Skip to content

2387. Median of a Row Wise Sorted Matrix 👍

  • Time: $O(\log(10^6) \cdot m\log n) = O(m\log n)$
  • Space: $O(1)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
class Solution {
 public:
  int matrixMedian(vector<vector<int>>& grid) {
    const int noGreaterThanMedianCount = grid.size() * grid[0].size() / 2 + 1;
    int l = 1;
    int r = 1e6;

    while (l < r) {
      const int m = (l + r) / 2;
      if (numsNoGreaterThan(grid, m) >= noGreaterThanMedianCount)
        r = m;
      else
        l = m + 1;
    }

    return l;
  }

 private:
  int numsNoGreaterThan(const vector<vector<int>>& grid, int m) {
    int count = 0;
    for (const vector<int>& row : grid)
      count += ranges::upper_bound(row, m) - row.begin();
    return count;
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
class Solution {
  public int matrixMedian(int[][] grid) {
    final int noGreaterThanMedianCount = grid.length * grid[0].length / 2 + 1;
    int l = 1;
    int r = (int) 1e6;

    while (l < r) {
      final int m = (l + r) / 2;
      if (numsNoGreaterThan(grid, m) >= noGreaterThanMedianCount)
        r = m;
      else
        l = m + 1;
    }

    return l;
  }

  private int numsNoGreaterThan(int[][] grid, int m) {
    int count = 0;
    for (int[] row : grid)
      count += firstGreater(row, m);
    return count;
  }

  private int firstGreater(int[] A, int target) {
    int l = 0;
    int r = A.length;
    while (l < r) {
      final int m = (l + r) / 2;
      if (A[m] > target)
        r = m;
      else
        l = m + 1;
    }
    return l;
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
class Solution:
  def matrixMedian(self, grid: List[List[int]]) -> int:
    noGreaterThanMedianCount = len(grid) * len(grid[0]) // 2 + 1
    l = 1
    r = int(1e6)

    while l < r:
      m = (l + r) // 2
      if sum(bisect_right(row, m) for row in grid) >= \
              noGreaterThanMedianCount:
        r = m
      else:
        l = m + 1

    return l