Leetcode #2713: Maximum Strictly Increasing Cells in a Matrix
In this guide, we solve Leetcode #2713 Maximum Strictly Increasing Cells in a Matrix in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a 1-indexed m x n integer matrix mat, you can select any cell in the matrix as your starting cell. From the starting cell, you can move to any other cell in the same row or column, but only if the value of the destination cell is strictly greater than the value of the current cell.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Memoization, Array, Hash Table, Binary Search, Dynamic Programming, Matrix, Ordered Set, Sorting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: mat = [[3,1],[3,4]]
Output: 2
Explanation: 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.
Python Solution
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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.