Sort an array according to the order defined by another array

3.6

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.

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 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 size of A1[] be m and size of A2[] be n.
1) Create a temporary array temp of size m and copy contents of A1[] to it.
2) 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[].
3) Sort temp[]
4) Initialize the output index ind as 0.
5) Do following for every element of A2[i] in A2[]
…..a) 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[]
6) Copy all unvisited elements from temp[] to A1[].

Time complexity: The steps 1 and 2 require O(m) time. Step 3 requires O(mLogm) time. Step 5 requires O(nLogm) time. Therefore overall time complexity is O(m + nLogm).

Thanks to vivek for suggesting this method. Following is the implementation of above algorithm.

C++

// A C++ program to sort an array according to the order defined
// by another array
#include <iostream>
#include <algorithm>
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 */
int first(int arr[], int low, int high, int x, int n)
{
    if (high >= low)
    {
        int mid = low + (high-low)/2;  /* (low + high)/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].
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);

    int 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 (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)
{
    for (int i=0; i<n; i++)
        cout << arr[i] << " ";
    cout << endl;
}

// Driver program to test above function.
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]);
    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.


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

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.
1) 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.
2) Initialize the output index ind as 0.
3) Do following for every element of A2[i] in A2[]
…..a) 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.
4) Do an inorder traversal of BST and copy all unvisited keys to A1[].

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

Method 3 (Use Hashing)
1. Loop through A1[], store the count of every number in a HashMap (key: number, value: count of number) .
2. 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.
3. Sort the rest of the numbers present in HashMap and put in output array.

Thanks to Anurag Sigh for suggesting this method.

The 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(pLogp) 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.
1. If num1 and num2 both are in A2 then number with lower index in A2 will be treated smaller than other.
2. 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.
3. If both are not in A2, then natural ordering will be taken.

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.

Following is C implementation of this method.

// 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];
int size = 5; // size of A2[]

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 comapre 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;
}

Output:

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

This method is based on comments by readers (Xinuo Chen, Pranay Doshi and javakurious) and compiled by Anurag Singh.

This article is compiled by Piyush. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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