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

• Last Updated : 13 May, 2021

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-ascendingint 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 codeint 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 approachimport 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 approachimport sys; # Function to check if it possible# to form rectangles with heights# as non-ascendingdef 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 rectanglesn = 3; # initialize n rectangles# with length and breadthL = [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 approachusing System; class GFG{     // Function to check if it possible// to form rectangles with heights// as non-ascendingstatic 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 codepublic 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



## Javascript


Output:
NO

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