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:
// Java code to sort the array// according to second elementimport java.io.*;import java.util.*; // User defined Pair classclass Pair { int x; int y; // Constructorpublic Pair(int x, int y) { this.x = x; this.y = y; }} // class to define user defined conparatorclass 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 classclass GFG { // Driver codepublic 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); }} |
Output:
3 1 1 2 4 3 10 8 10 20
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