LeetCode #764 — MEDIUM

Largest Plus Sign

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer n. You have an n x n binary grid grid with all values initially 1's except for some indices given in the array mines. The i^th element of the array mines is defined as mines[i] = [x_i, y_i] where grid[x_i][y_i] == 0.

Return the order of the largest axis-aligned plus sign of 1's contained in grid. If there is none, return 0.

An axis-aligned plus sign of 1's of order k has some center grid[r][c] == 1 along with four arms of length k - 1 going up, down, left, and right, and made of 1's. Note that there could be 0's or 1's beyond the arms of the plus sign, only the relevant area of the plus sign is checked for 1's.

Example 1:

Input: n = 5, mines = [[4,2]]
Output: 2
Explanation: In the above grid, the largest plus sign can only be of order 2. One of them is shown.

Example 2:

Input: n = 1, mines = [[0,0]]
Output: 0
Explanation: There is no plus sign, so return 0.

Constraints:

1 <= n <= 500
1 <= mines.length <= 5000
0 <= x_i, y_i < n
All the pairs (x_i, y_i) are unique.

Step 01

Brute Force Baseline

Problem summary: You are given an integer n. You have an n x n binary grid grid with all values initially 1's except for some indices given in the array mines. The ith element of the array mines is defined as mines[i] = [xi, yi] where grid[xi][yi] == 0. Return the order of the largest axis-aligned plus sign of 1's contained in grid. If there is none, return 0. An axis-aligned plus sign of 1's of order k has some center grid[r][c] == 1 along with four arms of length k - 1 going up, down, left, and right, and made of 1's. Note that there could be 0's or 1's beyond the arms of the plus sign, only the relevant area of the plus sign is checked for 1's.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

5
[[4,2]]

Example 2

1
[[0,0]]

Core Insight

What unlocks the optimal approach

For each direction such as "left", find left[r][c] = the number of 1s you will see before a zero starting at r, c and walking left. You can find this in N^2 time with a dp. The largest plus sign at r, c is just the minimum of left[r][c], up[r][c] etc.

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

Upper-end input sizes

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 #764: Largest Plus Sign
class Solution {
    public int orderOfLargestPlusSign(int n, int[][] mines) {
        int[][] dp = new int[n][n];
        for (var e : dp) {
            Arrays.fill(e, n);
        }
        for (var e : mines) {
            dp[e[0]][e[1]] = 0;
        }
        for (int i = 0; i < n; ++i) {
            int left = 0, right = 0, up = 0, down = 0;
            for (int j = 0, k = n - 1; j < n; ++j, --k) {
                left = dp[i][j] > 0 ? left + 1 : 0;
                right = dp[i][k] > 0 ? right + 1 : 0;
                up = dp[j][i] > 0 ? up + 1 : 0;
                down = dp[k][i] > 0 ? down + 1 : 0;
                dp[i][j] = Math.min(dp[i][j], left);
                dp[i][k] = Math.min(dp[i][k], right);
                dp[j][i] = Math.min(dp[j][i], up);
                dp[k][i] = Math.min(dp[k][i], down);
            }
        }
        return Arrays.stream(dp).flatMapToInt(Arrays::stream).max().getAsInt();
    }
}

// Accepted solution for LeetCode #764: Largest Plus Sign
func orderOfLargestPlusSign(n int, mines [][]int) (ans int) {
	dp := make([][]int, n)
	for i := range dp {
		dp[i] = make([]int, n)
		for j := range dp[i] {
			dp[i][j] = n
		}
	}
	for _, e := range mines {
		dp[e[0]][e[1]] = 0
	}
	for i := 0; i < n; i++ {
		var left, right, up, down int
		for j, k := 0, n-1; j < n; j, k = j+1, k-1 {
			left, right, up, down = left+1, right+1, up+1, down+1
			if dp[i][j] == 0 {
				left = 0
			}
			if dp[i][k] == 0 {
				right = 0
			}
			if dp[j][i] == 0 {
				up = 0
			}
			if dp[k][i] == 0 {
				down = 0
			}
			dp[i][j] = min(dp[i][j], left)
			dp[i][k] = min(dp[i][k], right)
			dp[j][i] = min(dp[j][i], up)
			dp[k][i] = min(dp[k][i], down)
		}
	}
	for _, e := range dp {
		ans = max(ans, slices.Max(e))
	}
	return
}

# Accepted solution for LeetCode #764: Largest Plus Sign
class Solution:
    def orderOfLargestPlusSign(self, n: int, mines: List[List[int]]) -> int:
        dp = [[n] * n for _ in range(n)]
        for x, y in mines:
            dp[x][y] = 0
        for i in range(n):
            left = right = up = down = 0
            for j, k in zip(range(n), reversed(range(n))):
                left = left + 1 if dp[i][j] else 0
                right = right + 1 if dp[i][k] else 0
                up = up + 1 if dp[j][i] else 0
                down = down + 1 if dp[k][i] else 0
                dp[i][j] = min(dp[i][j], left)
                dp[i][k] = min(dp[i][k], right)
                dp[j][i] = min(dp[j][i], up)
                dp[k][i] = min(dp[k][i], down)
        return max(max(v) for v in dp)

// Accepted solution for LeetCode #764: Largest Plus Sign
struct Solution;

impl Solution {
    fn order_of_largest_plus_sign(n: i32, mines: Vec<Vec<i32>>) -> i32 {
        let n = n as usize;
        let mut grid = vec![vec![1; n]; n];
        let mut left = vec![vec![0; n]; n];
        let mut top = vec![vec![0; n]; n];
        let mut right = vec![vec![0; n]; n];
        let mut bottom = vec![vec![0; n]; n];
        for mine in mines {
            let i = mine[0] as usize;
            let j = mine[1] as usize;
            grid[i][j] = 0;
        }
        for i in 0..n {
            for j in 0..n {
                if grid[i][j] == 1 {
                    if j > 0 {
                        left[i][j] = left[i][j - 1] + 1;
                    } else {
                        left[i][j] = 1;
                    }
                }
            }
        }
        for j in 0..n {
            for i in 0..n {
                if grid[i][j] == 1 {
                    if i > 0 {
                        top[i][j] = top[i - 1][j] + 1;
                    } else {
                        top[i][j] = 1;
                    }
                }
            }
        }
        for i in 0..n {
            for j in (0..n).rev() {
                if grid[i][j] == 1 {
                    if j + 1 < n {
                        right[i][j] = right[i][j + 1] + 1;
                    } else {
                        right[i][j] = 1;
                    }
                }
            }
        }
        for j in 0..n {
            for i in (0..n).rev() {
                if grid[i][j] == 1 {
                    if i + 1 < n {
                        bottom[i][j] = bottom[i + 1][j] + 1;
                    } else {
                        bottom[i][j] = 1;
                    }
                }
            }
        }
        let mut res = 0;
        for i in 0..n {
            for j in 0..n {
                let mut min = n;
                min = min.min(left[i][j]);
                min = min.min(right[i][j]);
                min = min.min(top[i][j]);
                min = min.min(bottom[i][j]);
                res = res.max(min);
            }
        }
        res as i32
    }
}

#[test]
fn test() {
    let n = 5;
    let mines = vec_vec_i32![[4, 2]];
    let res = 2;
    assert_eq!(Solution::order_of_largest_plus_sign(n, mines), res);
    let n = 2;
    let mines = vec_vec_i32![];
    let res = 1;
    assert_eq!(Solution::order_of_largest_plus_sign(n, mines), res);
    let n = 1;
    let mines = vec_vec_i32![[0, 0]];
    let res = 0;
    assert_eq!(Solution::order_of_largest_plus_sign(n, mines), res);
}

// Accepted solution for LeetCode #764: Largest Plus Sign
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #764: Largest Plus Sign
// class Solution {
//     public int orderOfLargestPlusSign(int n, int[][] mines) {
//         int[][] dp = new int[n][n];
//         for (var e : dp) {
//             Arrays.fill(e, n);
//         }
//         for (var e : mines) {
//             dp[e[0]][e[1]] = 0;
//         }
//         for (int i = 0; i < n; ++i) {
//             int left = 0, right = 0, up = 0, down = 0;
//             for (int j = 0, k = n - 1; j < n; ++j, --k) {
//                 left = dp[i][j] > 0 ? left + 1 : 0;
//                 right = dp[i][k] > 0 ? right + 1 : 0;
//                 up = dp[j][i] > 0 ? up + 1 : 0;
//                 down = dp[k][i] > 0 ? down + 1 : 0;
//                 dp[i][j] = Math.min(dp[i][j], left);
//                 dp[i][k] = Math.min(dp[i][k], right);
//                 dp[j][i] = Math.min(dp[j][i], up);
//                 dp[k][i] = Math.min(dp[k][i], down);
//             }
//         }
//         return Arrays.stream(dp).flatMapToInt(Arrays::stream).max().getAsInt();
//     }
// }

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(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

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.