LeetCode-in-Java
2209. Minimum White Tiles After Covering With Carpets
Hard
You are given a 0-indexed binary string floor, which represents the colors of tiles on a floor:
floor[i] = '0'denotes that theithtile of the floor is colored black.- On the other hand,
floor[i] = '1'denotes that theithtile of the floor is colored white.
You are also given numCarpets and carpetLen. You have numCarpets black carpets, each of length carpetLen tiles. Cover the tiles with the given carpets such that the number of white tiles still visible is minimum. Carpets may overlap one another.
Return the minimum number of white tiles still visible.
Example 1:
Input: floor = “10110101”, numCarpets = 2, carpetLen = 2
Output: 2
Explanation:
The figure above shows one way of covering the tiles with the carpets such that only 2 white tiles are visible.
No other way of covering the tiles with the carpets can leave less than 2 white tiles visible.
Example 2:
Input: floor = “11111”, numCarpets = 2, carpetLen = 3
Output: 0
Explanation:
The figure above shows one way of covering the tiles with the carpets such that no white tiles are visible.
Note that the carpets are able to overlap one another.
Constraints:
1 <= carpetLen <= floor.length <= 1000floor[i]is either'0'or'1'.1 <= numCarpets <= 1000
Solution
public class Solution {
public int minimumWhiteTiles(String floor, int numCarpets, int carpetLen) {
int len = floor.length();
int[][] dp = new int[numCarpets + 1][len + 1];
int[] prefix = new int[len];
int tiles = 0;
int total = 0;
for (int i = 0; i < len; i++) {
// calculate total no of Tiles within the Carpet Length Window
tiles += floor.charAt(i) - '0';
// start excluding tiles which are not in the Range anymore of the Carpet Length given
if (i - carpetLen >= 0) {
tiles -= floor.charAt(i - carpetLen) - '0';
}
// the total no of tiles covered within the Carpet Length range for current index
prefix[i] = tiles;
total += floor.charAt(i) - '0';
}
for (int i = 1; i <= numCarpets; i++) {
for (int j = 0; j < len; j++) {
// if we do not wish to cover current Tile
int doNot = dp[i][j];
// if we do wish to cover current tile
int doTake = dp[i - 1][Math.max(0, j - carpetLen + 1)] + prefix[j];
// we should go back the Carpet length & check for tiles not covered before j -
// carpet Length distance
dp[i][j + 1] = Math.max(doTake, doNot);
}
}
return total - dp[numCarpets][len];
}
}