LeetCode #3128 — MEDIUM

Right Triangles

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

The Problem

Problem Statement

You are given a 2D boolean matrix grid.

A collection of 3 elements of grid is a right triangle if one of its elements is in the same row with another element and in the same column with the third element. The 3 elements may not be next to each other.

Return an integer that is the number of right triangles that can be made with 3 elements of grid such that all of them have a value of 1.

Example 1:

0	1	0
0	1	1
0	1	0

0	1	0
0	1	1
0	1	0

0	1	0
0	1	1
0	1	0

Input: grid = [[0,1,0],[0,1,1],[0,1,0]]

Output: 2

Explanation:

There are two right triangles with elements of the value 1. Notice that the blue ones do not form a right triangle because the 3 elements are in the same column.

Example 2:

1	0	0
0	1	1
1	0	0

Input: grid = [[1,0,0,0],[0,1,0,1],[1,0,0,0]]

Output: 0

Explanation:

There are no right triangles with elements of the value 1. Notice that the blue ones do not form a right triangle.

Example 3:

1	0	1
1	0	0
1	0	0

1	0	1
1	0	0
1	0	0

Input: grid = [[1,0,1],[1,0,0],[1,0,0]]

Output: 2

Explanation:

There are two right triangles with elements of the value 1.

Constraints:

1 <= grid.length <= 1000
1 <= grid[i].length <= 1000
0 <= grid[i][j] <= 1

Step 01

Brute Force Baseline

Problem summary: You are given a 2D boolean matrix grid. A collection of 3 elements of grid is a right triangle if one of its elements is in the same row with another element and in the same column with the third element. The 3 elements may not be next to each other. Return an integer that is the number of right triangles that can be made with 3 elements of grid such that all of them have a value of 1.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math

Example 1

[[0,1,0],[0,1,1],[0,1,0]]

Example 2

[[1,0,0,0],[0,1,0,1],[1,0,0,0]]

Example 3

[[1,0,1],[1,0,0],[1,0,0]]

Step 02

Core Insight

What unlocks the optimal approach

If <code>grid[x][y]</code> is 1, it can form a right triangle with an element of <code>grid</code> with value 1 in the same row and an element of <code>grid</code> with value 1 in the same column.
So we just need to count the number of 1s in each row and column.
For each <code>x, y</code> with <code>grid[x][y] = 1</code> if there are <code>row[x]</code> 1s in the row <code>x</code> and <code>col[y]</code> 1s in column <code>y</code>, the answer should be added by <code>(row[x] - 1) * (col[y] - 1)</code>.

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 #3128: Right Triangles
class Solution {
    public long numberOfRightTriangles(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[] rows = new int[m];
        int[] cols = new int[n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                rows[i] += grid[i][j];
                cols[j] += grid[i][j];
            }
        }
        long ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    ans += (rows[i] - 1) * (cols[j] - 1);
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3128: Right Triangles
func numberOfRightTriangles(grid [][]int) (ans int64) {
	m, n := len(grid), len(grid[0])
	rows := make([]int, m)
	cols := make([]int, n)
	for i, row := range grid {
		for j, x := range row {
			rows[i] += x
			cols[j] += x
		}
	}
	for i, row := range grid {
		for j, x := range row {
			if x == 1 {
				ans += int64((rows[i] - 1) * (cols[j] - 1))
			}
		}
	}
	return
}

# Accepted solution for LeetCode #3128: Right Triangles
class Solution:
    def numberOfRightTriangles(self, grid: List[List[int]]) -> int:
        rows = [0] * len(grid)
        cols = [0] * len(grid[0])
        for i, row in enumerate(grid):
            for j, x in enumerate(row):
                rows[i] += x
                cols[j] += x
        ans = 0
        for i, row in enumerate(grid):
            for j, x in enumerate(row):
                if x:
                    ans += (rows[i] - 1) * (cols[j] - 1)
        return ans

// Accepted solution for LeetCode #3128: Right Triangles
// 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 #3128: Right Triangles
// class Solution {
//     public long numberOfRightTriangles(int[][] grid) {
//         int m = grid.length, n = grid[0].length;
//         int[] rows = new int[m];
//         int[] cols = new int[n];
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 rows[i] += grid[i][j];
//                 cols[j] += grid[i][j];
//             }
//         }
//         long ans = 0;
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 if (grid[i][j] == 1) {
//                     ans += (rows[i] - 1) * (cols[j] - 1);
//                 }
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3128: Right Triangles
function numberOfRightTriangles(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    const rows: number[] = Array(m).fill(0);
    const cols: number[] = Array(n).fill(0);
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            rows[i] += grid[i][j];
            cols[j] += grid[i][j];
        }
    }
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] === 1) {
                ans += (rows[i] - 1) * (cols[j] - 1);
            }
        }
    }
    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)

Space

O(m + n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.