LeetCode #2713 — HARD

Maximum Strictly Increasing Cells in a Matrix

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

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. You can repeat this process as many times as possible, moving from cell to cell until you can no longer make any moves.

Your task is to find the maximum number of cells that you can visit in the matrix by starting from some cell.

Return an integer denoting the maximum number of cells that can be visited.

Example 1:

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.

Example 2:

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

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: 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. You can repeat this process as many times as possible, moving from cell to cell until you can no longer make any moves. Your task is to find the maximum number of cells that you can visit in the matrix by starting from some cell. Return an integer denoting the maximum number of cells that can be visited.

Baseline thinking

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

Pattern signal: Array · Hash Map · Binary Search · Dynamic Programming · Segment Tree

Example 1

[[3,1],[3,4]]

Example 2

[[1,1],[1,1]]

Example 3

[[3,1,6],[-9,5,7]]

Related Problems

Number of Increasing Paths in a Grid (number-of-increasing-paths-in-a-grid)

Step 02

Core Insight

What unlocks the optimal approach

We can try to build the answer in a bottom-up fashion, starting from the smallest values and increasing to the larger values.
Going through the values in sorted order, we can store the maximum path we have seen so far for a row/column.
When we are at a cell, we check its row and column to find out the best previous smaller value that we’ve got so far, and we use it to increment the current value of the row and column.

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

Largest constraint values

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 #2713: Maximum Strictly Increasing Cells in a Matrix
class Solution {
    public int maxIncreasingCells(int[][] mat) {
        int m = mat.length, n = mat[0].length;
        TreeMap<Integer, List<int[]>> g = new TreeMap<>();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                g.computeIfAbsent(mat[i][j], k -> new ArrayList<>()).add(new int[] {i, j});
            }
        }
        int[] rowMax = new int[m];
        int[] colMax = new int[n];
        int ans = 0;
        for (var e : g.entrySet()) {
            var pos = e.getValue();
            int[] mx = new int[pos.size()];
            int k = 0;
            for (var p : pos) {
                int i = p[0], j = p[1];
                mx[k] = Math.max(rowMax[i], colMax[j]) + 1;
                ans = Math.max(ans, mx[k++]);
            }
            for (k = 0; k < mx.length; ++k) {
                int i = pos.get(k)[0], j = pos.get(k)[1];
                rowMax[i] = Math.max(rowMax[i], mx[k]);
                colMax[j] = Math.max(colMax[j], mx[k]);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2713: Maximum Strictly Increasing Cells in a Matrix
func maxIncreasingCells(mat [][]int) (ans int) {
	m, n := len(mat), len(mat[0])
	g := map[int][][2]int{}
	for i, row := range mat {
		for j, v := range row {
			g[v] = append(g[v], [2]int{i, j})
		}
	}
	nums := make([]int, 0, len(g))
	for k := range g {
		nums = append(nums, k)
	}
	sort.Ints(nums)
	rowMax := make([]int, m)
	colMax := make([]int, n)
	for _, k := range nums {
		pos := g[k]
		mx := make([]int, len(pos))
		for i, p := range pos {
			mx[i] = max(rowMax[p[0]], colMax[p[1]]) + 1
			ans = max(ans, mx[i])
		}
		for i, p := range pos {
			rowMax[p[0]] = max(rowMax[p[0]], mx[i])
			colMax[p[1]] = max(colMax[p[1]], mx[i])
		}
	}
	return
}

# Accepted solution for LeetCode #2713: Maximum Strictly Increasing Cells in a Matrix
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

// Accepted solution for LeetCode #2713: Maximum Strictly Increasing Cells in a Matrix
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2713: Maximum Strictly Increasing Cells in a Matrix
// class Solution {
//     public int maxIncreasingCells(int[][] mat) {
//         int m = mat.length, n = mat[0].length;
//         TreeMap<Integer, List<int[]>> g = new TreeMap<>();
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 g.computeIfAbsent(mat[i][j], k -> new ArrayList<>()).add(new int[] {i, j});
//             }
//         }
//         int[] rowMax = new int[m];
//         int[] colMax = new int[n];
//         int ans = 0;
//         for (var e : g.entrySet()) {
//             var pos = e.getValue();
//             int[] mx = new int[pos.size()];
//             int k = 0;
//             for (var p : pos) {
//                 int i = p[0], j = p[1];
//                 mx[k] = Math.max(rowMax[i], colMax[j]) + 1;
//                 ans = Math.max(ans, mx[k++]);
//             }
//             for (k = 0; k < mx.length; ++k) {
//                 int i = pos.get(k)[0], j = pos.get(k)[1];
//                 rowMax[i] = Math.max(rowMax[i], mx[k]);
//                 colMax[j] = Math.max(colMax[j], mx[k]);
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2713: Maximum Strictly Increasing Cells in a Matrix
function maxIncreasingCells(mat: number[][]): number {
    const m = mat.length;
    const n = mat[0].length;
    const g: { [key: number]: [number, number][] } = {};

    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            if (!g[mat[i][j]]) {
                g[mat[i][j]] = [];
            }
            g[mat[i][j]].push([i, j]);
        }
    }

    const rowMax = Array(m).fill(0);
    const colMax = Array(n).fill(0);
    let ans = 0;

    const sortedKeys = Object.keys(g)
        .map(Number)
        .sort((a, b) => a - b);

    for (const key of sortedKeys) {
        const pos = g[key];
        const mx: number[] = [];

        for (const [i, j] of pos) {
            mx.push(1 + Math.max(rowMax[i], colMax[j]));
            ans = Math.max(ans, mx[mx.length - 1]);
        }

        for (let k = 0; k < pos.length; k++) {
            const [i, j] = pos[k];
            rowMax[i] = Math.max(rowMax[i], mx[k]);
            colMax[j] = Math.max(colMax[j], mx[k]);
        }
    }

    return 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(m × n × log(m × n)

Space

O(m × n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.