Store duplicate keys-values pair and sort the key-value pair by key
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:
C++
// CPP code to implement the approach #include <bits/stdc++.h> using namespace std; // 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[n1]; int R[n2]; int Lk[n1]; int Rk[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 */ void printArray( int arr[], int n) { for ( int i = 0; i < n; ++i) cout << arr[i] << " " ; cout << endl; } // Driver code int main() { // Size of Array int n = 10; // array of keys int arrk[] = { 5, 1, 4, 6, 8, 0, 6, 6, 5, 5 }; // size of array int N = sizeof (arrk) / sizeof (arrk[0]); // array of values int arrv[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }; sort(arrk, 0, n - 1, arrv); cout << "Keys: " ; printArray(arrk, N); cout << endl; cout << "Values: " ; printArray(arrv, N); } // This code is contributed by sanjoy_62. |
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> |
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)
Space Complexity: O(N)