LeetCode 1580: Put Boxes Into the Warehouse II
LeetCode 1580 Solution Explanation
Explanation:
To solve this problem, we can use a two-pointer approach. We will sort both the box sizes and the warehouse heights in non-increasing order. Then, we will iterate through the boxes and warehouse from the largest to the smallest, and try to fit as many boxes as possible into the warehouse.
- Sort the box sizes and warehouse heights in non-increasing order.
- Initialize pointers
ifor boxes andjfor warehouse at the 0th index. - While both pointers are within bounds, check if the current box can fit in the current warehouse height.
- If it fits, increment both pointers.
- If it doesn't fit, only increment the box pointer.
- Repeat until all boxes are processed.
- The number of boxes that can fit into the warehouse is the number of times the box pointer is incremented.
Time Complexity: O(n log n) where n is the maximum of number of boxes or warehouse heights. Space Complexity: O(1)
:
LeetCode 1580 Solutions in Java, C++, Python
class Solution {
public int maxBoxesInWarehouse(int[] boxes, int[] warehouse) {
Arrays.sort(boxes);
Arrays.sort(warehouse);
int i = 0, j = 0;
int count = 0;
while (i < boxes.length && j < warehouse.length) {
if (boxes[i] <= warehouse[j]) {
count++;
i++;
}
j++;
}
return count;
}
}Interactive Code Editor for LeetCode 1580
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