Sort an array of pairs using Java Pair and Comparator

Given an array of pairs of integers. The task is to sort the array with respect to second element of the pair.

Examples:

Input: [(1, 2), (3, 5), (2, 6), (1, 7)]
Output: [(1, 2), (3, 5), (2, 6), (1, 7)]

Input: [(10, 20), (20, 30), (5, 6), (2, 5)]
Output: [(2, 5), (5, 6), (10, 20), (20, 30)]

Approach:

  • Store the pairs in an array using a user defined Pair class.
  • Override the comparator method to sort the array according to the first element.
  • Sort the array according to the first element.

Below is the implementation of above approach:

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// Java code to sort the array
// according to second element
import java.io.*;
import java.util.*;
  
// User defined Pair class
class Pair {
    int x;
    int y;
  
    // Constructor
public Pair(int x, int y)
    {
        this.x = x;
        this.y = y;
    }
}
  
// class to define user defined conparator
class Compare {
  
    static void compare(Pair arr[], int n)
    {
        // Comparator to sort the pair according to second element
        Arrays.sort(arr, new Comparator<Pair>() {
            @Override public int compare(Pair p1, Pair p2)
            {
                return p1.y - p2.y;
            }
        });
  
        for (int i = 0; i < n; i++) {
            System.out.print(arr[i].x + " " + arr[i].y + " ");
        }
        System.out.println();
    }
}
  
// Driver class
class GFG {
  
    // Driver code
public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
  
        // length of array
        int n = 5;
  
        // Array of Pair
        Pair arr[] = new Pair[n];
  
        arr[0] = new Pair(10, 20);
        arr[1] = new Pair(1, 2);
        arr[2] = new Pair(3, 1);
        arr[3] = new Pair(10, 8);
        arr[4] = new Pair(4, 3);
  
        Compare obj = new Compare();
  
        obj.compare(arr, n);
    }
}

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Output:


3 1 1 2 4 3 10 8 10 20







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