How do you solve Find Valid Matrix Given Row and Column Sums on LeetCode?

Find Valid Matrix Given Row and Column Sums (LeetCode #1605) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The sum of row sums must equal the sum of column sums.

What is the brute force solution for Find Valid Matrix Given Row and Column Sums?

The brute force approach for Find Valid Matrix Given Row and Column Sums is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves filling the matrix by iterating through each cell and assigning values based on the current row and column sums. This is straightforward but can be inefficient as it may require multiple passes to satisfy all conditions. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Valid Matrix Given Row and Column Sums?

Find Valid Matrix Given Row and Column Sums uses the following concepts: Array, Greedy, Matrix. The recommended patterns to study are: Greedy, Matrix Manipulation.

What are the interview tips for Find Valid Matrix Given Row and Column Sums?

Clarify the problem requirements and constraints before coding. Think about edge cases, such as when row or column sums are zero. Explain your thought process while coding to demonstrate your understanding.

What are common mistakes when solving Find Valid Matrix Given Row and Column Sums?

Not checking if the row and column sums match initially. Overcomplicating the filling process instead of using a straightforward greedy approach.

#1605

Find Valid Matrix Given Row and Column Sums

Medium

ArrayGreedyMatrixGreedyMatrix Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach is essentially the same as the brute-force method, but it emphasizes the greedy nature of filling the matrix. By always placing the minimum of the remaining row and column sums, we ensure that we efficiently meet the requirements without unnecessary iterations.

⚙️

Algorithm

3 steps

1Step 1: Initialize a 2D matrix of size rowSum.length x colSum.length with all zeros.
2Step 2: Iterate through each cell in the matrix using two nested loops.
3Step 3: For each cell, assign the minimum of the current rowSum and colSum, then update both rowSum and colSum.

solution.py12 lines

1# Full working Python code
2
3def fillMatrix(rowSum, colSum):
4    m, n = len(rowSum), len(colSum)
5    matrix = [[0] * n for _ in range(m)]
6    for i in range(m):
7        for j in range(n):
8            value = min(rowSum[i], colSum[j])
9            matrix[i][j] = value
10            rowSum[i] -= value
11            colSum[j] -= value
12    return matrix

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but the approach is optimal in terms of efficiency since it directly satisfies the conditions without unnecessary checks. The space complexity is O(n) for storing the matrix.

1The sum of row sums must equal the sum of column sums.
2Greedily choosing the minimum of row and column sums ensures we meet the requirements efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.