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Store duplicate keys-values pair and sort the key-value pair by key

  • Difficulty Level : Basic
  • Last Updated : 03 Jun, 2021

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: 

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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




// 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);
    }
}

Python3




# Python program to Store duplicate
# keys-values pair and sort the
# key-value pair by key
 
# Merges two subarrays of arr[].
# First subarray is arr[l..m]
# Second subarray is arr[m+1..r]
def merge(arrk, l, m, r, arrv):
     
    # Sizes of two subarrays
    # that are to be merged
    n1 = m - l + 1;
    n2 = r - m;
     
    #Create temporary arrays
    L = [0]*n1
    R = [0]*n2
    Lk = [0]*n1
    Rk = [0]*n2
     
    #Copy data to temporary arrays
    for i in range(n1):
        L[i] = arrk[l + i];
        Lk[i] = arrv[l + i];
         
    for j in range(n2):
        R[j] = arrk[m + 1 + j];
        Rk[j] = arrv[m + 1 + j];
     
    # Initial indexes of
    # first and second subarrays   
    a = 0
    b = 0   
    k = l
     
    while(a < n1 and b < n2):
        if (L[a] <= R[b]):       
            arrk[k] = L[a];
            arrv[k] = Lk[a];
  
            a += 1;
         
        else:
         
            arrk[k] = R[b];
            arrv[k] = Rk[b];
            b += 1;       
        k += 1;
         
    # Copy remaining elements of L[] if any
    while (a < n1):
        arrk[k] = L[a];
        arrv[k] = Lk[a];
  
        a += 1;
        k += 1;
     
    # Copy remaining elements of R[] if any   
    while (b < n2):
        arrk[k] = R[b];
        arrv[k] = Rk[b];
  
        b += 1;
        k += 1;
         
# Function that sorts arr[l..r] using merge()
def sort(arrk, l, r, arrv):
    if (l < r):
       
        # Find the middle point
        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       
def printArray(arr):
     
    n = len(arr)
     
    for i in range(n):
        print(arr[i], end = " ")
     
    print()
 
#  Driver code
 
# Size of Array
n = 10;
 
# array of keys
arrk = [5, 1, 4, 6, 8,
                   0, 6, 6, 5, 5 ]
 
# array of values                  
arrv = [0, 1, 2, 3, 4,
                   5, 6, 7, 8, 9]
 
sort(arrk, 0, n - 1, arrv)
print("Keys: ", end = '')
printArray(arrk);
print()
print("Values: ", end = "")
printArray(arrv);
     
# This code is contributed by avanitrachhadiya2155

C#




// C# program to Store duplicate
// keys-values pair and sort the
// key-value pair by key
using System;
 
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 a = 0 , b = 0;
 
    int k = l;
 
    while (a < n1 && b < n2)
    {
        if (L[a] <= R[b])
        {
            arrk[k] = L[a];
            arrv[k] = Lk[a];
 
            a++;
        }
        else
        {
            arrk[k] = R[b];
            arrv[k] = Rk[b];
 
            b++;
        }
        k++;
    }
 
    // Copy remaining elements of L[] if any
    while (a < n1)
    {
        arrk[k] = L[a];
        arrv[k] = Lk[a];
 
        a++;
        k++;
    }
 
    // Copy remaining elements of R[] if any
    while (b < n2)
    {
        arrk[k] = R[b];
        arrv[k] = Rk[b];
 
        b++;
        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)
        Console.Write(arr[i] + " ");
 
    Console.WriteLine();
}
 
// Driver code
public static void Main(string[] args)
{
     
    // 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);
 
    Console.Write("Keys: ");
    printArray(arrk);
 
    Console.WriteLine();
 
    Console.Write("Values: ");
    printArray(arrv);
}
}
 
// This code is contributed by ukasp

Javascript




<script>
 
// Javascript program to Store duplicate
// keys-values pair and sort the
// key-value pair by key
 
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
function merge(arrk, l, m, r, arrv)
{
    // Sizes of two subarrays
    // that are to be merged
    var n1 = m - l + 1;
    var n2 = r - m;
     
    /* Create temporary arrays */
    var L = Array(n1).fill(0);
    var R = Array(n2).fill(0);
    var Lk = Array(n1).fill(0);
    var Rk = Array(n2).fill(0);
     
    /* Copy data to temporary arrays */
    for (var i = 0; i < n1; ++i) {
        L[i] = arrk[l + i];
        Lk[i] = arrv[l + i];
    }
    for (var j = 0; j < n2; ++j) {
        R[j] = arrk[m + 1 + j];
        Rk[j] = arrv[m + 1 + j];
    }
    // Initial indexes of
    // first and second subarrays
    var i = 0, j = 0;
    var 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()
function sort(arrk, l, r, arrv)
{
    if (l < r) {
        // Find the middle point
        var m = parseInt((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 */
function printArray(arr)
{
    var n = arr.length;
    for (var i = 0; i < n; ++i)
        document.write(arr[i] + " ");
    document.write("<br>");
}
// Driver code
 
// Size of Array
var n = 10;
// array of keys
var arrk = [ 5, 1, 4, 6, 8, 0, 6, 6, 5, 5 ]
// array of values
var arrv = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
sort(arrk, 0, n - 1, arrv);
document.write("Keys: ");
printArray(arrk);
document.write("<br>");
document.write("Values: ");
printArray(arrv);
 
</script>
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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