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 <= 1000floor[i] is either '0' or '1'.1 <= numCarpets <= 1000This approach involves using dynamic programming to solve the problem by defining a dp array. The value of dp[i] represents the minimum number of white tiles visible when considering the first i tiles and utilizing a certain number of carpets.
To compute dp[i], you iterate over each possible position of the last carpet and calculate the number of visible white tiles, updating the dp array to store the minimum value found.
The Python solution defines a function min_white_tiles which uses dynamic programming. The prefix_white array helps compute the number of white tiles efficiently, while the dp 2D array stores the minimum number of visible white tiles for each scenario. The solution updates the dp array based on whether the current position is covered by a new carpet or not.
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Time Complexity: O(n * numCarpets), where n is the length of the floor string.
Space Complexity: O(n * numCarpets), used by the dp array.
A different approach uses sliding windows combined with a prefix sum to determine the maximum number of white tiles that can be covered with a rug. Calculate the number of visible white tiles by subtracting the maximum covered tiles from the total number of white tiles.
Move the window and keep track of maximum white tiles covered within the carpet length and the number of carpets available.
This C++ solution takes advantage of prefix sums to quickly calculate the number of white tiles in any window of size carpetLen. It slides a window across the string to determine the configuration that covers the maximum number of white tiles, then subtracts this from the total number of white tiles.
Time Complexity: O(n), where n is the length of the floor string due to the prefix sum and single pass window technique.
Space Complexity: O(n) for the prefix sum array.
Instead of using dynamic programming, a greedy approach can be used for certain cases, leveraging precomputation of white tiles in segments. Sum the number of white tiles visible across overlapping segments, and cover the ones with the highest values first.
This solution uses a greedy strategy to iteratively select and cover the segment with the most uncovered white tiles using the prefix sum array for quick counting, maintaining an auxiliary array to store quantities of carpets covering each tile to avoid double counting.
Time Complexity: O(n * numCarpets) since we try to cover each high white count segment per carpet.
Space Complexity: O(n) due to the prefix sum and covered arrays.
| Approach | Complexity |
|---|---|
| Dynamic Programming Approach | Time Complexity: O(n * numCarpets), where n is the length of the floor string. |
| Sliding Window with Prefix Sum Approach | Time Complexity: O(n), where n is the length of the floor string due to the prefix sum and single pass window technique. |
| Greedy with Precomputation | Time Complexity: Space Complexity: |
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