Question

2711. Difference of Number of Distinct Values on Diagonals

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Description

Given a 0-indexed 2D grid of size m x n, you should find the matrix answer of size m x n.

The value of each cell (r, c) of the matrix answer is calculated in the following way:

• Let topLeft[r][c] be the number of distinct values in the top-left diagonal of the cell (r, c) in the matrix grid.
• Let bottomRight[r][c] be the number of distinct values in the bottom-right diagonal of the cell (r, c) in the matrix grid.

Then answer[r][c] = |topLeft[r][c] - bottomRight[r][c]|.

Return _the matrix_ answer.

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 reaching the matrix's end.

A cell (r1, c1) belongs to the top-left diagonal of the cell (r, c), if both belong to the same diagonal and r1 < r. Similarly is defined bottom-right diagonal.

 

Example 1:

Input: grid = [[1,2,3],[3,1,5],[3,2,1]]
Output: [[1,1,0],[1,0,1],[0,1,1]]
Explanation: The 1st diagram denotes the initial grid. 
The 2nd diagram denotes a grid for cell (0,0), where blue-colored cells are cells on its bottom-right diagonal.
The 3rd diagram denotes a grid for cell (1,2), where red-colored cells are cells on its top-left diagonal.
The 4th diagram denotes a grid for cell (1,1), where blue-colored cells are cells on its bottom-right diagonal and red-colored cells are cells on its top-left diagonal.
- The cell (0,0) contains [1,1] on its bottom-right diagonal and [] on its top-left diagonal. The answer is |1 - 0| = 1.
- The cell (1,2) contains [] on its bottom-right diagonal and [2] on its top-left diagonal. The answer is |0 - 1| = 1.
- The cell (1,1) contains [1] on its bottom-right diagonal and [1] on its top-left diagonal. The answer is |1 - 1| = 0.
The answers of other cells are similarly calculated.

Example 2:

Input: grid = [[1]]
Output: [[0]]
Explanation: - The cell (0,0) contains [] on its bottom-right diagonal and [] on its top-left diagonal. The answer is |0 - 0| = 0.

 

Constraints:

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

Solutions

Solution 1

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

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;
  }
}

class Solution {
public:
  vector<vector<int>> differenceOfDistinctValues(vector<vector<int>>& grid) {
    int m = grid.size(), n = grid[0].size();
    vector<vector<int>> ans(m, vector<int>(n));
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        int x = i, y = j;
        unordered_set<int> s;
        while (x > 0 && y > 0) {
          s.insert(grid[--x][--y]);
        }
        int tl = s.size();
        x = i;
        y = j;
        s.clear();
        while (x < m - 1 && y < n - 1) {
          s.insert(grid[++x][++y]);
        }
        int br = s.size();
        ans[i][j] = 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
}

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;
}