#2713

Hard

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Maximum Strictly Increasing Cells in a Matrix

Find the maximum number of cells that can be visited in a matrix by following strictly increasing values from a starting cell.

array hash table binary search dynamic programming memoization

Problem Statement

Given a 1-indexed m x n integer matrix, you are tasked with finding the maximum number of cells you can visit in a strictly increasing path, starting from any cell. You can move to any cell in the same row or column, provided the value of the destination cell is strictly greater than the current cell. The goal is to determine the maximum number of cells that can be visited by starting from any cell in the matrix.

The matrix's dimensions are large, with values constrained between -105 and 105, and the number of cells can be as high as 105. Efficient algorithms are necessary to compute the longest possible path while maintaining optimal performance. You are encouraged to explore dynamic programming combined with hash lookups to solve this problem.

Examples

Example 1

Input: mat = [[3,1],[3,4]]

Output: 2

The image shows how we can visit 2 cells starting from row 1, column 2. It can be shown that we cannot visit more than 2 cells no matter where we start from, so the answer is 2.

Example 2

Input: mat = [[1,1],[1,1]]

Output: 1

Since the cells must be strictly increasing, we can only visit one cell in this example.

Example 3

Input: mat = [[3,1,6],[-9,5,7]]

Output: 4

The image above shows how we can visit 4 cells starting from row 2, column 1. It can be shown that we cannot visit more than 4 cells no matter where we start from, so the answer is 4.

Constraints

m == mat.length
n == mat[i].length
1 <= m, n <= 105
1 <= m * n <= 105
-105 <= mat[i][j] <= 105

Solution Approach

Dynamic Programming Approach

Use dynamic programming to store the maximum path length starting from each cell. By scanning the matrix in a bottom-up manner, starting from the smallest values and progressively moving to larger values, we ensure that when processing each cell, the results for its neighbors have already been computed. This approach helps in minimizing recomputation and avoids redundant checks.

Hash Table Lookup Optimization

To further optimize the solution, use a hash table to store the longest paths calculated for each cell. This allows quick lookups during the traversal process, preventing redundant calculations and speeding up the overall process. With this setup, each cell's neighboring cells are checked only when necessary.

Binary Search for Efficiency

For additional optimization, binary search can be applied in cases where matrix values are sorted or can be mapped to indices. This ensures that transitions between cells follow the strictly increasing path condition, making the approach more efficient for larger matrices by reducing the number of checks needed.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the matrix traversal method, dynamic programming optimization, and hash table lookups. The space complexity is determined by the size of the matrix and the additional data structures used for storing the longest paths for each cell. Given the constraints, the approach is designed to handle the problem within these limits efficiently.

What Interviewers Usually Probe

Understanding the importance of bottom-up dynamic programming for efficiently computing path lengths.
Demonstrating familiarity with hash table optimizations for matrix traversal.
Being able to handle large inputs by applying efficient traversal techniques such as binary search.

Common Pitfalls or Variants

Common pitfalls

Not leveraging dynamic programming effectively, resulting in redundant calculations.
Failing to implement hash lookups or forgetting to store already computed paths.
Not optimizing for large input sizes, leading to time limit exceeded errors.

Follow-up variants

Consider variations where you are limited to only vertical or horizontal moves.
Explore the case where you need to find the longest path from a specified starting point.
Handle matrices where values are guaranteed to be sorted, requiring additional optimizations.

FAQ

What is the main algorithmic pattern for the Maximum Strictly Increasing Cells in a Matrix problem?

The problem follows a bottom-up dynamic programming approach combined with hash lookups to compute the longest path efficiently.

How can I optimize my solution for large matrices in this problem?

By using dynamic programming and hash tables, and considering binary search where applicable, you can ensure that the solution scales efficiently for large matrices.

What data structure is recommended for storing already computed longest paths in the matrix?

A hash table is recommended for storing the computed longest paths, enabling quick lookups and avoiding redundant calculations.

What is the time complexity of this solution?

The time complexity depends on the matrix size and the traversal method used. The dynamic programming approach with hash lookups typically yields a time complexity of O(m * n), where m and n are the matrix dimensions.

Can binary search be applied in the Maximum Strictly Increasing Cells in a Matrix problem?

Yes, binary search can be used to optimize transitions between matrix cells, especially when matrix values are sorted or mapped to indices.

terminal

Solution

Solution 1: Sorting + Dynamic Programming

Based on the problem description, the value of the cells we move through in sequence must strictly increase. Therefore, we can use a hash table $g$ to record the positions of all cells corresponding to each value, and then traverse from the smallest to the largest value.

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class Solution:
    def maxIncreasingCells(self, mat: List[List[int]]) -> int:
        m, n = len(mat), len(mat[0])
        g = defaultdict(list)
        for i in range(m):
            for j in range(n):
                g[mat[i][j]].append((i, j))
        rowMax = [0] * m
        colMax = [0] * n
        ans = 0
        for _, pos in sorted(g.items()):
            mx = []
            for i, j in pos:
                mx.append(1 + max(rowMax[i], colMax[j]))
                ans = max(ans, mx[-1])
            for k, (i, j) in enumerate(pos):
                rowMax[i] = max(rowMax[i], mx[k])
                colMax[j] = max(colMax[j], mx[k])
        return ans

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