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Store duplicate keys-values pair and sort the key-value pair by key
  • Last Updated : 28 Apr, 2020

Given N key-value pairs that contain duplicate keys and values, the task is to store these pairs and sort the pairs by key.

Examples:

Input :
N : 10
Keys : 5 1 4 6 8 0 6 6 5 5
values: 0 1 2 3 4 5 6 7 8 9
Output :
Keys : 0 1 4 5 5 5 6 6 6 8
values: 5 1 2 0 8 9 3 6 7 4
Explanation:
We have given 10 key, value pairs which contain duplicate keys and values.
The key value pair is {(5, 0), (1, 1), (4, 2), (6, 3), (8, 4), (0, 5), (6, 6),
(6, 7), (5, 8), (5, 9)} and we want to store these key value pair and sort these
key value pair by keys. So, the expected output is {(0, 5), (1, 1), (4, 2), (5, 0),
(5, 8), (5, 9), (6, 3), (6, 6), (6, 7), (8, 4)}. Because the sorted
increasing order of keys is {0, 1, 4, 5, 5, 5, 6, 6, 6, 8}.

Approach:

To solve the problem mentioned above we can use separate arrays to store keys and values and then we can simply sort keys by Merge sort algorithm. We can use any sorting Algorithm but Mergesort is the fastest standard sort algorithm and parallelly we can perform the same operations on values array which is performed on the keys so that the key-value pair will stay at the same index on both the array.



Below is the implementation of the above approach:




// Java program to Store duplicate
// keys-values pair and sort the
// key-value pair by key
import java.util.*;
import java.lang.*;
import java.io.*;
  
class Solution {
  
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    void merge(int arrk[], int l, int m,
            int r, int arrv[])
    {
        // Sizes of two subarrays
        // that are to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
  
        /* Create temporary arrays */
        int L[] = new int[n1];
        int R[] = new int[n2];
        int Lk[] = new int[n1];
        int Rk[] = new int[n2];
  
        /* Copy data to temporary arrays */
        for (int i = 0; i < n1; ++i) {
            L[i] = arrk[l + i];
            Lk[i] = arrv[l + i];
        }
        for (int j = 0; j < n2; ++j) {
            R[j] = arrk[m + 1 + j];
            Rk[j] = arrv[m + 1 + j];
        }
  
        // Initial indexes of
        // first and second subarrays
        int i = 0, j = 0;
  
        int k = l;
  
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) {
                arrk[k] = L[i];
                arrv[k] = Lk[i];
  
                i++;
            }
            else {
                arrk[k] = R[j];
                arrv[k] = Rk[j];
  
                j++;
            }
  
            k++;
        }
  
        /* Copy remaining elements of L[] if any */
        while (i < n1) {
            arrk[k] = L[i];
            arrv[k] = Lk[i];
  
            i++;
            k++;
        }
  
        /* Copy remaining elements of R[] if any */
        while (j < n2) {
            arrk[k] = R[j];
            arrv[k] = Rk[j];
  
            j++;
            k++;
        }
    }
  
    // Function that sorts arr[l..r] using merge()
    void sort(int arrk[], int l, int r, int arrv[])
    {
        if (l < r) {
            // Find the middle point
            int m = (l + r) / 2;
  
            // Sort first and second halves
            sort(arrk, l, m, arrv);
            sort(arrk, m + 1, r, arrv);
  
            // Merge the sorted halves
            merge(arrk, l, m, r, arrv);
        }
    }
  
    /* Function to print array of size n */
    static void printArray(int arr[])
    {
        int n = arr.length;
  
        for (int i = 0; i < n; ++i)
  
            System.out.print(arr[i] + " ");
  
        System.out.println();
    }
  
    // Driver code
    public static void main(String[] args)
                    throws java.lang.Exception
    {
  
        // Size of Array
        int n = 10;
  
        // array of keys
        int[] arrk = { 5, 1, 4, 6, 8, 0
                                    6, 6, 5, 5 };
  
        // array of values
        int[] arrv = { 0, 1, 2, 3, 4, 5
                                    6, 7, 8, 9 };
  
        Solution ob = new Solution();
  
        ob.sort(arrk, 0, n - 1, arrv);
  
        System.out.print("Keys: ");
        printArray(arrk);
  
        System.out.println();
  
        System.out.print("Values: ");
        printArray(arrv);
    }
}


Output:

Keys: 0 1 4 5 5 5 6 6 6 8 

Values: 5 1 2 0 8 9 3 6 7 4

Time Complexity: O(N*logN)

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