Given two arrays, A[] and B[], the task is to check if they are equal or not. Arrays are considered equal if any permutation of array B equates to array A.
Examples:
Input: A[] = [2, 4, 5, 7, 5, 6] and B[] = [4, 2, 5, 5, 6, 7]
Output: Yes
Explanation: All the elements in array A are present in array B and same number of times.Input: A[] = [2, 5, 8, 9, 78] and B[] = [5, 2, 7, 78, 8]
Output: No
Explanation: In array A there is a 9 and in array B there is a 7
Approach: The problem can be solved using hashmap based on the below idea:
Two arrays will be equal only if the frequency of the respective elements in both arrays are equal.
Follow the steps to solve the problem:
- If the sizes of both arrays are not equal, return NO.
- Maintain a hashmap and count the frequency of both the array elements
- If for any element the frequency is not the same, then return NO
- Otherwise, return YES
Below is the implementation of the above approach.
C++
// C++ code to implement the approachÂ
#include <bits/stdc++.h>using namespace std;bool isEqual(vector<int> a, vector<int> b, int n, int m){    // If size is not same return false    if (n != m) {        return 0;    }Â
    // Create 2 unordered maps to store    // the frequency    unordered_map<int, int> mp1, mp2;    for (int i : a) {        mp1[i]++;    }    for (int i : b) {        mp2[i]++;    }Â
    // Compare the frequency    for (auto i = mp1.begin(); i != mp1.end(); i++) {Â
        // If frequency not same return false        if (mp2[i->first] != i->second) {            return 0;        }    }    return 1;}Â
// Drivers codeint main(){    vector<int> a = { 2, 4, 5, 7, 5, 6 },                b = { 4, 2, 5, 5, 6, 7 };    int n = a.size(), m = b.size();    bool flag = isEqual(a, b, n, m);Â
    // Return 1 if equal    if (flag) {        cout << "YES\n";    }    else {        cout << "NO\n";    }    return 0;} |
Java
// Java code to implement the approachimport java.io.*;import java.util.*;Â
class GFG {    public static boolean isEqual(int a[], int b[], int n,                                  int m)    {        // If size is not same return false        if (n != m) {            return false;        }Â
        // Create 2 unordered maps to store        // the frequency        HashMap<Integer, Integer> mp1            = new HashMap<Integer, Integer>();        HashMap<Integer, Integer> mp2            = new HashMap<Integer, Integer>();        for (int i : a) {            if (mp1.get(i) != null)                mp1.put(i, mp1.get(i) + 1);            else                mp1.put(i, 1);        }        for (int i : b) {            if (mp2.get(i) != null)                mp2.put(i, mp2.get(i) + 1);            else                mp2.put(i, 1);        }Â
        // Compare the frequency        for (Map.Entry<Integer, Integer> i :             mp1.entrySet()) {            Integer key = i.getKey();            // If frequency not same return false            if (mp2.get(key) != i.getValue()) {                return false;            }        }        return true;    }Â
    // Driver Code    public static void main(String[] args)    {        int a[] = { 2, 4, 5, 7, 5, 6 };        int b[] = { 4, 2, 5, 5, 6, 7 };        int n = a.length, m = b.length;        boolean flag = isEqual(a, b, n, m);Â
        // Return 1 if equal        if (flag == true) {            System.out.print("YES\n");        }        else {            System.out.print("NO\n");        }    }}Â
// This code is contributed by Rohit Pradhan |
Python3
# Python3 code to implement the approachÂ
def isEqual(a, b, n, m) :Â
    # If size is not same return false    if (n != m) :        return 0Â
    # Create 2 unordered maps to store    # the frequency    mp1 = dict.fromkeys(a, 0);    mp2 = dict.fromkeys(b, 0);         for i in a :        if i in mp1 :            mp1[i] += 1;        else:            mp1[i] = 0             for i in b :        if i in mp2 :            mp2[i] += 1;        else :            mp2[i] = 0Â
    # Compare the frequency    for i in mp1 :Â
        # If frequency not same return false        if (mp2[i] != mp1[i]) :            return 0;Â
    return 1;Â
# Drivers codeif __name__ == "__main__" :Â
    a = [ 2, 4, 5, 7, 5, 6 ];    b = [ 4, 2, 5, 5, 6, 7 ];    n = len(a);    m = len(b);    flag = isEqual(a, b, n, m);Â
    # Return 1 if equal    if (flag) :        print("YES");Â
    else :        print("NO");     # This code is contributed by AnkThon |
C#
// C# code to implement the approachusing System;using System.Collections.Generic;Â
public class GFG {  public static bool isEqual(int []a, int []b, int n,                             int m)  {    // If size is not same return false    if (n != m) {      return false;    }Â
    // Create 2 unordered maps to store    // the frequency    Dictionary<int, int> mp1      = new Dictionary<int, int>();    Dictionary<int, int> mp2      = new Dictionary<int, int>();    foreach (int i in a) {      if (mp1.ContainsKey(i))        mp1[i] = mp1[i] + 1;      else        mp1.Add(i, 1);    }    foreach (int i in b) {      if (mp2.ContainsKey(i))        mp2[i] = mp2[i] + 1;      else        mp2.Add(i, 1);    }Â
    // Compare the frequency    foreach (KeyValuePair<int, int> i in             mp1) {      int key = i.Key;             // If frequency not same return false      if (mp2[key] != i.Value) {        return false;      }    }    return true;  }Â
  // Driver Code  public static void Main(String[] args)  {    int []a = { 2, 4, 5, 7, 5, 6 };    int []b = { 4, 2, 5, 5, 6, 7 };    int n = a.Length, m = b.Length;    bool flag = isEqual(a, b, n, m);Â
    // Return 1 if equal    if (flag == true) {      Console.Write("YES\n");    }    else {      Console.Write("NO\n");    }  }}Â
// This code is contributed by shikhasingrajput |
Javascript
<script>// Javascript code to implement the approachfunction isEqual(a, b, n, m){Â
    // If size is not same return false    if (n != m) {        return false;    }Â
    // Create 2 unordered maps to store    // the frequency    let mp1 = new Map();    let mp2 = new Map();    for (let i of a) {        if (mp1.get(i) != null)            mp1.set(i, mp1.get(i) + 1);        else            mp1.set(i, 1);    }    for (let i of b) {        if (mp2.get(i) != null)            mp2.set(i, mp2.get(i) + 1);        else            mp2.set(i, 1);    }Â
    // Compare the frequency    for (let i of mp1.keys()) {        // If frequency not same return false        if (mp2.get(i) != mp1.get(i)) {            return false;        }    }    return true;}Â
// Driver CodeÂ
let a = [2, 4, 5, 7, 5, 6];let b = [4, 2, 5, 5, 6, 7];let n = a.length, m = b.length;let flag = isEqual(a, b, n, m);Â
// Return 1 if equalif (flag == true) {Â Â Â Â document.write("YES<br>");}else {Â Â Â Â document.write("NO<br>");}Â
// This code is contributed by Saurabh Jaiswal</script> |
Output
YES
Time complexity:Â O(N) in average case and O(N2) in worst case.
Auxiliary Space:Â O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
