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Make N pairs from Array as (X, Y) coordinate point that are enclosed inside a minimum area rectangle

  • Last Updated : 02 Jul, 2021

Given a number N, and an array A[] of size 2N, the task is to make N pairs of these array elements and place them on an X-Y coordinate plane, such that they are enclosed inside a minimum area rectangle( with sides parallel to the X-axis and Y-axis) and print the area of the rectangle.

Examples:

Input: N = 4, A = {1, 4, 2, 5, 3, 6, 7, 8}
Output: 9
Explanation: One possible way of making N pairs to get minimum rectangle area is {(1, 5), (2, 7), (3, 6), (4, 8)}
The minimum area rectangle has been shown in the following diagram:



Note: There maybe other ways to make N pairs such that the area remains minimum, but the minimum area remains 9.

Input: N = 3, A = {1, 3, 1, 1, 2, 1}
Output: 0

 

Approach: The area of the rectangle with the bottom-left corner in (X1, Y1) and top-right corner in (X2, Y2) would be (X2 – X1)*(Y2 – Y1). Thus, the task can be presented as partitioning the array A into two N-sized partitions say X and Y, such that (Max(X) – Min(X)) * (Max(Y) – Min(Y)) is minimized. Here, X represents the X-coordinates of the pairs and Y represents the Y-coordinates.

After sorting A, the minimum would be A1, and the maximum would be A2N. Now, there can be the following two cases:

  • Both A1 and A2N are present in the same partition, say X.The area of the rectangle would be (A2N – A1) * (Max(Y) – Min(Y)). Then the task would be to minimize Max(Y) – Min(Y). Also, if i is the index of Min(Y) and j is the index of Max(Y), then j – i >= N – 1, as there has to be at least N elements in Y. Thus, it is optimal to use a segment of size N for Y(barring A1 and A2N, as they already have been taken),
  • A1 and A2N are present in different partitions. For this case, it would be optimal to use a prefix of size N, and a suffix of size N as the partitions, i.e placing the first N elements in one partition and the last N elements in the other.       

Follow the steps below to solve the problem:

  • Initialize a variable say, ans to store the minimum area of the rectangle.
  • Sort the array A[] in ascending order.
  • For the first case:
    • Update ans as (A[N – 1] – A[0]) * (A[2 * N – 1] – A[N]).
  • For the second case:
    • Iterate in the range [1, N-1] using the variable i: 
      • Update ans as the minimum of ans and (A[2 * N – 1] – A[0]) * (A[i + N – 1] – A[i]).
  • Finally, return ans.

Below is an implementation of the above code:

C++




// C++ program for tha above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to make N pairs of coordinates such that they
// are enclosed in a minimum area rectangle with sides
// parallel to the X and Y axes
int minimumRectangleArea(int A[], int N)
{
    // A variable to store the answer
    int ans;
 
    sort(A, A + 2 * N);
 
    // For the case where the maximum
    // and minimum are in different partitions
    ans = (A[N - 1] - A[0]) * (A[2 * N - 1] - A[N]);
 
    // For the case where the maximum and
    // minimum are in the same partition
    for (int i = 1; i < N; i++)
        ans = min(ans, (A[2 * N - 1] - A[0])
                           * (A[i + N - 1] - A[i]));
 
    // Return the answer
    return ans;
}
 
// Driver code
int main()
{
    // Given Input
    int A[] = { 2, 4, 1, 5, 3, 6, 7, 8 };
    int N = sizeof(A) / sizeof(A[0]);
    N /= 2;
 
    // Function call
    cout << minimumRectangleArea(A, N) << endl;
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.Arrays;
 
class GFG{
     
// Function to make N pairs of coordinates
// such that they are enclosed in a minimum
// area rectangle with sides parallel to
// the X and Y axes
public static int minimumRectangleArea(int A[], int N)
{
     
    // A variable to store the answer
    int ans;
 
    Arrays.sort(A);
 
    // For the case where the maximum
    // and minimum are in different partitions
    ans = (A[N - 1] - A[0]) * (A[2 * N - 1] - A[N]);
 
    // For the case where the maximum and
    // minimum are in the same partition
    for(int i = 1; i < N; i++)
        ans = Math.min(ans, (A[2 * N - 1] - A[0]) *
                            (A[i + N - 1] - A[i]));
 
    // Return the answer
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given Input
    int A[] = { 2, 4, 1, 5, 3, 6, 7, 8 };
    int N = A.length;
    N = (int)N / 2;
 
    // Function call
    System.out.println(minimumRectangleArea(A, N));
}
}
 
// This code is contributed by lokeshpotta20

Python3




# Python3 program for tha above approach
 
# Function to make N pairs of coordinates
# such that they are enclosed in a minimum
# area rectangle with sides parallel to the
# X and Y axes
def minimumRectangleArea(A, N):
     
    # A variable to store the answer
    ans = 0
 
    A.sort()
 
    # For the case where the maximum
    # and minimum are in different partitions
    ans = (A[N - 1] - A[0]) * (A[2 * N - 1] - A[N])
 
    # For the case where the maximum and
    # minimum are in the same partition
    for i in range(1, N, 1):
        ans = min(ans, (A[2 * N - 1] - A[0]) *
                       (A[i + N - 1] - A[i]))
 
    # Return the answer
    return ans
 
# Driver code
if __name__ == '__main__':
     
    # Given Input
    A = [ 2, 4, 1, 5, 3, 6, 7, 8 ]
    N = len(A)
    N //= 2
 
    # Function call
    print(minimumRectangleArea(A, N))
     
# This code is contributed by ipg2016107

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function to make N pairs of coordinates
// such that they are enclosed in a minimum
// area rectangle with sides parallel to
// the X and Y axes
public static int minimumRectangleArea(int []A, int N)
{
     
    // A variable to store the answer
    int ans;
 
    Array.Sort(A);
 
    // For the case where the maximum
    // and minimum are in different partitions
    ans = (A[N - 1] - A[0]) * (A[2 * N - 1] - A[N]);
 
    // For the case where the maximum and
    // minimum are in the same partition
    for(int i = 1; i < N; i++)
        ans = Math.Min(ans, (A[2 * N - 1] - A[0]) *
                            (A[i + N - 1] - A[i]));
 
    // Return the answer
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Given Input
    int []A = { 2, 4, 1, 5, 3, 6, 7, 8 };
    int N = A.Length;
    N = (int)N / 2;
 
    // Function call
    Console.Write(minimumRectangleArea(A, N));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript




<script>
 
// JavaScript program for tha above approach
 
 
// Function to make N pairs of coordinates such that they
// are enclosed in a minimum area rectangle with sides
// parallel to the X and Y axes
function minimumRectangleArea(A, N) {
// A variable to store the answer
    let ans;
 
    A.sort();
 
    // For the case where the maximum
    // and minimum are in different partitions
    ans = (A[N - 1] - A[0]) * (A[2 * N - 1] - A[N]);
 
    // For the case where the maximum and
    // minimum are in the same partition
    for (let i = 1; i < N; i++)
        ans = Math.min(ans, (A[2 * N - 1] - A[0])
            * (A[i + N - 1] - A[i]));
 
    // Return the answer
    return ans;
}
 
// Driver code
 
// Given Input
let A = [2, 4, 1, 5, 3, 6, 7, 8];
let N = A.length
N = Math.floor(N / 2);
 
// Function call
document.write(minimumRectangleArea(A, N) + "<br>");
 
</script>
Output
9

Time Complexity: O(NLogN)
Auxiliary Space: O(1)

 




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