LeetCode #2711 — MEDIUM

Difference of Number of Distinct Values on Diagonals

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

The Problem

Problem Statement

Given a 2D grid of size m x n, you should find the matrix answer of size m x n.

The cell answer[r][c] is calculated by looking at the diagonal values of the cell grid[r][c]:

Let leftAbove[r][c] be the number of distinct values on the diagonal to the left and above the cell grid[r][c] not including the cell grid[r][c] itself.
Let rightBelow[r][c] be the number of distinct values on the diagonal to the right and below the cell grid[r][c], not including the cell grid[r][c] itself.
Then answer[r][c] = |leftAbove[r][c] - rightBelow[r][c]|.

A matrix diagonal is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until the end of the matrix is reached.

For example, in the below diagram the diagonal is highlighted using the cell with indices (2, 3) colored gray:
- Red-colored cells are left and above the cell.
- Blue-colored cells are right and below the cell.

Return the matrix answer.

Example 1:

Input: grid = [[1,2,3],[3,1,5],[3,2,1]]

Output: Output: [[1,1,0],[1,0,1],[0,1,1]]

Explanation:

To calculate the answer cells:

answer	left-above elements	leftAbove	right-below elements	rightBelow	\|leftAbove - rightBelow\|
[0][0]	[]	0	[grid[1][1], grid[2][2]]	\|{1, 1}\| = 1	1
[0][1]	[]	0	[grid[1][2]]	\|{5}\| = 1	1
[0][2]	[]	0	[]	0	0
[1][0]	[]	0	[grid[2][1]]	\|{2}\| = 1	1
[1][1]	[grid[0][0]]	\|{1}\| = 1	[grid[2][2]]	\|{1}\| = 1	0
[1][2]	[grid[0][1]]	\|{2}\| = 1	[]	0	1
[2][0]	[]	0	[]	0	0
[2][1]	[grid[1][0]]	\|{3}\| = 1	[]	0	1
[2][2]	[grid[0][0], grid[1][1]]	\|{1, 1}\| = 1	[]	0	1

Example 2:

Input: grid = [[1]]

Output: Output: [[0]]

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n, grid[i][j] <= 50

Step 01

Brute Force Baseline

Problem summary: Given a 2D grid of size m x n, you should find the matrix answer of size m x n. The cell answer[r][c] is calculated by looking at the diagonal values of the cell grid[r][c]: Let leftAbove[r][c] be the number of distinct values on the diagonal to the left and above the cell grid[r][c] not including the cell grid[r][c] itself. Let rightBelow[r][c] be the number of distinct values on the diagonal to the right and below the cell grid[r][c], not including the cell grid[r][c] itself. Then answer[r][c] = |leftAbove[r][c] - rightBelow[r][c]|. A matrix diagonal is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until the end of the matrix is reached. For example, in the below diagram the diagonal is highlighted using the cell with indices (2, 3) colored gray: Red-colored cells are left and above the cell.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[[1,2,3],[3,1,5],[3,2,1]]

Example 2

[[1]]

Step 02

Core Insight

What unlocks the optimal approach

Use the set to count the number of distinct elements on diagonals.

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 #2711: Difference of Number of Distinct Values on Diagonals
class Solution {
    public int[][] differenceOfDistinctValues(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[][] ans = new int[m][n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int x = i, y = j;
                Set<Integer> s = new HashSet<>();
                while (x > 0 && y > 0) {
                    s.add(grid[--x][--y]);
                }
                int tl = s.size();
                x = i;
                y = j;
                s.clear();
                while (x < m - 1 && y < n - 1) {
                    s.add(grid[++x][++y]);
                }
                int br = s.size();
                ans[i][j] = Math.abs(tl - br);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2711: Difference of Number of Distinct Values on Diagonals
func differenceOfDistinctValues(grid [][]int) [][]int {
	m, n := len(grid), len(grid[0])
	ans := make([][]int, m)
	for i := range grid {
		ans[i] = make([]int, n)
		for j := range grid[i] {
			x, y := i, j
			s := map[int]bool{}
			for x > 0 && y > 0 {
				x, y = x-1, y-1
				s[grid[x][y]] = true
			}
			tl := len(s)
			x, y = i, j
			s = map[int]bool{}
			for x+1 < m && y+1 < n {
				x, y = x+1, y+1
				s[grid[x][y]] = true
			}
			br := len(s)
			ans[i][j] = abs(tl - br)
		}
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #2711: Difference of Number of Distinct Values on Diagonals
class Solution:
    def differenceOfDistinctValues(self, grid: List[List[int]]) -> List[List[int]]:
        m, n = len(grid), len(grid[0])
        ans = [[0] * n for _ in range(m)]
        for i in range(m):
            for j in range(n):
                x, y = i, j
                s = set()
                while x and y:
                    x, y = x - 1, y - 1
                    s.add(grid[x][y])
                tl = len(s)
                x, y = i, j
                s = set()
                while x + 1 < m and y + 1 < n:
                    x, y = x + 1, y + 1
                    s.add(grid[x][y])
                br = len(s)
                ans[i][j] = abs(tl - br)
        return ans

// Accepted solution for LeetCode #2711: Difference of Number of Distinct Values on Diagonals
/**
 * [2711] Difference of Number of Distinct Values on Diagonals
 */
pub struct Solution {}

// submission codes start here
use std::collections::HashMap;
use std::collections::HashSet;

impl Solution {
    pub fn difference_of_distinct_values(grid: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        let m = grid.len();
        let n = grid[0].len();
        let mut result = vec![vec![0; n]; m];

        Self::iterate_diagonal(&grid, &mut result, (0..m).rev().map(|i| (i, 0)));
        Self::iterate_diagonal(&grid, &mut result, (0..n).rev().map(|i| (0, i)));

        result
    }

    fn iterate_diagonal<T>(grid: &Vec<Vec<i32>>, result: &mut Vec<Vec<i32>>, iter: T)
    where
        T: Iterator<Item = (usize, usize)>,
    {
        let m = grid.len();
        let n = grid[0].len();
        let max_length = m.min(n);

        let mut length = 1;
        for (i, j) in iter {
            let mut left_set = HashSet::with_capacity(length);
            // 右下对角线因为涉及到添加和删除
            // 使用哈希表存储对应数字的出现次数
            let mut right_set = HashMap::with_capacity(length);

            // 首先遍历右下
            let (mut x, mut y) = (i + 1, j + 1);

            for _ in 1..length {
                let entry = right_set.entry(grid[x][y]).or_insert(0);
                *entry += 1;

                x += 1;
                y += 1;
            }

            // 然后开始遍历
            let (mut x, mut y) = (i, j);

            for k in 0..length {
                // 删除右下对角线
                if k != 0 {
                    let entry = right_set.get_mut(&grid[x][y]).unwrap();
                    *entry -= 1;
                    if *entry == 0 {
                        right_set.remove(&grid[x][y]);
                    }
                }

                result[x][y] = left_set.len().abs_diff(right_set.len()) as i32;

                // 插入左上对角线
                left_set.insert(grid[x][y]);
                x += 1;
                y += 1;
            }

            length = max_length.min(length + 1);
        }
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2711() {
        assert_eq!(
            vec![vec![1, 1, 0], vec![1, 0, 1], vec![0, 1, 1]],
            Solution::difference_of_distinct_values(vec![
                vec![1, 2, 3],
                vec![3, 1, 5],
                vec![3, 2, 1]
            ])
        );
        assert_eq!(
            vec![vec![0]],
            Solution::difference_of_distinct_values(vec![vec![1]])
        );
    }
}

// Accepted solution for LeetCode #2711: Difference of Number of Distinct Values on Diagonals
function differenceOfDistinctValues(grid: number[][]): number[][] {
    const m = grid.length;
    const n = grid[0].length;
    const ans: number[][] = Array(m)
        .fill(0)
        .map(() => Array(n).fill(0));
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            let [x, y] = [i, j];
            const s = new Set<number>();
            while (x && y) {
                s.add(grid[--x][--y]);
            }
            const tl = s.size;
            [x, y] = [i, j];
            s.clear();
            while (x + 1 < m && y + 1 < n) {
                s.add(grid[++x][++y]);
            }
            const br = s.size;
            ans[i][j] = Math.abs(tl - br);
        }
    }
    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 × \min(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.