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# Check if it is possible to rearrange rectangles in a non-ascending order of breadths

Given n number of rectangles with it’s L-length and B-Breadth. We can turn any rectangle by 90 degrees. In other words, after turning them, the breadth will become length and length will be breadth.
The task is to check if there is a way to make the rectangles go in order of non-ascending breadth. That is, a breadth of every rectangle has to be not greater than the breadth of the previous rectangle.
Note: You can not change the order of the rectangles.
Examples:

Input:
l = 3, b = 4
l1 = 4, b1 = 6
l2 = 3, b2 = 5
Output: YES
The given breadths are [ 4, 6, 5 ] we can rotate the second and the third rectangle so that the breadths will satisfy the above condition [ 4, 4, 3 ] ( 3 is not greater than 4 and 4 is not greater than 4 ) which is why we print YES.
Input:
1 60
70 55
56 80
Output: NO
The breadths are [ 60, 55, 80 ] as 55 55 or 56 > 55. So it’s not possible to arrange the breadths in non-ascending order, which is why we’ll print NO.

Approach: Below is the step by step algorithm to solve this problem:

1. Initialize n rectangle with their lengths and breadths.
2. Iterate over the rectangle from left to right.
3. Turn each rectangle in such a way that it’s breadth is as big as possible but not greater than the previous rectangle.
4. If on some iteration there is no such way to place the rectangle, the answer is “NO”

Below is the implementation of above approach:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `// Function to check if it possible to form``// rectangles with heights as non-ascending``int` `rotateRec(``int` `n, ``int` `L[], ``int` `B[])``{` `    ``// set maximum``    ``int` `m = INT_MAX;` `    ``for` `(``int` `i = 0; i < n; i++) {``        ``// replace the maximum with previous maximum``        ``if` `(max(L[i], B[i]) <= m)``            ``m = max(L[i], B[i]);` `        ``// replace the minimum with previous minimum``        ``else` `if` `(min(L[i], B[i]) <= m)``            ``m = min(L[i], B[i]);` `        ``// print NO if the above``        ``// two conditions fail at least once``        ``else` `{``            ``return` `0;``        ``}``    ``}``    ``return` `1;``}` `// Driver code``int` `main()``{``    ``// initialize the number of rectangles``    ``int` `n = 3;` `    ``// initialize n rectangles with length and breadth``    ``int` `L[] = { 5, 5, 6 };``    ``int` `B[] = { 6, 7, 8 };``    ``rotateRec(n, L, B) == 1 ? cout << ``"YES"``                            ``: cout << ``"NO"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of above approach``import` `java.io.*;` `class` `GFG {``   ` `// Function to check if it possible to form``// rectangles with heights as non-ascending`` ``static` `int` `rotateRec(``int` `n, ``int` `L[], ``int` `B[])``{` `    ``// set maximum``    ``int` `m = Integer.MAX_VALUE;` `    ``for` `(``int` `i = ``0``; i < n; i++) {``        ``// replace the maximum with previous maximum``        ``if` `(Math.max(L[i], B[i]) <= m)``            ``m = Math.max(L[i], B[i]);` `        ``// replace the minimum with previous minimum``        ``else` `if` `(Math.min(L[i], B[i]) <= m)``            ``m = Math.min(L[i], B[i]);` `        ``// print NO if the above``        ``// two conditions fail at least once``        ``else` `{``            ``return` `0``;``        ``}``    ``}``    ``return` `1``;``}` `// Driver code` `    ``public` `static` `void` `main (String[] args) {``    ``// initialize the number of rectangles``    ``int` `n = ``3``;` `    ``// initialize n rectangles with length and breadth``    ``int` `L[] = { ``5``, ``5``, ``6` `};``    ``int` `B[] = { ``6``, ``7``, ``8` `};``    ``if``(rotateRec(n, L, B) == ``1``)``     ``System.out.println( ``"YES"``);``     ``else``     ``System.out.println( ``"NO"``);``    ``}``}`` ``// This Code is contributed by inder_verma..`

## Python3

 `# Python3 implementation of above approach``import` `sys;` `# Function to check if it possible``# to form rectangles with heights``# as non-ascending``def` `rotateRec(n, L, B):` `    ``# set maximum``    ``m ``=` `sys.maxsize;` `    ``for` `i ``in` `range``(n):` `        ``# replace the maximum with``        ``# previous maximum``        ``if` `(``max``(L[i], B[i]) <``=` `m):``            ``m ``=` `max``(L[i], B[i]);` `        ``# replace the minimum``        ``# with previous minimum``        ``elif` `(``min``(L[i], B[i]) <``=` `m):``            ``m ``=` `min``(L[i], B[i]);` `        ``# print NO if the above two``        ``# conditions fail at least once``        ``else``:``            ``return` `0``;``    ` `    ``return` `1``;` `# Driver code` `# initialize the number``# of rectangles``n ``=` `3``;` `# initialize n rectangles``# with length and breadth``L ``=` `[``5``, ``5``, ``6``];``B ``=` `[ ``6``, ``7``, ``8` `];``if``(rotateRec(n, L, B) ``=``=` `1``):``    ``print``(``"YES"``);``else``:``    ``print``(``"NO"``);` `# This code is contributed by mits`

## C#

 `// C# implementation of above approach``using` `System;` `class` `GFG``{``    ` `// Function to check if it possible``// to form rectangles with heights``// as non-ascending``static` `int` `rotateRec(``int` `n, ``int` `[]L,``                            ``int` `[]B)``{` `    ``// set maximum``    ``int` `m = ``int``.MaxValue ;` `    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``// replace the maximum with``        ``// previous maximum``        ``if` `(Math.Max(L[i], B[i]) <= m)``            ``m = Math.Max(L[i], B[i]);` `        ``// replace the minimum with``        ``// previous minimum``        ``else` `if` `(Math.Min(L[i], B[i]) <= m)``            ``m = Math.Min(L[i], B[i]);` `        ``// print NO if the above``        ``// two conditions fail``        ``// at least once``        ``else``        ``{``            ``return` `0;``        ``}``    ``}``    ``return` `1;``}` `// Driver code``public` `static` `void` `Main ()``{``    ``// initialize the number``    ``// of rectangles``    ``int` `n = 3;``    ` `    ``// initialize n rectangles with``    ``// length and breadth``    ``int` `[]L = { 5, 5, 6 };``    ``int` `[]B = { 6, 7, 8 };``    ``if``(rotateRec(n, L, B) == 1)``    ``Console.WriteLine(``"YES"``);``    ``else``    ``Console.WriteLine(``"NO"``);``}``}` `// This code is contributed``// by inder_verma`

## PHP

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## Javascript

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Output:

`NO`

Time Complexity: O(n), since there runs a loop for n times.
Auxiliary Space: O(1), since no extra space has been taken.

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