**Rectangular Overlap - Amazon Top Interview Questions**

### Problem Statement :

You are given two lists of integers rect0 and rect1 representing coordinates (x0, y0, x1, y1) of two rectangles. (x0, y0) is the bottom left coordinate and (x1, y1) is the upper right coordinate. Return whether the two rectangles overlap. Note: if two rectangles touch only in their perimeters, they don't overlap. Example 1 Input rect0 = [0, 0, 10, 10] rect1 = [10, 10, 30, 30] Output False Example 2 Input rect0 = [0, 0, 10, 10] rect1 = [5, 5, 30, 30] Output True

### Solution :

` ````
Solution in C++ :
bool solve(vector<int>& rect0, vector<int>& rect1) {
return !(rect0[3] <= rect1[1] || rect1[3] <= rect0[1] || rect0[2] <= rect1[0] ||
rect1[2] <= rect0[0]);
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public boolean solve(int[] rect0, int[] rect1) {
int x1 = Math.max(rect0[0], rect1[0]);
int y1 = Math.max(rect0[1], rect1[1]);
int x2 = Math.min(rect0[2], rect1[2]);
int y2 = Math.min(rect0[3], rect1[3]);
if (x1 >= x2 || y1 >= y2)
return false;
return true;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, rect0, rect1):
return not (
rect1[0] >= rect0[2]
or rect0[0] >= rect1[2]
or rect1[1] >= rect0[3]
or rect0[1] >= rect1[3]
)
```

## View More Similar Problems

## Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

View Solution →## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

View Solution →## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func

View Solution →## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

View Solution →## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

View Solution →