Welcome to Subscribe On Youtube

2271. Maximum White Tiles Covered by a Carpet

Description

You are given a 2D integer array tiles where tiles[i] = [li, ri] represents that every tile j in the range li <= j <= ri is colored white.

You are also given an integer carpetLen, the length of a single carpet that can be placed anywhere.

Return the maximum number of white tiles that can be covered by the carpet.

 

Example 1:

Input: tiles = [[1,5],[10,11],[12,18],[20,25],[30,32]], carpetLen = 10
Output: 9
Explanation: Place the carpet starting on tile 10. 
It covers 9 white tiles, so we return 9.
Note that there may be other places where the carpet covers 9 white tiles.
It can be shown that the carpet cannot cover more than 9 white tiles.

Example 2:

Input: tiles = [[10,11],[1,1]], carpetLen = 2
Output: 2
Explanation: Place the carpet starting on tile 10. 
It covers 2 white tiles, so we return 2.

 

Constraints:

  • 1 <= tiles.length <= 5 * 104
  • tiles[i].length == 2
  • 1 <= li <= ri <= 109
  • 1 <= carpetLen <= 109
  • The tiles are non-overlapping.

Solutions

  • class Solution {
    public int maximumWhiteTiles(int[][] tiles, int carpetLen) {
        Arrays.sort(tiles, (a, b) -> a[0] - b[0]);
        int n = tiles.length;
        int s = 0, ans = 0;
        for (int i = 0, j = 0; i < n; ++i) {
            while (j < n && tiles[j][1] - tiles[i][0] + 1 <= carpetLen) {
                s += tiles[j][1] - tiles[j][0] + 1;
                ++j;
            }
            if (j < n && tiles[i][0] + carpetLen > tiles[j][0]) {
                ans = Math.max(ans, s + tiles[i][0] + carpetLen - tiles[j][0]);
            } else {
                ans = Math.max(ans, s);
            }
            s -= (tiles[i][1] - tiles[i][0] + 1);
        }
        return ans;
    }
}

  • class Solution {
public:
    int maximumWhiteTiles(vector<vector<int>>& tiles, int carpetLen) {
        sort(tiles.begin(), tiles.end());
        int s = 0, ans = 0, n = tiles.size();
        for (int i = 0, j = 0; i < n; ++i) {
            while (j < n && tiles[j][1] - tiles[i][0] + 1 <= carpetLen) {
                s += tiles[j][1] - tiles[j][0] + 1;
                ++j;
            }
            if (j < n && tiles[i][0] + carpetLen > tiles[j][0]) {
                ans = max(ans, s + tiles[i][0] + carpetLen - tiles[j][0]);
            } else {
                ans = max(ans, s);
            }
            s -= (tiles[i][1] - tiles[i][0] + 1);
        }
        return ans;
    }
};

  • class Solution:
    def maximumWhiteTiles(self, tiles: List[List[int]], carpetLen: int) -> int:
        tiles.sort()
        n = len(tiles)
        s = ans = j = 0
        for i, (li, ri) in enumerate(tiles):
            while j < n and tiles[j][1] - li + 1 <= carpetLen:
                s += tiles[j][1] - tiles[j][0] + 1
                j += 1
            if j < n and li + carpetLen > tiles[j][0]:
                ans = max(ans, s + li + carpetLen - tiles[j][0])
            else:
                ans = max(ans, s)
            s -= ri - li + 1
        return ans


  • func maximumWhiteTiles(tiles [][]int, carpetLen int) int {
	sort.Slice(tiles, func(i, j int) bool { return tiles[i][0] < tiles[j][0] })
	n := len(tiles)
	s, ans := 0, 0
	for i, j := 0, 0; i < n; i++ {
		for j < n && tiles[j][1]-tiles[i][0]+1 <= carpetLen {
			s += tiles[j][1] - tiles[j][0] + 1
			j++
		}
		if j < n && tiles[i][0]+carpetLen > tiles[j][0] {
			ans = max(ans, s+tiles[i][0]+carpetLen-tiles[j][0])
		} else {
			ans = max(ans, s)
		}
		s -= (tiles[i][1] - tiles[i][0] + 1)
	}
	return ans
}

All Problems

All Solutions