Given two arrays A1[] and A2[], sort A1 in such a way that the relative order among the elements will be same as those are in A2. For the elements not present in A2, append them at last in sorted order.
Example:
Input: A1[] = {2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8}
A2[] = {2, 1, 8, 3}
Output: A1[] = {2, 2, 1, 1, 8, 8, 3, 5, 6, 7, 9}Input: A1[] = {4, 5, 1, 1, 3, 2}
A2[] = {3, 1}
Output: A1[] = {3, 1, 1, 2, 4, 5}
We strongly recommend that you click here and practice it, before moving on to the solution.
Sort an array according to the order defined by another array using Sorting and Binary Search:
The idea is to sort the A1[] array and then according to A2[] store the elements.
Let the size of A1[] be m and the size of A2[] be n.
- Create a temporary array temp of size m and copy the contents of A1[] to it.
- Create another array visited[] and initialize all entries in it as false. visited[] is used to mark those elements in temp[] which are copied to A1[].
- Sort temp[]
- Initialize the output index ind as 0.
- Do following for every element of A2[i] in A2[]
- Binary search for all occurrences of A2[i] in temp[], if present then copy all occurrences to A1[ind] and increment ind. Also mark the copied elements visited[]
- Copy all unvisited elements from temp[] to A1[]
Below image is a dry run of the above approach:
Below is the implementation of the above approach:
C++
// A C++ program to sort an array according to the order// defined by another array#include <bits/stdc++.h>using namespace std;// A Binary Search based function to find index of FIRST// occurrence of x in arr[]. If x is not present, then it// returns -1// The same can be done using the lower_bound// function in C++ STLint first(int arr[], int low, int high, int x, int n){ // Checking condition if (high >= low) { // FInd the mid element int mid = low + (high - low) / 2; // Check if the element is the extreme left // in the left half of the array if ((mid == 0 || x > arr[mid - 1]) && arr[mid] == x) return mid; // If the element lies on the right half if (x > arr[mid]) return first(arr, (mid + 1), high, x, n); // Check for element in the left half return first(arr, low, (mid - 1), x, n); } // ELement not found return -1;}// Sort A1[0..m-1] according to the order defined by// A2[0..n-1].void sortAccording(int A1[], int A2[], int m, int n){ // The temp array is used to store a copy of A1[] and // visited[] is used mark the visited elements in // temp[]. int temp[m], visited[m]; for (int i = 0; i < m; i++) { temp[i] = A1[i]; visited[i] = 0; } // Sort elements in temp sort(temp, temp + m); // for index of output which is sorted A1[] int ind = 0; // Consider all elements of A2[], find them in temp[] // and copy to A1[] in order. for (int i = 0; i < n; i++) { // Find index of the first occurrence of A2[i] in // temp int f = first(temp, 0, m - 1, A2[i], m); // If not present, no need to proceed if (f == -1) continue; // Copy all occurrences of A2[i] to A1[] for (int j = f; (j < m && temp[j] == A2[i]); j++) { A1[ind++] = temp[j]; visited[j] = 1; } } // Now copy all items of temp[] // which are not present in A2[] for (int i = 0; i < m; i++) if (visited[i] == 0) A1[ind++] = temp[i];}// Utility function to print an arrayvoid printArray(int arr[], int n){ // Iterate in the array for (int i = 0; i < n; i++) cout << arr[i] << " "; cout << endl;}// Driver Codeint main(){ int A1[] = { 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8 }; int A2[] = { 2, 1, 8, 3 }; int m = sizeof(A1) / sizeof(A1[0]); int n = sizeof(A2) / sizeof(A2[0]); // Prints the sorted array cout << "Sorted array is \n"; sortAccording(A1, A2, m, n); printArray(A1, m); return 0;} |
Java
// A JAVA program to sort an array according// to the order defined by another arrayimport java.io.*;import java.util.Arrays;class GFG { /* A Binary Search based function to find index of FIRST occurrence of x in arr[]. If x is not present, then it returns -1 */ static int first(int arr[], int low, int high, int x, int n) { if (high >= low) { /* (low + high)/2; */ int mid = low + (high - low) / 2; if ((mid == 0 || x > arr[mid - 1]) && arr[mid] == x) return mid; if (x > arr[mid]) return first(arr, (mid + 1), high, x, n); return first(arr, low, (mid - 1), x, n); } return -1; } // Sort A1[0..m-1] according to the order // defined by A2[0..n-1]. static void sortAccording(int A1[], int A2[], int m, int n) { // The temp array is used to store a copy // of A1[] and visited[] is used to mark the // visited elements in temp[]. int temp[] = new int[m], visited[] = new int[m]; for (int i = 0; i < m; i++) { temp[i] = A1[i]; visited[i] = 0; } // Sort elements in temp Arrays.sort(temp); // for index of output which is sorted A1[] int ind = 0; // Consider all elements of A2[], find them // in temp[] and copy to A1[] in order. for (int i = 0; i < n; i++) { // Find index of the first occurrence // of A2[i] in temp int f = first(temp, 0, m - 1, A2[i], m); // If not present, no need to proceed if (f == -1) continue; // Copy all occurrences of A2[i] to A1[] for (int j = f; (j < m && temp[j] == A2[i]); j++) { A1[ind++] = temp[j]; visited[j] = 1; } } // Now copy all items of temp[] which are // not present in A2[] for (int i = 0; i < m; i++) if (visited[i] == 0) A1[ind++] = temp[i]; } // Utility function to print an array static void printArray(int arr[], int n) { for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); System.out.println(); } // Driver program to test above function. public static void main(String args[]) { int A1[] = { 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8 }; int A2[] = { 2, 1, 8, 3 }; int m = A1.length; int n = A2.length; System.out.println("Sorted array is "); sortAccording(A1, A2, m, n); printArray(A1, m); }}/*This code is contributed by Nikita Tiwari.*/ |
Python3
"""A Python 3 program to sort an array according to the order defined by another array""""""A Binary Search based function to find index of FIRST occurrence of x in arr[].If x is not present, then it returns -1 """def first(arr, low, high, x, n): if (high >= low): mid = low + (high - low) // 2 # (low + high)/2 if ((mid == 0 or x > arr[mid-1]) and arr[mid] == x): return mid if (x > arr[mid]): return first(arr, (mid + 1), high, x, n) return first(arr, low, (mid - 1), x, n) return -1# Sort A1[0..m-1] according to the order# defined by A2[0..n-1].def sortAccording(A1, A2, m, n): """The temp array is used to store a copy of A1[] and visited[] is used mark the visited elements in temp[].""" temp = [0] * m visited = [0] * m for i in range(0, m): temp[i] = A1[i] visited[i] = 0 # Sort elements in temp temp.sort() # for index of output which is sorted A1[] ind = 0 """Consider all elements of A2[], find them in temp[] and copy to A1[] in order.""" for i in range(0, n): # Find index of the first occurrence # of A2[i] in temp f = first(temp, 0, m-1, A2[i], m) # If not present, no need to proceed if (f == -1): continue # Copy all occurrences of A2[i] to A1[] j = f while (j < m and temp[j] == A2[i]): A1[ind] = temp[j] ind = ind + 1 visited[j] = 1 j = j + 1 # Now copy all items of temp[] which are # not present in A2[] for i in range(0, m): if (visited[i] == 0): A1[ind] = temp[i] ind = ind + 1# Utility function to print an arraydef printArray(arr, n): for i in range(0, n): print(arr[i], end=" ") print("")# Driver program to test above function.A1 = [2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8]A2 = [2, 1, 8, 3]m = len(A1)n = len(A2)print("Sorted array is ")sortAccording(A1, A2, m, n)printArray(A1, m)# This code is contributed by Nikita Tiwari. |
C#
// A C# program to sort an array according// to the order defined by another arrayusing System;class GFG { /* A Binary Search based function to find index of FIRST occurrence of x in arr[]. If x is not present, then it returns -1 */ static int first(int[] arr, int low, int high, int x, int n) { if (high >= low) { /* (low + high)/2; */ int mid = low + (high - low) / 2; if ((mid == 0 || x > arr[mid - 1]) && arr[mid] == x) return mid; if (x > arr[mid]) return first(arr, (mid + 1), high, x, n); return first(arr, low, (mid - 1), x, n); } return -1; } // Sort A1[0..m-1] according to the order // defined by A2[0..n-1]. static void sortAccording(int[] A1, int[] A2, int m, int n) { // The temp array is used to store a copy // of A1[] and visited[] is used to mark // the visited elements in temp[]. int[] temp = new int[m]; int[] visited = new int[m]; for (int i = 0; i < m; i++) { temp[i] = A1[i]; visited[i] = 0; } // Sort elements in temp Array.Sort(temp); // for index of output which is // sorted A1[] int ind = 0; // Consider all elements of A2[], find // them in temp[] and copy to A1[] in // order. for (int i = 0; i < n; i++) { // Find index of the first occurrence // of A2[i] in temp int f = first(temp, 0, m - 1, A2[i], m); // If not present, no need to proceed if (f == -1) continue; // Copy all occurrences of A2[i] to A1[] for (int j = f; (j < m && temp[j] == A2[i]); j++) { A1[ind++] = temp[j]; visited[j] = 1; } } // Now copy all items of temp[] which are // not present in A2[] for (int i = 0; i < m; i++) if (visited[i] == 0) A1[ind++] = temp[i]; } // Utility function to print an array static void printArray(int[] arr, int n) { for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); Console.WriteLine(); } // Driver program to test above function. public static void Main() { int[] A1 = { 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8 }; int[] A2 = { 2, 1, 8, 3 }; int m = A1.Length; int n = A2.Length; Console.WriteLine("Sorted array is "); sortAccording(A1, A2, m, n); printArray(A1, m); }}// This code is contributed by nitin mittal. |
Javascript
<script>// A JavaScript program to sort an array according// to the order defined by another array /* A Binary Search based function to find index of FIRST occurrence of x in arr[]. If x is not present, then it returns -1 */ function first(arr,low,high,x,n) { if (high >= low) { // (low + high)/2; let mid = low + Math.floor((high - low) / 2); if ((mid == 0 || x > arr[mid - 1]) && arr[mid] == x) return mid; if (x > arr[mid]) return first(arr, (mid + 1), high,x, n); return first(arr, low, (mid - 1), x, n); } return -1; } // Sort A1[0..m-1] according to the order // defined by A2[0..n-1]. function sortAccording(A1,A2,m,n) { // The temp array is used to store a copy // of A1[] and visited[] is used to mark the // visited elements in temp[]. let temp=[]; let visited=[]; for (let i = 0; i < m; i++) { temp[i] = A1[i]; visited[i] = 0; } // Sort elements in temp temp.sort(function(a, b){return a-b}); // for index of output which is sorted A1[] let ind = 0; // Consider all elements of A2[], find them // in temp[] and copy to A1[] in order. for (let i = 0; i < n; i++) { // Find index of the first occurrence // of A2[i] in temp let f = first(temp, 0, m - 1, A2[i], m); // If not present, no need to proceed if (f == -1) { continue; } // Copy all occurrences of A2[i] to A1[] for (let j = f; (j < m && temp[j] == A2[i]);j++) { A1[ind++] = temp[j]; visited[j] = 1; } } // Now copy all items of temp[] which are // not present in A2[] for (let i = 0; i < m; i++) { if (visited[i] == 0) A1[ind++] = temp[i]; } } // Utility function to print an array function printArray(arr,n) { for (let i = 0; i < n; i++) { document.write(arr[i] + " "); } document.write("<br>"); } // Driver program to test above function. let A1=[2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8 ]; let A2=[2, 1, 8, 3 ]; let m = A1.length; let n = A2.length; document.write("Sorted array is <br>"); sortAccording(A1, A2, m, n); printArray(A1, m); // This code is contributed by avanitrachhadiya2155 </script> |
PHP
<?php // A PHP program to sort an array according // to the order defined by another array/* A Binary Search based function to find index of FIRST occurrence of x in arr[]. If x is not present, then it returns -1 */function first(&$arr, $low, $high, $x, $n){ if ($high >= $low) { $mid = intval($low + ($high - $low) / 2); if (($mid == 0 || $x > $arr[$mid - 1]) && $arr[$mid] == $x) return $mid; if ($x > $arr[$mid]) return first($arr, ($mid + 1), $high, $x, $n); return first($arr, $low, ($mid - 1), $x, $n); } return -1;}// Sort A1[0..m-1] according to the order // defined by A2[0..n-1].function sortAccording(&$A1, &$A2, $m, $n){ // The temp array is used to store a copy // of A1[] and visited[] is used mark the // visited elements in temp[]. $temp = array_fill(0, $m, NULL); $visited = array_fill(0, $m, NULL); for ($i = 0; $i < $m; $i++) { $temp[$i] = $A1[$i]; $visited[$i] = 0; } // Sort elements in temp sort($temp); $ind = 0; // for index of output which is sorted A1[] // Consider all elements of A2[], find // them in temp[] and copy to A1[] in order. for ($i = 0; $i < $n; $i++) { // Find index of the first occurrence // of A2[i] in temp $f = first($temp, 0, $m - 1, $A2[$i], $m); // If not present, no need to proceed if ($f == -1) continue; // Copy all occurrences of A2[i] to A1[] for ($j = $f; ($j < $m && $temp[$j] == $A2[$i]); $j++) { $A1[$ind++] = $temp[$j]; $visited[$j] = 1; } } // Now copy all items of temp[] which // are not present in A2[] for ($i = 0; $i < $m; $i++) if ($visited[$i] == 0) $A1[$ind++] = $temp[$i];}// Utility function to print an arrayfunction printArray(&$arr, $n){ for ($i = 0; $i < $n; $i++) echo $arr[$i] . " "; echo "\n";}// Driver Code$A1 = array(2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8);$A2 = array(2, 1, 8, 3);$m = sizeof($A1);$n = sizeof($A2);echo "Sorted array is \n";sortAccording($A1, $A2, $m, $n);printArray($A1, $m);// This code is contributed by ita_c?> |
Output
Sorted array is 2 2 1 1 8 8 3 5 6 7 9
Time complexity: O(M Log M + N Log M), Sorting Arr1[] of size M i.e M log M and searching of Arr2[] elements of size N in Arr1[] i.e N log M
Auxiliary Space: O(M), visited and temp array of size M for storing Arr1[].
Sort an array according to the order defined by another array using Self-Balancing Binary Search Tree:
We can also use a self-balancing BST like AVL Tree, Red Black Tree, etc. Following are detailed steps.
- Create a self-balancing BST of all elements in A1[]. In every node of BST, also keep track of the count of occurrences of the key and a bool field visited which is initialized as false for all nodes.
- Initialize the output index ind as 0.
- Do the following for every element of A2[i] in A2[]
- Search for A2[i] in the BST, if present then copy all occurrences to A1[ind] and increment ind. Also, mark the copied elements visited in the BST node.
- Do an in-order traversal of BST and copy all unvisited keys to A1[].
Time complexity: O(M Log M + N Log M), M Log M for making self balancing bst of arr1[] of size M.
Auxiliary Space: O(M), space for making self balancing bst for Arr1[] of size M.
Sort an array according to the order defined by another array using Hashing:
The idea is to use hashing. Store the frequency of A1[] and decrement the frequency in the A2[] order.
Follow the steps to solve the problem:
- Loop through A1[], store the count of every number in a HashMap (key: number, value: count of number)
- Loop through A2[], check if it is present in HashMap, if so, put in output array that many times and remove the number from HashMap.
- Sort the rest of the numbers present in HashMap and put in the output array.
Below is the implementation of the above approach:
C++
// A C++ program to sort an array according to the order// defined by another array#include <bits/stdc++.h>using namespace std;// function to sort A1 according to A2 using hash map in C++void sortA1ByA2(int A1[], int N, int A2[], int M, int ans[]){ map<int, int> mp; // indexing for answer = ans array int ind = 0; // initially storing frequency of each element of A1 in // map [ key, value ] = [ A1[i] , frequency[ A1[i] ] ] for (int i = 0; i < N; i++) { mp[A1[i]] += 1; } // traversing each element of A2, first come first serve for (int i = 0; i < M; i++) { // checking if current element of A2 is present in // A1 or not if not present go to next iteration // else store number of times it is appearing in A1 // in ans array if (mp[A2[i]] != 0) { // mp[ A2[i] ] = frequency of A2[i] element in // A1 array for (int j = 1; j <= mp[A2[i]]; j++) ans[ind++] = A2[i]; } // to avoid duplicate storing of same element of A2 // in ans array mp.erase(A2[i]); } // store the remaining elements of A1 in sorted order in // ans array for (auto it : mp) { // it.second = frequency of remaining elements for (int j = 1; j <= it.second; j++) ans[ind++] = it.first; }}// Utility function to print an arrayvoid printArray(int arr[], int n){ // Iterate in the array for (int i = 0; i < n; i++) cout << arr[i] << " "; cout << endl;}// Driver Codeint main(){ int A1[] = { 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8 }; int A2[] = { 2, 1, 8, 3 }; int n = sizeof(A1) / sizeof(A1[0]); int m = sizeof(A2) / sizeof(A2[0]); // The ans array is used to store the final sorted array int ans[n]; sortA1ByA2(A1, n, A2, m, ans); // Prints the sorted array cout << "Sorted array is \n"; printArray(ans, n); return 0;} |
Java
// A Java program to sort an array according to the order// defined by another arrayimport java.io.*;import java.util.Arrays;import java.util.HashMap;class GFG { // function to sort A1 according to A2 using hash map in // C++ static void sortA1ByA2(int A1[], int N, int A2[], int M, int ans[]) { HashMap<Integer, Integer> mp = new HashMap<>(); // indexing for answer = ans array int ind = 0; // initially storing frequency of each element of A1 // in map [ key, value ] = [ A1[i] , frequency[ // A1[i] ] ] for (int i = 0; i < N; i++) { if (!mp.containsKey(A1[i])) mp.put(A1[i], 1); else mp.put(A1[i], mp.get(A1[i]) + 1); } // traversing each element of A2, first come first // serve for (int i = 0; i < M; i++) { // checking if current element of A2 is present // in A1 or not if not present go to next // iteration else store number of times it is // appearing in A1 in ans array if (mp.containsKey(A2[i])) { // mp[ A2[i] ] = frequency of A2[i] element // in A1 array for (int j = 1; j <= mp.get(A2[i]); j++) ans[ind++] = A2[i]; } // to avoid duplicate storing of same element of // A2 in ans array mp.remove(A2[i]); } int idx = ind; // store the remaining elements of A1 in sorted // order in ans array for (HashMap.Entry<Integer, Integer> it : mp.entrySet()) { // it.second = frequency of remaining elements for (int j = 1; j <= it.getValue(); j++) ans[ind++] = it.getKey(); } Arrays.sort(ans, idx, N); } // Utility function to print an array static void printArray(int arr[], int n) { for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); System.out.println(); } // Driver code public static void main(String[] args) { int A1[] = { 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8 }; int A2[] = { 2, 1, 8, 3 }; int n = A1.length; int m = A2.length; // The ans array is used to store the final sorted // array int ans[] = new int[n]; sortA1ByA2(A1, n, A2, m, ans); System.out.println("Sorted array is "); printArray(ans, n); }}// This code is contributed by Pushpesh Raj. |
Python3
from collections import Counter# Function to sort arr1# according to arr2def solve(arr1, arr2): # Our output array res = [] # Counting Frequency of each # number in arr1 f = Counter(arr1) # Iterate over arr2 and append all # occurrences of element of # arr2 from arr1 for e in arr2: # Appending element 'e', # f[e] number of times res.extend([e]*f[e]) # Count of 'e' after appending is zero f[e] = 0 # Remaining numbers in arr1 in sorted # order (Numbers with non-zero frequency) rem = list(sorted(filter( lambda x: f[x] != 0, f.keys()))) # Append them also for e in rem: res.extend([e]*f[e]) return res# Driver Codeif __name__ == "__main__": arr1 = [2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8] arr2 = [2, 1, 8, 3] print(*solve(arr1, arr2)) |
C#
// A C# program to sort an array according to the order// defined by another arrayusing System;using System.Collections.Generic;class GFG { // Utility function to print an array static void printArray(int[] arr, int n) { // Iterate in the array for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); Console.WriteLine(); } // function to sort A1 according to A2 using hash map in // C++ static void sortA1ByA2(int[] A1, int N, int[] A2, int M, int[] ans) { Dictionary<int, int> mp = new Dictionary<int, int>(); // indexing for answer = ans array int ind = 0; // initially storing frequency of each element of A1 // in map [ key, value ] = [ A1[i] , frequency[ // A1[i] ] ] for (int i = 0; i < N; i++) { if (mp.ContainsKey(A1[i])) { var val = mp[A1[i]]; mp.Remove(A1[i]); mp.Add(A1[i], val + 1); } else mp.Add(A1[i], 1); } // traversing each element of A2, first come first // serve for (int i = 0; i < M; i++) { // checking if current element of A2 is present // in A1 or not if not present go to next // iteration else store number of times it is // appearing in A1 in ans array if (mp[A2[i]] != 0) { // mp[ A2[i] ] = frequency of A2[i] element // in A1 array for (int j = 1; j <= mp[A2[i]]; j++) ans[ind++] = A2[i]; } // to avoid duplicate storing of same element of // A2 in ans array mp.Remove(A2[i]); } // store the remaining elements of A1 in sorted // order in ans array foreach(KeyValuePair<int, int> it in mp) { // it.second = frequency of remaining elements for (int j = 1; j <= it.Value; j++) ans[ind++] = it.Key; } } static void Main() { int[] A1 = { 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8 }; int[] A2 = { 2, 1, 8, 3 }; int n = A1.Length; int m = A2.Length; // The ans array is used to store the final sorted // array int[] ans = new int[n]; sortA1ByA2(A1, n, A2, m, ans); // Prints the sorted array Console.WriteLine("Sorted array is "); printArray(ans, n); }}// This code is contributed by garg28harsh. |
Javascript
<script>// A JavaScript program to sort an array according to the order// defined by another array// function to sort A1 according to A2 using hash map in C++function sortA1ByA2(A1, N, A2, M, ans){ let mp = new Map(); // indexing for answer = ans array let ind = 0; // initially storing frequency of each element of A1 in // map [ key, value ] = [ A1[i] , frequency[ A1[i] ] ] for (let i = 0; i < N; i++) { if(!mp.has(A1[i])){ mp.set(A1[i],1); } else mp.set(A1[i],mp.get(A1[i]) + 1); } // traversing each element of A2, first come first serve for (let i = 0; i < M; i++) { // checking if current element of A2 is present in // A1 or not if not present go to next iteration // else store number of times it is appearing in A1 // in ans array if (mp.has(A2[i])) { // mp[ A2[i] ] = frequency of A2[i] element in // A1 array for (let j = 1; j <= mp.get(A2[i]); j++) ans[ind++] = A2[i]; } // to avoid duplicate storing of same element of A2 // in ans array mp.delete(A2[i]); } // store the remaining elements of A1 in sorted order in // ans array mp = new Map([...mp.entries()].sort()); for (let [key,value] of mp) { // it.second = frequency of remaining elements for (let j = 1; j <= value; j++) ans[ind++] = key; }}// Utility function to print an arrayfunction printArray(arr, n){ // Iterate in the array for (let i = 0; i <n; i++) document.write(arr[i]," "); document.write("</br>");}// Driver Codelet A1 = [ 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8 ];let A2 = [ 2, 1, 8, 3 ];let n = A1.length;let m = A2.length;// The ans array is used to store the final sorted arraylet ans = new Array(n);sortA1ByA2(A1, n, A2, m, ans);// Prints the sorted arraydocument.write("Sorted array is ");printArray(ans, n);// This code is contributed by shinjanpatra</script> |
Output
Sorted array is 2 2 1 1 8 8 3 5 6 7 9
Time Complexity: O(M + N), Traversing over both the array
Auxiliary Space: O(M), Space for storing frequency of arr1[] of size M.
Sort an array according to the order defined by another array By Writing a Customized Comparator Method:
The idea is to make a customized comparator.
Follow the steps below to solve the problem:
- If num1 and num2 both are in A2 then the number with a lower index in A2 will be treated smaller than others.
- If only one of num1 or num2 present in A2, then that number will be treated smaller than the other which doesn’t present in A2.
- If both are not in A2, then the natural ordering will be taken.
The time complexity of this method is O(mnLogm) if we use a O(nLogn) time complexity sorting algorithm. We can improve time complexity to O(mLogm) by using a Hashing instead of doing linear search.
Below is the implementation of the above approach:
C++
// A C++ program to sort an array according to the order// defined by another array#include <bits/stdc++.h>using namespace std;// function that sorts the first array based on order of// them in second arrayvoid sortA1ByA2(vector<int>& arr1, vector<int>& arr2){ // map to store the indices of second array // so that we can easily judge the position of two // elements in first array unordered_map<int, int> index; for (int i = 0; i < arr2.size(); i++) { // assigning i+1 // because by default value of map is zero // Consider only first occurrence of element if (index[arr2[i]] == 0) { index[arr2[i]] = i + 1; } } // comparator function that sorts arr1 based on order // defined in arr2 auto comp = [&](int a, int b) { // if indices of two elements are equal // we need to sort them in increasing order if (index[a] == 0 && index[b] == 0) return a < b; // if a not present in arr2 then b should come // before it if (index[a] == 0) return false; // if b not present in arr2 then no swap if (index[b] == 0) return true; // sorting in increasing order return index[a] < index[b]; }; sort(arr1.begin(), arr1.end(), comp);}int main(){ vector<int> arr1{ 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8, 7, 5, 6, 9, 7, 5 }; vector<int> arr2{ 2, 1, 8, 3, 4, 1 }; sortA1ByA2(arr1, arr2); // printing the array cout << "Sorted array is \n"; for (auto i : arr1) { cout << i << " "; } return 0;} |
C
// A C++ program to sort an array according to the order// defined by another array#include <stdio.h>#include <stdlib.h>// A2 is made global here so that it can be accessed by// compareByA2() The syntax of qsort() allows only two// parameters to compareByA2()int A2[5];// size of A2[]int size = 5;int search(int key){ int i = 0, idx = 0; for (i = 0; i < size; i++) if (A2[i] == key) return i; return -1;}// A custom compare method to compare elements of A1[]// according to the order defined by A2[].int compareByA2(const void* a, const void* b){ int idx1 = search(*(int*)a); int idx2 = search(*(int*)b); if (idx1 != -1 && idx2 != -1) return idx1 - idx2; else if (idx1 != -1) return -1; else if (idx2 != -1) return 1; else return (*(int*)a - *(int*)b);}// This method mainly uses qsort to sort A1[] according to// A2[]void sortA1ByA2(int A1[], int size1){ qsort(A1, size1, sizeof(int), compareByA2);}// Driver program to test above functionint main(int argc, char* argv[]){ int A1[] = { 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8, 7, 5, 6, 9, 7, 5 }; // A2[] = {2, 1, 8, 3, 4}; A2[0] = 2; A2[1] = 1; A2[2] = 8; A2[3] = 3; A2[4] = 4; int size1 = sizeof(A1) / sizeof(A1[0]); sortA1ByA2(A1, size1); printf("Sorted Array is "); int i; for (i = 0; i < size1; i++) printf("%d ", A1[i]); return 0;} |
Java
// A Java program to sort an array according to the order// defined by another arrayimport java.io.*;import java.util.Arrays;import java.util.HashMap;class GFG { static void sortAccording(int[] A1, int[] A2, int m, int n) { // HashMap to store the indices of elements in // the second array HashMap<Integer, Integer> index = new HashMap<>(); for (int i = 0; i < n; i++) { // Consider only first occurrence of element if (!index.containsKey(A2[i])) // Assign value of i+1 index.put(A2[i], i + 1); } // Since Java does not support custom comparators on // primitive data types, we box the elements in // wrapper classes. // Sorted values are stored in a temporary // array. int[] tmp = Arrays.stream(A1) .boxed() .sorted((p1, p2) -> { int idx1 = index.getOrDefault(p1, 0); int idx2 = index.getOrDefault(p2, 0); // If both p1 and p2 are not present // in the second array, // sort them in ascending order if (idx1 == 0 && idx2 == 0) return p1 - p2; // If only p2 is present in the second // array, p2 comes before p1 if (idx1 == 0) return 1; // If only p1 is present in the second // array, p1 comes before p2 (no swap) if (idx2 == 0) return -1; // If both p1 and p2 are present in // the second array, sort them // according to their respective // indices return idx1 - idx2; }) .mapToInt(i -> i) .toArray(); // Sorted values are copied to the original // array for (int i = 0; i < m; i++) { A1[i] = tmp[i]; } } // Driver program to test the above function public static void main(String[] args) { int A1[] = { 2, 1, 2, 5, 7, 1, 9, 9, 3, 6, 8, 8 }; int A2[] = { 2, 1, 8, 3, 1 }; int m = A1.length; int n = A2.length; sortAccording(A1, A2, m, n); System.out.println("Sorted array is "); System.out.println(Arrays.toString(A1)); }}// This code is contributed by anonymouscegian |
Python3
# A python program to sort an array according to the order# defined by another arrayimport functools# function that sorts the first array based on order of# them in second arraydef sortA1ByA2(arr1, arr2): # map to store the indices of second array # so that we can easily judge the position of two # elements in first array index = {} for i in range(len(arr2)): # Consider only first occurrence of element if arr2[i] not in index.keys(): # Assign value of i+1 index[arr2[i]] = i + 1 def cmp(a, b): # If both a and b are not present in the second array, # sort them in ascending order if a not in index.keys() and b not in index.keys(): return a-b # If only b is present in the second array, b comes before a if a not in index.keys(): return 1 # If only a is present in the second array, a comes before b if b not in index.keys(): return -1 # If both a and b are present in the second array, # sort them according to their respective indices return index[a]-index[b] arr1.sort(key=functools.cmp_to_key(cmp))arr1 = [ 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8, 7, 5, 6, 9, 7, 5 ]arr2 = [ 2, 1, 8, 3, 4, 1 ]sortA1ByA2(arr1, arr2)# printing the arrayprint("Sorted array is ")for i in range(len(arr1)): print(arr1[i], end = " ") # The code is contributed by Nidhi goel. |
C#
using System;using System.Collections.Generic;class GFG{ // function that sorts the first array based on order of // them in second array static void sortA1ByA2(ref int[] arr1, int[] arr2) { // dictionary to store the indices of second array // so that we can easily judge the position of two // elements in first array Dictionary<int, int> index = new Dictionary<int, int>(); for (int i = 0; i < arr2.Length; i++) { // assigning i+1 // because by default value of dictionary is zero // Consider only first occurrence of element if (!index.ContainsKey(arr2[i])) { index[arr2[i]] = i + 1; } } // comparator function that sorts arr1 based on order // defined in arr2 Array.Sort(arr1, (a, b) => { // if indices of two elements are equal // we need to sort them in increasing order if (!index.ContainsKey(a) && !index.ContainsKey(b)) { return a.CompareTo(b); } // if a not present in arr2 then b should come // before it if (!index.ContainsKey(a)) { return 1; } // if b not present in arr2 then no swap if (!index.ContainsKey(b)) { return -1; } // sorting in increasing order return index[a].CompareTo(index[b]); }); } static void Main(string[] args) { int[] arr1 = new int[] { 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8, 7, 5, 6, 9, 7, 5 }; int[] arr2 = new int[] { 2, 1, 8, 3, 4, 1 }; sortA1ByA2(ref arr1, arr2); // printing the array Console.WriteLine("Sorted array is"); foreach (int i in arr1) { Console.Write(i + " "); } Console.WriteLine(); }}// This code is contributed by Shivam Tiwari |
Javascript
function sortAccording(A1, A2) { // Map to store the indices of elements in the second array let index = new Map(); for (let i = 0; i < A2.length; i++) { // Consider only first occurrence of element if (!index.has(A2[i])) { // Assign value of i+1 index.set(A2[i], i + 1); } } A1.sort((a, b) => { let idx1 = index.get(a) || 0; let idx2 = index.get(b) || 0; // If both a and b are not present in the second array, sort them in ascending order if (idx1 === 0 && idx2 === 0) { return a - b; } // If only b is present in the second array, b comes before a if (idx1 === 0) { return 1; } // If only a is present in the second array, a comes before b if (idx2 === 0) { return -1; } // If both a and b are present in the second array, sort them according to their respective indices return idx1 - idx2; });}// Driver Codelet A1 = [2, 1, 2, 5, 7, 1, 9, 9, 3, 6, 8, 8];let A2 = [2, 1, 8, 3, 1];sortAccording(A1, A2);console.log("Sorted array is ");console.log(A1);// This code is contributed by Shivam Tiwari |
Output
Sorted array is 2 2 1 1 8 8 3 5 5 5 6 6 7 7 7 9 9
Time complexity: O(MLogM + N), MLogM for sorting Arr1[] of size M and N for iterating over Arr2[] of size N.
Auxiliary Space: O(N), Storing first occurrence of every element of arr2[] of size N.
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