Map elements of an array to elements of another array

Last Updated : 08 May, 2023

Given two arrays A and B of positive integers, elements of array B can be mapped to elements of array A only if both the elements have same value. The task is to compute the positions in array A to which elements of array B will be mapped. Print NA if mapping for a particular element cannot be done.

Note: For one position only one integer can be mapped.

Examples:

Input: A[] = {1, 5, 2, 4, 4, 3}, B[] = {1, 2, 5, 1}
Output: 0 2 1
NA B[0], B[1] and B[2] can be mapped to A[0], A[2] and A[1] respectively but B[3] cannot be mapped to any element of A because the only '1' in A has already been mapped

Input: A[] = {2, 1, 2, 3, 3, 4, 2, 4, 1}, B[] = {1, 2, 5, 1, 2, 4, 2, 3, 2, 1}
Output: 1 0 NA 8 2 5 6 3 NA NA

The idea is to use a hash table where keys are elements of A[] and values are indexes of these elements. Since there can be more than one occurrences of an element, we use a list of items as values in the hash table. Below is the implementation of the above problem:

Implementation:

CPP

// C++ program to map elements of an array
// to equal elements of another array
#include <bits/stdc++.h>
using namespace std;

// Function to print the mapping of elements
void printMapping(int A[], int B[], int N, int M)
{

// Create a hash table where all indexes are
// stored for a given value
unordered_map<int, list<int>> m;
for (int i=0; i<N; i++)
    m[A[i]].push_back(i);

// Traverse through B[]
for (int i=0; i<M; i++)
{
    // If a mapping is found, print the mapping and
    // remove the index from hash table so that the
    // same element of A[] is not mapped again.
    if (m.find(B[i]) != m.end() && m[B[i]].size() > 0)
    {
        cout << m[B[i]].front() << " ";
        m[B[i]].pop_front();
    }

    else // No mapping found
    {
        cout << "NA ";
    }
}
}

// Driver code
int main()
{
    int A[] = {2, 1, 2, 3, 3, 4, 2, 4, 1};
    int N = sizeof(A) / sizeof(A[0]);
    int B[] = {1, 2, 5, 1, 2, 4, 2, 3, 2, 1};
    int M = sizeof(B) / sizeof(B[0]);
    printMapping(A, B, N, M);
        return 0;
}

Java

/*package whatever //do not write package name here */

import java.util.*;

class GFG {
  
  // Function to print the mapping of elements
static void printMapping(int A[], int B[], int N, int M)
{

// Create a hash table where all indexes are
// stored for a given value
HashMap<Integer,List<Integer>> m = new HashMap<>();
  
  for(int i:A){
      m.put(i,new ArrayList<Integer>());
  }
  
for (int i=0; i<N; i++)
    m.get(A[i]).add(i);

// Traverse through B[]
for (int i=0; i<M; i++)
{
    // If a mapping is found, print the mapping and
    // remove the index from hash table so that the
    // same element of A[] is not mapped again.
    if (m.containsKey(B[i]) && m.get(B[i]).size() > 0)
    {
        System.out.print(m.get(B[i]).get(0) + " ");
        m.get(B[i]).remove(0);
    }

    else // No mapping found
    {
        System.out.print("NA" + " ");
    }
}
}
    public static void main (String[] args) {
            int A[] = {2, 1, 2, 3, 3, 4, 2, 4, 1};
    int N = A.length;
    int B[] = {1, 2, 5, 1, 2, 4, 2, 3, 2, 1};
    int M = B.length;
    printMapping(A, B, N, M);
    
    }
}

Python3

# Python3 program to map elements of an array
# to equal elements of another array

# Function to print the mapping of elements
def printMapping(A, B):

    # Create a hash table where all indexes are
    # stored for a given value
    m = {}
    for i in range(len(A)):
        if A[i] in m:
            m[A[i]].append(i)
        else:
            m[A[i]] = [i]

    # Traverse through B[]
    for i in range(len(B)):
    
        # If a mapping is found, print the mapping and
        # remove the index from hash table so that the
        # same element of A[] is not mapped again.
        if B[i] in m and len(m[B[i]]) > 0:
            print(m[B[i]].pop(0),end=" ")
        else:
            print("NA",end=" ")

# Driver code
A = [2, 1, 2, 3, 3, 4, 2, 4, 1]
B = [1, 2, 5, 1, 2, 4, 2, 3, 2, 1]
printMapping(A, B)

# This code is contributed by akashish__

using System;
using System.Collections.Generic;

public class GFG {

  // Function to print the mapping of elements
  static void printMapping(int[] A, int[] B, int N, int M)
  {

    // Create a hash table where all indexes are
    // stored for a given value
    Dictionary<int, List<int> > m
      = new Dictionary<int, List<int> >();

    for (int i = 0; i < A.Length; i++) {
      if (!m.ContainsKey(A[i]))
        m.Add(A[i], new List<int>());
    }

    for (int i = 0; i < N; i++)
      m[A[i]].Add(i);

    // Traverse through B[]
    for (int i = 0; i < M; i++) {
      // If a mapping is found, print the mapping and
      // remove the index from hash table so that the
      // same element of A[] is not mapped again.
      if (m.ContainsKey(B[i]) && m[B[i]].Count > 0) {
        Console.Write(m[B[i]][0] + " ");
        m[B[i]].RemoveAt(0);
      }

      else // No mapping found
      {
        Console.Write("NA"
                      + " ");
      }
    }
  }
  static public void Main()
  {

    int[] A = { 2, 1, 2, 3, 3, 4, 2, 4, 1 };
    int N = A.Length;
    int[] B = { 1, 2, 5, 1, 2, 4, 2, 3, 2, 1 };
    int M = B.Length;
    printMapping(A, B, N, M);
  }
}

// This code is contributed by akashish__

JavaScript

// JavaScript program to map elements of an array
// to equal elements of another array

// Function to print the mapping of elements
function printMapping(A, B)
{

    // Create a hash table where all indexes are
    // stored for a given value
    let m = new Map();
    for (let i = 0; i < A.length; i++) {
        if (m.has(A[i])) {
            m.get(A[i]).push(i);
        } else {
            m.set(A[i], [i]);
        }
    }

    // Traverse through B[]
    for (let i = 0; i < B.length; i++) 
    {
    
        // If a mapping is found, print the mapping and
        // remove the index from hash table so that the
        // same element of A[] is not mapped again.
        if (m.has(B[i]) && m.get(B[i]).length > 0) {
            console.log(m.get(B[i]).shift());
        } else {
            console.log("NA");
        }
    }
}

// Driver code
let A = [2, 1, 2, 3, 3, 4, 2, 4, 1];
let B = [1, 2, 5, 1, 2, 4, 2, 3, 2, 1];
printMapping(A, B);

// This code is contributed by akashish__

Output

1 0 NA 8 2 5 6 3 NA NA

Time Complexity: O(N+M)

Auxiliary Space: O(N)

Map elements of an array to elements of another array

Kushdeep_Mittal

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Map elements of an array to elements of another array

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