How do you solve Maximum Strictly Increasing Cells in a Matrix on LeetCode?

Maximum Strictly Increasing Cells in a Matrix (LeetCode #2713) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n log(m * n)) time complexity and O(m * n) space complexity. Key insight: Sorting the cells allows us to efficiently compute paths based on previously computed results.

What is the time complexity of Maximum Strictly Increasing Cells in a Matrix?

The optimal solution for Maximum Strictly Increasing Cells in a Matrix runs in O(m * n log(m * n)) time and uses O(m * n) space. The time complexity is dominated by the sorting step, which is O(m * n log(m * n)), while the space complexity is O(m * n) due to the DP array.

What is the brute force solution for Maximum Strictly Increasing Cells in a Matrix?

The brute force approach for Maximum Strictly Increasing Cells in a Matrix is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves starting from each cell in the matrix and exploring all possible paths to count how many cells can be visited. This is straightforward but inefficient as it checks every possible starting point. The optimal Optimal Solution approach improves this to O(m * n log(m * n)).

What are the interview tips for Maximum Strictly Increasing Cells in a Matrix?

Clarify the movement rules and constraints before diving into coding. Think about edge cases, such as matrices with all equal values. Consider the efficiency of your approach and be ready to discuss optimizations.

What are common mistakes when solving Maximum Strictly Increasing Cells in a Matrix?

Not considering all possible paths from a cell, leading to incorrect counts. Failing to reset visited states in the brute force approach, causing incorrect results.

#2713

Maximum Strictly Increasing Cells in a Matrix

Hard

ArrayHash TableBinary SearchDynamic ProgrammingMemoizationSortingMatrixOrdered SetDynamic ProgrammingSortingGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(m * n log(m * n))Space O(m * n)

The optimal solution sorts the cells by their values and processes them in increasing order. This allows us to efficiently calculate the maximum path length using dynamic programming, leveraging previously computed results.

⚙️

Algorithm

3 steps

1Step 1: Create a list of all cells with their values and coordinates, then sort this list by values.
2Step 2: Initialize a DP array to store the maximum path length ending at each cell.
3Step 3: Iterate through the sorted list, and for each cell, check all cells that can be reached (those with greater values) and update the DP array accordingly.

solution.py20 lines

1# Full working Python code
2from typing import List
3
4def maxIncreasingCells(mat: List[List[int]]) -> int:
5    m, n = len(mat), len(mat[0])
6    cells = [(mat[i][j], i, j) for i in range(m) for j in range(n)]
7    cells.sort()
8    dp = [[1] * n for _ in range(m)]
9    max_count = 1
10
11    for value, x, y in cells:
12        for i in range(m):
13            if mat[i][y] > value:
14                dp[i][y] = max(dp[i][y], dp[x][y] + 1)
15        for j in range(n):
16            if mat[x][j] > value:
17                dp[x][j] = max(dp[x][j], dp[x][y] + 1)
18        max_count = max(max_count, dp[x][y])
19
20    return max_count

ℹ

Complexity note: The time complexity is dominated by the sorting step, which is O(m * n log(m * n)), while the space complexity is O(m * n) due to the DP array.

1Sorting the cells allows us to efficiently compute paths based on previously computed results.
2Dynamic programming helps to store intermediate results, avoiding redundant calculations.

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