LeetCode 2711 - Difference of Number of Distinct Values on Diagonals

The problem is asking us to compute a new matrix answer based on a given m x n grid, where each cell in answer represents the absolute difference between the number of distinct elements on the diagonal above and to the left of the current cell, and the number of distinct…

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LeetCode Problem 2711

Difficulty: 🟡 Medium
Topics: Array, Hash Table, Matrix

Solution

Problem Understanding

The problem is asking us to compute a new matrix answer based on a given m x n grid, where each cell in answer represents the absolute difference between the number of distinct elements on the diagonal above and to the left of the current cell, and the number of distinct elements on the diagonal below and to the right. Specifically, for a cell (r, c) in grid, leftAbove[r][c] counts all distinct values along the diagonal starting from cells in the top-left direction (not including (r, c)), and rightBelow[r][c] counts all distinct values along the diagonal extending to the bottom-right direction (again excluding (r, c)).

The input grid can be as small as 1 x 1 or as large as 50 x 50, and the cell values are guaranteed to be positive integers up to 50. The small constraint on m and n suggests that solutions with quadratic complexity in terms of the total number of cells are feasible.

Important edge cases include a grid with a single element, grids where all elements are identical, and grids with varying diagonal lengths. The problem guarantees a rectangular matrix with at least one row and column.

Approaches

Brute Force

A brute-force approach would iterate over every cell (r, c) and explicitly collect all distinct values along its top-left diagonal for leftAbove and along its bottom-right diagonal for rightBelow. For each cell, this requires scanning up to min(m, n) cells in each diagonal direction. While this would produce the correct result, the time complexity is high: for each of the m x n cells, we may iterate up to O(min(m, n)) elements twice. For a 50 x 50 grid, this is feasible but inefficient, with complexity roughly O(m * n * min(m, n)).

Optimal Approach

The key observation is that all cells along a diagonal share the same diagonal index d = r - c. Using this, we can traverse each diagonal from top-left to bottom-right once, maintaining a running set of distinct elements encountered so far for leftAbove, and then traverse in reverse for rightBelow. This allows computing distinct counts for all cells in a diagonal in linear time relative to the diagonal length, reducing redundant work.

Approach Time Complexity Space Complexity Notes
Brute Force O(m * n * min(m, n)) O(min(m, n)) Scans diagonals for every cell independently
Optimal O(m * n) O(m + n) Uses hash sets per diagonal and two linear passes

Algorithm Walkthrough

  1. Compute leftAbove matrix: iterate over each diagonal identified by r - c. For each diagonal, traverse cells from top-left to bottom-right, maintaining a set seen of elements encountered so far. For each cell, store the size of seen in leftAbove[r][c] and then add the current element to seen.
  2. Compute rightBelow matrix: iterate over each diagonal again but traverse from bottom-right to top-left. Maintain a separate set seen for the diagonal and update rightBelow[r][c] similarly, storing the number of distinct elements seen so far before adding the current element.
  3. Construct the answer matrix by taking the absolute difference between leftAbove[r][c] and rightBelow[r][c] for every cell.

Why it works: Each diagonal is uniquely identified by r - c, ensuring that every element along the diagonal is considered exactly once. By traversing in two directions, we separate the "leftAbove" and "rightBelow" contributions correctly. Using sets ensures only distinct elements are counted, matching the problem requirements.

Python Solution

from typing import List

class Solution:
    def differenceOfDistinctValues(self, grid: List[List[int]]) -> List[List[int]]:
        m, n = len(grid), len(grid[0])
        leftAbove = [[0] * n for _ in range(m)]
        rightBelow = [[0] * n for _ in range(m)]
        
        # Compute leftAbove
        diagonals = {}
        for r in range(m):
            for c in range(n):
                d = r - c
                if d not in diagonals:
                    diagonals[d] = set()
                leftAbove[r][c] = len(diagonals[d])
                diagonals[d].add(grid[r][c])
        
        # Compute rightBelow
        diagonals.clear()
        for r in reversed(range(m)):
            for c in reversed(range(n)):
                d = r - c
                if d not in diagonals:
                    diagonals[d] = set()
                rightBelow[r][c] = len(diagonals[d])
                diagonals[d].add(grid[r][c])
        
        # Construct answer
        answer = [[abs(leftAbove[r][c] - rightBelow[r][c]) for c in range(n)] for r in range(m)]
        return answer

This solution first initializes leftAbove and rightBelow matrices. It iterates diagonally by using the key r - c and uses sets to track distinct elements. Finally, it constructs the result using element-wise absolute differences.

Go Solution

func differenceOfDistinctValues(grid [][]int) [][]int {
    m, n := len(grid), len(grid[0])
    leftAbove := make([][]int, m)
    rightBelow := make([][]int, m)
    for i := 0; i < m; i++ {
        leftAbove[i] = make([]int, n)
        rightBelow[i] = make([]int, n)
    }

    diagonals := map[int]map[int]bool{}
    
    // Compute leftAbove
    for r := 0; r < m; r++ {
        for c := 0; c < n; c++ {
            d := r - c
            if diagonals[d] == nil {
                diagonals[d] = make(map[int]bool)
            }
            leftAbove[r][c] = len(diagonals[d])
            diagonals[d][grid[r][c]] = true
        }
    }

    // Compute rightBelow
    diagonals = map[int]map[int]bool{}
    for r := m - 1; r >= 0; r-- {
        for c := n - 1; c >= 0; c-- {
            d := r - c
            if diagonals[d] == nil {
                diagonals[d] = make(map[int]bool)
            }
            rightBelow[r][c] = len(diagonals[d])
            diagonals[d][grid[r][c]] = true
        }
    }

    // Construct answer
    answer := make([][]int, m)
    for r := 0; r < m; r++ {
        answer[r] = make([]int, n)
        for c := 0; c < n; c++ {
            diff := leftAbove[r][c] - rightBelow[r][c]
            if diff < 0 {
                diff = -diff
            }
            answer[r][c] = diff
        }
    }
    return answer
}

In Go, we use maps to simulate sets for tracking distinct elements on each diagonal. The reversed iteration handles the bottom-right calculation, and absolute differences are computed manually.

Worked Examples

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

Step 1: Compute leftAbove

Diagonal keys r - c:

  • d=0 -> (0,0),(1,1),(2,2)
  • d=-1 -> (0,1),(1,2)
  • d=-2 -> (0,2)
  • d=1 -> (1,0),(2,1)
  • d=2 -> (2,0)

Tracking sets along each diagonal:

Cell leftAbove set Count
(0,0) {} 0
(1,1) {1} 1
(2,2) {1,1} -> {1} 1
(0,1) {} 0
(1,2) {3} 1
(0,2) {} 0
(1,0) {} 0
(2,1) {3} 1
(2,0) {} 0

Step 2: Compute rightBelow (reverse traversal along diagonals)

  • Similarly, compute counts for each cell.

Step 3: Compute answer

  • answer[r][c] = |leftAbove[r][c] - rightBelow[r][c]|

Final matrix: [[1,1,0],[1,0,1],[0,1,1]]

Example 2: grid = [[1]]

  • Single element, no diagonals exist.
  • leftAbove = [[0]], rightBelow = [[0]], answer = [[0]]

Complexity Analysis

Measure Complexity Explanation
Time O(m * n) Each cell is visited twice, once for leftAbove and once for rightBelow, with constant-time set operations for at most 50 elements
Space O(m + n) Sets per diagonal, at most one set per diagonal,