Hard
You are given a 0-indexed binary string floor
, which represents the colors of tiles on a floor:
floor[i] = '0'
denotes that the ith
tile of the floor is colored black.floor[i] = '1'
denotes that the ith
tile 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 <= 1000
floor[i]
is either '0'
or '1'
.1 <= numCarpets <= 1000
class Solution {
fun minimumWhiteTiles(floor: String, numCarpets: Int, carpetLen: Int): Int {
val len = floor.length
val dp = Array(numCarpets + 1) { IntArray(len + 1) }
val prefix = IntArray(len)
var tiles = 0
var total = 0
for (i in 0 until len) {
// calculate total no of Tiles within the Carpet Length Window
tiles += floor[i].code - '0'.code
// start excluding tiles which are not in the Range anymore of the Carpet Length given
if (i - carpetLen >= 0) {
tiles -= floor[i - carpetLen].code - '0'.code
}
// the total no of tiles covered within the Carpet Length range for current index
prefix[i] = tiles
total += floor[i].code - '0'.code
}
for (i in 1..numCarpets) {
for (j in 0 until len) {
// if we do not wish to cover current Tile
val doNot = dp[i][j]
// if we do wish to cover current tile
val 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]
}
}