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Sort an array according to the order defined by another array

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  • Difficulty Level : Medium
  • Last Updated : 23 Jun, 2022

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}

The code should handle all cases like the number of elements in A2[] may be more or less compared to A1[]. A2[] may have some elements which may not be there in A1[] and vice versa is also possible.

Source: Amazon Interview | Set 110 (On-Campus) 

We strongly recommend that you click here and practice it, before moving on to the solution.

Method 1 (Using Sorting and Binary Search) 
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++ STL
int 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 array
void printArray(int arr[], int n)
{
 
    // Iterate in the array
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    cout << endl;
}
 
// Driver Code
int 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 array
import 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 array
def 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 array
using 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.


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 array
function 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
?>


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>


Output

Sorted array is 
2 2 1 1 8 8 3 5 6 7 9 

 Time complexity: The steps 1 and 2 require O(m) time. Step 3 requires O(M * Log M) time. Step 5 requires O(N Log M) time. Therefore overall time complexity is O(M Log M + N Log M).

Method 2 (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 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 following for every element of A2[i] in A2[] 
    1. 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 inorder traversal of BST and copy all unvisited keys to A1[].

Time Complexity of this method is the same as the previous method. Note that in a self-balancing Binary Search Tree, all operations require logm time.

Method 3 (Use Hashing):

  • 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 array
void printArray(int arr[], int n)
{
    // Iterate in the array
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    cout << endl;
}
 
// Driver Code
int 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 array
 
import 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]);
    }
  
    // 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();
    }
}
     
     
    // 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 arr2
def 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 Code
if __name__ == "__main__":
    arr1 = [2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8]
    arr2 = [2, 1, 8, 3]
    print(*solve(arr1, arr2))


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 array
function printArray(arr, n)
{
    // Iterate in the array
    for (let i = 0; i <n; i++)
        document.write(arr[i]," ");
    document.write("</br>");
}
 
// Driver Code
 
let 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 array
let ans = new Array(n);
sortA1ByA2(A1, n, A2, m, ans);
 
// Prints the sorted array
document.write("Sorted array is ");
printArray(ans, n);
 
// This code is contributed by shinjanpatra
</script>


Output

2 2 1 1 8 8 3 5 6 7 9

Steps 1 and 2 on average take O(m+n) time under the assumption that we have a good hashing function that takes O(1) time for insertion and search on average. The third step takes O(p Log p) time where p is the number of elements remained after considering elements of A2[].

Method 4 (By Writing a Customized Compare Method):

We can also customize compare method of a sorting algorithm to solve the above problem. For example, qsort() in C allows us to pass our own customized compare method

  • 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 array
void 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 accesed 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 function
int 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 array
 
import 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


Output

Sorted Array is 2 2 1 1 8 8 3 5 5 5 6 6 7 7 7 9 9 

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