution for LeetCode #1253: Reconstruct a 2-Row Binary Matrix
struct Solution;

impl Solution {
    fn reconstruct_matrix(mut upper: i32, mut lower: i32, colsum: Vec<i32>) -> Vec<Vec<i32>> {
        let n = colsum.len();
        let mut res = vec![vec![0; n]; 2];
        for j in 0..n {
            match colsum[j] {
                2 => {
                    res[0][j] = 1;
                    res[1][j] = 1;
                    upper -= 1;
                    lower -= 1;
                }
                1 => {
                    if upper >= lower {
                        res[0][j] = 1;
                        upper -= 1;
                    } else {
                        res[1][j] = 1;
                        lower -= 1;
                    }
                }
                _ => {}
            }
        }
        if upper == 0 && lower == 0 {
            res
        } else {
            vec![]
        }
    }
}

#[test]
fn test() {
    let upper = 2;
    let lower = 1;
    let colsum = vec![1, 1, 1];
    let res = vec_vec_i32![[1, 1, 0], [0, 0, 1]];
    assert_eq!(Solution::reconstruct_matrix(upper, lower, colsum), res);
    let upper = 2;
    let lower = 3;
    let colsum = vec![2, 2, 1, 1];
    let res = vec_vec_i32![];
    assert_eq!(Solution::reconstruct_matrix(upper, lower, colsum), res);
    let upper = 5;
    let lower = 5;
    let colsum = vec![2, 1, 2, 0, 1, 0, 1, 2, 0, 1];
    let res = vec_vec_i32![
        [1, 1, 1, 0, 0, 0, 1, 1, 0, 0],
        [1, 0, 1, 0, 1, 0, 0, 1, 0, 1]
    ];
    assert_eq!(Solution::reconstruct_matrix(upper, lower, colsum), res);
}

// Accepted solution for LeetCode #1253: Reconstruct a 2-Row Binary Matrix
function reconstructMatrix(upper: number, lower: number, colsum: number[]): number[][] {
    const n = colsum.length;
    const ans: number[][] = Array(2)
        .fill(0)
        .map(() => Array(n).fill(0));
    for (let j = 0; j < n; ++j) {
        if (colsum[j] === 2) {
            ans[0][j] = ans[1][j] = 1;
            upper--;
            lower--;
        } else if (colsum[j] === 1) {
            if (upper > lower) {
                ans[0][j] = 1;
                upper--;
            } else {
                ans[1][j] = 1;
                lower--;
            }
        }
        if (upper < 0 || lower < 0) {
            break;
        }
    }
    return upper || lower ? [] : 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(n log n)

Space

O(1)

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.