Difference of Number of Distinct Values on Diagonals solution and explanation
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The problem
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 cellgrid[r][c]not including the cellgrid[r][c]itself. - Let
rightBelow[r][c]be the number of distinct values on the diagonal to the right and below the cellgrid[r][c], not including the cellgrid[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.lengthn == grid[i].length1 <= m, n, grid[i][j] <= 50
Approach overview
Problem Overview: You are given an m x n grid. For every cell (i, j), compute the absolute difference between the number of distinct values on the top-left diagonal before that cell and the number of distinct values on the bottom-right diagonal after it. The result is stored in a matrix of the same size.
Approach 1: Diagonal Traversal with Precomputation (O(m*n) time, O(m*n) space)
Each cell belongs to exactly one top-left → bottom-right diagonal. Traverse every diagonal and compute two pieces of information: the count of distinct values seen before each position and the count of distinct values after it. Use a hash set while walking the diagonal from top-left to bottom-right to track distinct values encountered so far, storing prefix counts. Then walk the same diagonal in reverse to collect suffix distinct counts. The result for each cell is abs(prefixDistinct - suffixDistinct). This approach processes each element a constant number of times, giving O(m*n) time. Extra arrays for prefix/suffix information lead to O(m*n) space.
Diagonal traversal is a common pattern in matrix problems where elements share structural relationships. Using a hash table or set ensures distinct checks remain O(1) on average.
Approach 2: Simultaneous Diagonal Pass Method (O(m*n) time, O(min(m,n)) space)
Instead of storing prefix and suffix arrays, process each diagonal in a single controlled pass using two hash sets. Move along the diagonal and maintain a set of values already seen (top-left side). Before updating it, compute the number of distinct values remaining on the bottom-right side by maintaining another set or pre-counted structure. As you advance, shift elements from the future set to the past set and update the answer in place. Because each diagonal length is at most min(m, n), the auxiliary memory remains bounded by the diagonal size rather than the whole matrix.
This method keeps the same O(m*n) runtime but reduces auxiliary memory usage. It relies heavily on fast set operations and careful ordering of updates while iterating along the diagonal.
Problems like this combine array traversal with diagonal indexing tricks. The key insight: treat each diagonal independently instead of recalculating distinct counts from every cell.
Recommended for interviews: The diagonal traversal with precomputation approach is the safest explanation during interviews. It clearly demonstrates how you decompose the matrix into diagonals and track distinct values using hash sets. Once the interviewer sees the O(m*n) logic, mentioning the simultaneous-pass optimization shows deeper understanding of space tradeoffs.
Approach 1: Diagonal Traversal with Precomputation
This approach involves precomputing diagonal elements so that when we calculate distinct values for each diagonal, we can directly access those precomputed results instead of recalculating them from scratch for every cell.
def difference_of_number_of_distinct_values(grid):
m, n = len(grid), len(grid[0])
answer = [[0] * n for _ in range(m)]
left_above = [set() for _ in range(m + n)]
right_below = [set() for _ in range(m + n)]
for r in range(m):
for c in range(n):
answer[r][c] = abs(len(left_above[r - c + n]) - len(right_below[r - c + n]))
left_above[r - c + n].add(grid[r][c])
for r in range(m - 1, -1, -1):
for c in range(n - 1, -1, -1):
right_below[r - c + n].add(grid[r][c])
return answer
# Example usage:
grid = [[1, 2, 3], [3, 1, 5], [3, 2, 1]]
result = difference_of_number_of_distinct_values(grid)
for row in result:
print(row)import java.util.*;
public class DiagonalDifference {
public static int[][] differenceOfNumberOfDistinctValues(int[][] grid) {
int m = grid.length, n = grid[0].length;
int[][] answer = new int[m][n];
List<Set<Integer>> leftAbove = new ArrayList<>(m + n);
List<Set<Integer>> rightBelow = new ArrayList<>(m + n);
for (int i = 0; i < m + n; i++) {
leftAbove.add(new HashSet<>());
rightBelow.add(new HashSet<>());
}
for (int r = 0; r < m; r++) {
for (int c = 0; c < n; c++) {
answer[r][c] = Math.abs(leftAbove.get(r - c + n).size() - rightBelow.get(r - c + n).size());
leftAbove.get(r - c + n).add(grid[r][c]);
}
}
for (int r = m - 1; r >= 0; r--) {
for (int c = n - 1; c >= 0; c--) {
rightBelow.get(r - c + n).add(grid[r][c]);
}
}
return answer;
}
public static void main(String[] args) {
int[][] grid = {{1, 2, 3}, {3, 1, 5}, {3, 2, 1}};
int[][] result = differenceOfNumberOfDistinctValues(grid);
for (int[] row : result) {
System.out.println(Arrays.toString(row));
}
}
}#include <iostream>
#include <vector>
#include <set>
using namespace std;
vector<vector<int>> differenceOfNumberOfDistinctValues(vector<vector<int>>& grid) {
int m = grid.size();
int n = grid[0].size();
vector<vector<int>> answer(m, vector<int>(n));
vector<set<int>> leftAboveDiag(m + n), rightBelowDiag(m + n);
for (int d = -(m - 1); d < n; ++d) {
set<int> leftAbove, rightBelow;
for (int r = 0; r < m; ++r) {
int c = r + d;
if (c >= 0 && c < n) {
answer[r][c] = abs((int)leftAbove.size() - (int)rightBelow.size());
leftAbove.insert(grid[r][c]);
}
}
for (int r = m - 1; r >= 0; --r) {
int c = r + d;
if (c >= 0 && c < n) {
rightBelow.insert(grid[r][c]);
}
}
}
return answer;
}
int main() {
vector<vector<int>> grid = {{1, 2, 3}, {3, 1, 5}, {3, 2, 1}};
vector<vector<int>> result = differenceOfNumberOfDistinctValues(grid);
for (const auto& row : result) {
for (int val : row) {
cout << val << ' ';
}
cout << endl;
}
return 0;
}function differenceOfNumberOfDistinctValues(grid) {
const m = grid.length, n = grid[0].length;
const answer = Array.from({ length: m }, () => Array(n).fill(0));
const leftAbove = Array(m + n).fill(0).map(() => new Set());
const rightBelow = Array(m + n).fill(0).map(() => new Set());
for (let r = 0; r < m; r++) {
for (let c = 0; c < n; c++) {
const diagIndex = r - c + n;
answer[r][c] = Math.abs(leftAbove[diagIndex].size - rightBelow[diagIndex].size);
leftAbove[diagIndex].add(grid[r][c]);
}
}
for (let r = m - 1; r >= 0; r--) {
for (let c = n - 1; c >= 0; c--) {
const diagIndex = r - c + n;
rightBelow[diagIndex].add(grid[r][c]);
}
}
return answer;
}
// Example usage:
const grid = [[1, 2, 3], [3, 1, 5], [3, 2, 1]];
const result = differenceOfNumberOfDistinctValues(grid);
console.log(result.map(row => row.join(', ')).join('\n'));#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_M 50
#define MAX_N 50
int countDistinct(int* array, int size) {
int count = 0;
int set[1001] = {0}; // Assume grid values are bound [1, 50]
for (int i = 0; i < size; i++) {
if (!set[array[i]]) {
set[array[i]] = 1;
count++;
}
}
return count;
}
int** differenceOfNumberOfDistinctValues(int grid[MAX_M][MAX_N], int m, int n, int* returnSize, int** returnColumnSizes) {
int** answer = (int**)malloc(m * sizeof(int*));
*returnSize = m;
*returnColumnSizes = (int*)malloc(m * sizeof(int));
for (int i = 0; i < m; i++) {
answer[i] = (int*)malloc(n * sizeof(int));
(*returnColumnSizes)[i] = n;
}
int leftAbove[MAX_M][MAX_N], rightBelow[MAX_M][MAX_N];
memset(leftAbove, 0, sizeof(leftAbove));
memset(rightBelow, 0, sizeof(rightBelow));
for (int start = 0; start < m + n - 1; start++) {
int pointsLeftAbove[MAX_M], sizeLA = 0;
int pointsRightBelow[MAX_M], sizeRB = 0;
for (int r = 0; r < m; r++) {
int c = start - r;
if (c >= 0 && c < n) {
leftAbove[r][c] = countDistinct(pointsLeftAbove, sizeLA);
rightBelow[r][c] = countDistinct(pointsRightBelow, sizeRB);
pointsLeftAbove[sizeLA++] = grid[r][c];
}
}
for (int r = m - 1; r >= 0; r--) {
int c = start - r;
if (c >= 0 && c < n) {
pointsRightBelow[sizeRB++] = grid[r][c];
}
}
}
for (int r = 0; r < m; r++) {
for (int c = 0; c < n; c++) {
answer[r][c] = abs(leftAbove[r][c] - rightBelow[r][c]);
}
}
return answer;
}
int main() {
int grid[MAX_M][MAX_N] = {{1, 2, 3}, {3, 1, 5}, {3, 2, 1}};
int m = 3, n = 3;
int returnSize;
int* returnColumnSizes;
int** result = differenceOfNumberOfDistinctValues(grid, m, n, &returnSize, &returnColumnSizes);
return 0;
}using System;
using System.Collections.Generic;
public class DiagonalDifference {
public static int[][] DifferenceOfNumberOfDistinctValues(int[][] grid) {
int m = grid.Length, n = grid[0].Length;
var answer = new int[m][];
var leftAbove = new List<HashSet<int>>(m + n);
var rightBelow = new List<HashSet<int>>(m + n);
for (int i = 0; i < m + n; i++) {
leftAbove.Add(new HashSet<int>());
rightBelow.Add(new HashSet<int>());
}
for (int i = 0; i < m; i++) answer[i] = new int[n];
for (int r = 0; r < m; r++) {
for (int c = 0; c < n; c++) {
answer[r][c] = Math.Abs(leftAbove[r - c + n].Count - rightBelow[r - c + n].Count);
leftAbove[r - c + n].Add(grid[r][c]);
}
}
for (int r = m - 1; r >= 0; r--) {
for (int c = n - 1; c >= 0; c--) {
rightBelow[r - c + n].Add(grid[r][c]);
}
}
return answer;
}
public static void Main() {
int[][] grid = new int[][] {
new int[] { 1, 2, 3 },
new int[] { 3, 1, 5 },
new int[] { 3, 2, 1 }
};
var result = DifferenceOfNumberOfDistinctValues(grid);
foreach (var row in result) {
Console.WriteLine(string.Join(", ", row));
}
}
}In Python, the solution employs dictionary-based set collections to capture distinct values across both diagonal directions. It operates by differentiating diagonal indices to cache precomputed distinct results.
Time Complexity: O(m * n) - Each matrix cell is involved in two set operations.
Space Complexity: O(m + n) - Efficiently uses space for storing diagonal sets of elements.
Approach 2: Simultaneous Diagonal Pass Method
This strategy employs a simultaneous diagonal pass from both directions using a two-pointer strategy to dynamically compute and update diagonal distinct values while traversing the grid matrix.
def difference_of_number_of_distinct_values(grid):
m, n = len(grid), len(grid[0])
answer = [[0] * n for _ in range(m)]
for d in range(m + n - 1):
left_seen, right_seen = {}, {}
for r in range(max(0, d - n + 1), min(m, d + 1)):
c = d - r
left_seen[grid[r][c]] = left_seen.get(grid[r][c], 0) + 1
answer[r][c] = abs(len(left_seen) - len(right_seen))
for r in range(m - 1, -1, -1):
c = d - r
if c < 0: continue
right_seen[grid[r][c]] = right_seen.get(grid[r][c], 0) + 1
return answer
# Example usage:
grid = [[1, 2, 3], [3, 1, 5], [3, 2, 1]]
result = difference_of_number_of_distinct_values(grid)
for row in result:
print(row)import java.util.*;
public class DiagonalDifferenceSimultaneous {
public static int[][] differenceOfNumberOfDistinctValues(int[][] grid) {
int m = grid.length, n = grid[0].length;
int[][] answer = new int[m][n];
for (int d = 0; d < m + n - 1; d++) {
Map<Integer, Integer> leftCount = new HashMap<>();
Map<Integer, Integer> rightCount = new HashMap<>();
for (int r = (d < n) ? 0 : d - n + 1; r < m && d - r >= 0; r++) {
int c = d - r;
leftCount.put(grid[r][c], leftCount.getOrDefault(grid[r][c], 0) + 1);
answer[r][c] = Math.abs(leftCount.size() - rightCount.size());
}
for (int r = m - 1; r >= 0 && d - r < n; r--) {
int c = d - r;
if (c < 0) continue;
rightCount.put(grid[r][c], rightCount.getOrDefault(grid[r][c], 0) + 1);
}
}
return answer;
}
public static void main(String[] args) {
int[][] grid = {{1, 2, 3}, {3, 1, 5}, {3, 2, 1}};
int[][] result = differenceOfNumberOfDistinctValues(grid);
for (int[] row : result) {
System.out.println(Arrays.toString(row));
}
}
}#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
vector<vector<int>> differenceOfNumberOfDistinctValues(vector<vector<int>>& grid) {
int m = grid.size();
int n = grid[0].size();
vector<vector<int>> answer(m, vector<int>(n));
unordered_map<int, int> leftCount, rightCount;
for (int d = 0; d < m + n - 1; d++) {
leftCount.clear();
rightCount.clear();
for (int r = (d < n) ? 0 : (d - n + 1); r < m && (d - r) >= 0; r++) {
int c = d - r;
leftCount[grid[r][c]]++;
answer[r][c] = abs(leftCount.size() - rightCount.size());
}
for (int r = m - 1; r >= 0 && (d - r) < n; r--) {
int c = d - r;
if (c < 0) continue;
rightCount[grid[r][c]]++;
}
}
return answer;
}
int main() {
vector<vector<int>> grid = {{1, 2, 3}, {3, 1, 5}, {3, 2, 1}};
vector<vector<int>> result = differenceOfNumberOfDistinctValues(grid);
for (auto& row : result) {
for (int val : row) cout << val << ' ';
cout << endl;
}
return 0;
}function differenceOfNumberOfDistinctValues(grid) {
const m = grid.length, n = grid[0].length;
const answer = Array.from({ length: m }, () => Array(n).fill(0));
for (let d = 0; d < m + n - 1; d++) {
const leftSeen = new Map();
const rightSeen = new Map();
for (let r = Math.max(0, d - n + 1); r < Math.min(m, d + 1); r++) {
const c = d - r;
leftSeen.set(grid[r][c], (leftSeen.get(grid[r][c]) || 0) + 1);
answer[r][c] = Math.abs(leftSeen.size - rightSeen.size);
}
for (let r = m - 1; r >= 0 && d - r < n; r--) {
const c = d - r;
rightSeen.set(grid[r][c], (rightSeen.get(grid[r][c]) || 0) + 1);
}
}
return answer;
}
// Example usage:
const grid = [[1, 2, 3], [3, 1, 5], [3, 2, 1]];
const result = differenceOfNumberOfDistinctValues(grid);
console.log(result.map(row => row.join(", ")).join("\n"));#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int** differenceOfNumberOfDistinctValues(int** grid, int gridSize, int* gridColSize, int* returnSize, int** returnColumnSizes) {
int m = gridSize, n = gridColSize[0];
int** answer = (int**)malloc(m * sizeof(int*));
*returnSize = m;
*returnColumnSizes = malloc(m * sizeof(int));
int leftCount[1001] = {0};
int rightCount[1001] = {0};
for (int i = 0; i < m; i++) {
answer[i] = (int*)malloc(n * sizeof(int));
(*returnColumnSizes)[i] = n;
}
for (int d = 0; d < m + n - 1; d++) {
int leftIdx = 0, rightIdx = 0;
int maxLength = m + n - abs(m-1) - 1;
int* leftDiagonal = (int*)malloc(maxLength * sizeof(int));
int* rightDiagonal = (int*)malloc(maxLength * sizeof(int));
for (int r = (d < n) ? 0 : d - n + 1; r < m && d - r >= 0; r++) {
int c = d - r;
int val = grid[r][c];
if (leftCount[val] == 0) leftIdx++;
leftCount[val]++;
leftDiagonal[--leftIdx] = val;
}
for (int r = m - 1; r >= (d < n ? 0 : d - n + 1); r--) {
int c = d - r;
if (c < 0) break;
int val = grid[r][c];
if (rightCount[val] == 0) rightIdx++;
rightCount[val]++;
rightDiagonal[--rightIdx] = val;
}
for (int r = (d < n) ? 0 : d - n + 1; r < m && d - r >= 0; r++) {
int c = d - r;
answer[r][c] = abs(leftIdx - rightIdx);
}
for (int i = 0; i < maxLength; i++) {
leftCount[leftDiagonal[i]] = 0;
rightCount[rightDiagonal[i]] = 0;
}
free(leftDiagonal);
free(rightDiagonal);
}
return answer;
}
int main() {
int m = 3, n = 3;
int* grid[3];
int row0[3] = {1, 2, 3};
int row1[3] = {3, 1, 5};
int row2[3] = {3, 2, 1};
grid[0] = row0;
grid[1] = row1;
grid[2] = row2;
int colSize = 3;
int returnSize;
int* returnColumnSizes;
int** result = differenceOfNumberOfDistinctValues(grid, m, &colSize, &returnSize, &returnColumnSizes);
return 0;
}using System;
using System.Collections.Generic;
public class DiagonalDifferenceSimultaneous {
public static int[][] DifferenceOfNumberOfDistinctValues(int[][] grid) {
int m = grid.Length, n = grid[0].Length;
int[][] answer = new<int>[m][];
for (int i = 0; i < m; i++) answer[i] = new int[n];
for (int d = 0; d < m + n - 1; d++) {
var leftCount = new Dictionary<int, int>();
var rightCount = new Dictionary<int, int>();
for (int r = Math.Max(0, d - n + 1); r < Math.Min(m, d + 1); r++) {
int c = d - r;
if (!leftCount.ContainsKey(grid[r][c])) leftCount[grid[r][c]] = 0;
leftCount[grid[r][c]]++;
answer[r][c] = Math.Abs(leftCount.Count - rightCount.Count);
}
for (int r = m - 1; r >= 0 && d - r < n; r--) {
int c = d - r;
if (c < 0) continue;
if (!rightCount.ContainsKey(grid[r][c])) rightCount[grid[r][c]] = 0;
rightCount[grid[r][c]]++;
}
}
return answer;
}
public static void Main() {
int[][] grid = {
new[] {1, 2, 3},
new[] {3, 1, 5},
new[] {3, 2, 1}
};
var result = DifferenceOfNumberOfDistinctValues(grid);
foreach (var row in result) {
Console.WriteLine(string.Join(", ", row));
}
}
}The Python solution follows a twin traversal technique with dictionary/maps to intermittently achieve diagonal index results. As the simultaneous pass unfolds, the distinction is adequately addressed.
Time Complexity: O(m * n) - Subsequent for clean diagonal allocation.
Space Complexity: O(m + n) - Both state maps resize expeditiously per diagonal handling.
Approach 3: Simulation
We can simulate the process described in the problem statement, calculating the number of distinct values on the top-left diagonal tl and the bottom-right diagonal br for each cell, then compute their difference |tl - br|.
The time complexity is O(m times n times min(m, n)), and the space complexity is O(m times n).
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;
}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
}Complexity comparison
| Approach | Complexity |
|---|---|
| Diagonal Traversal with Precomputation | Time Complexity: O(m * n) - Each matrix cell is involved in two set operations. |
| Simultaneous Diagonal Pass Method | Time Complexity: O(m * n) - Subsequent for clean diagonal allocation. |
| Simulation | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Diagonal Traversal with Precomputation | O(m*n) | O(m*n) | Best for clarity and interviews; easy to reason about prefix and suffix distinct counts per diagonal |
| Simultaneous Diagonal Pass Method | O(m*n) | O(min(m,n)) | When memory usage matters and you want to avoid storing extra matrices |
Frequently asked questions
Is Difference of Number of Distinct Values on Diagonals easy or hard?
The problem is generally classified as Medium difficulty. The challenge is recognizing that each cell's computation depends only on its diagonal and avoiding repeated scans. Once the diagonal traversal pattern is identified, the solution becomes a straightforward hash set application.
Difference of Number of Distinct Values on Diagonals Python/Java solution
Implementations typically iterate through each diagonal and maintain sets for values seen before and after a cell. Languages like Python use the built-in set, while Java and C++ use HashSet or unordered_set. The algorithm structure remains the same and runs in O(m*n) time.
How to solve Difference of Number of Distinct Values on Diagonals in O(m*n)?
Iterate through every diagonal of the matrix. Maintain a set of values seen before each position and compute distinct counts on the remaining part of the diagonal. Store the absolute difference between these counts for each cell. Because each element belongs to one diagonal and is processed only a few times, the runtime stays O(m*n).
What is the best approach for Difference of Number of Distinct Values on Diagonals?
The most reliable approach is diagonal traversal with prefix and suffix distinct counts. Treat each top-left to bottom-right diagonal independently, track values with a hash set, and compute the absolute difference for each cell. This method runs in O(m*n) time and is straightforward to implement in most languages.
Is Difference of Number of Distinct Values on Diagonals asked at Google/Amazon/Meta?
Matrix traversal and diagonal-processing problems frequently appear in technical interviews at companies like Amazon, Google, and Meta. While this exact problem may vary, the underlying skills—hash sets, matrix indexing, and diagonal traversal—are common interview patterns.
What data structure is used in Difference of Number of Distinct Values on Diagonals?
The primary data structure is a hash set (or hash table) used to track distinct values along a diagonal. It provides O(1) average-time insert and lookup operations, which keeps the overall algorithm linear with respect to the matrix size.
What is the time complexity of Difference of Number of Distinct Values on Diagonals?
The optimal solutions run in O(m*n) time, where m and n are the grid dimensions. Each element is processed a constant number of times while traversing diagonals. Hash set operations for inserting and checking distinct values take O(1) average time.