Boxes All the Way Down - Amazon Top Interview Questions
Problem Statement :
You are given a two-dimensional list of integers boxes. Each list contains two integers [width, height] which represent the width and height of a box. Given that you can put a box in another box if both of its width and height are smaller than the other box, return the maximum number of boxes you can fit into a box.
You cannot rotate the boxes.
Constraints
n ≤ 100,000 where n is the length of boxes
Example 1
Input
matrix = [
[10, 10],
[9, 9],
[5, 5],
[4, 9]
]
Output
3
Explanation
We can fit the box [5, 5] into [9, 9] which we can fit into the [10, 10] box.
Solution :
Solution in C++ :
struct SegTree {
int n;
vector<int> arr;
vector<int> tree;
SegTree(int n1, vector<int>& a) {
n = n1;
arr = a;
int s = 1;
while (s < 2 * n) {
s = s << 1;
}
tree.resize(s);
fill(tree.begin(), tree.end(), 0);
build(0, n - 1, 1);
}
void build(int start, int end, int index) {
if (start == end) {
arr[start] = tree[index];
return;
}
int mid = (start + end) / 2;
build(start, mid, index * 2);
build(mid + 1, end, index * 2 + 1);
tree[index] = max(tree[2 * index], tree[2 * index + 1]);
}
void update(int start, int end, int index, int idx, int value) {
if (start == end) {
arr[start] = max(arr[start], value);
tree[index] = arr[start];
return;
}
int mid = (start + end) / 2;
if (idx <= mid)
update(start, mid, index * 2, idx, value);
else
update(mid + 1, end, index * 2 + 1, idx, value);
tree[index] = max(tree[2 * index], tree[2 * index + 1]);
}
int query(int start, int end, int index, int left, int right) {
if (start > right || end < left) return 0;
if (start >= left && end <= right) return tree[index];
int mid = (start + end) / 2;
int val1 = query(start, mid, index * 2, left, right);
int val2 = query(mid + 1, end, index * 2 + 1, left, right);
return max(val1, val2);
}
};
bool compare(vector<int>& a, vector<int>& b) {
return a[0] < b[0];
}
int solve(vector<vector<int>>& matrix) {
map<int, int> compression;
int n = matrix.size();
for (int i = 0; i < n; i++) {
compression[matrix[i][1]] = 0;
}
int last = 1;
for (auto& i : compression) {
i.second = last++;
}
for (int i = 0; i < n; i++) {
matrix[i][1] = compression[matrix[i][1]];
}
sort(matrix.begin(), matrix.end(), compare);
vector<vector<int>> points;
int lastX = -1;
vector<int> prev = {};
for (int i = 0; i < n; i++) {
if (matrix[i][0] != lastX) {
if (prev.size() != 0) {
points.push_back(prev);
prev.clear();
}
prev.push_back(matrix[i][1]);
lastX = matrix[i][0];
} else {
prev.push_back(matrix[i][1]);
}
}
if (prev.size() != 0) {
points.push_back(prev);
}
vector<int> ans(n + 1, 0);
SegTree sg = SegTree(n + 1, ans);
for (auto i : points) {
vector<int> toUpdate;
for (auto j : i) {
int value = sg.query(0, n, 1, 0, j - 1);
toUpdate.push_back(value + 1);
}
for (int j = 0; j < i.size(); j++) {
sg.update(0, n, 1, i[j], toUpdate[j]);
}
}
return sg.query(0, n, 1, 0, n);
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] A) {
Arrays.sort(A, (a, b) -> {
if (a[0] == b[0])
return b[1] - a[1];
return a[0] - b[0];
});
int B[] = new int[A.length];
for (int i = 0; i < A.length; i++) {
B[i] = A[i][1];
}
return lis(B);
}
public int lis(int[] A) {
int dp[] = new int[A.length];
Arrays.fill(dp, Integer.MAX_VALUE);
for (int i = 0; i < A.length; i++) {
bin(dp, A[i]);
}
int res = 0;
for (int i = 0; i < dp.length; i++) {
if (dp[i] != Integer.MAX_VALUE)
res = i + 1;
}
return res;
}
public void bin(int dp[], int val) {
int l = 0, r = dp.length - 1;
int index = -1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (val <= dp[mid]) {
index = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
dp[index] = val;
}
}
Solution in Python :
class Solution:
def solve(self, a):
# double strict
a.sort(key=lambda x: [x[0], -x[1]])
ret = []
for w, h in a:
i = bisect_left(ret, h)
ret[i : i + 1] = [h]
return len(ret)
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