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Check if Arrays can be made equal by Replacing elements with their number of Digits

Given two arrays A[] and B[] of length N, the task is to check if both arrays can be made equal by performing the following operation at most K times: 

  • Choose any index i and either change Ai to the number of digits Ai have or change Bi to the number of digits Bi have.

Examples:

Input: N = 4, K = 1, A = [1, 2, 3, 4], B = [1, 2, 3, 5]
Output: Unequal
Explanation: It is impossible to make both the array equal by performing the following operation: 
1st operation on A will change A to [1, 2, 3, 1] as number of digits in 4 is 1. So both the arrays are unequal .

Input: N = 3, K = 3, A = [2, 9, 3], B = [1, 100, 9].
Output: Equal 
Explanation: It is possible to make both the array equal by performing the following operations: 
1st operation on A will change A to [1, 9, 3] as number of digits in 2 is 1.
2nd operation on B will change B to [1, 3, 9] as number of digits in 100 is 3.
So array A is equal to array B after the operations.

Approach: To solve the problem follow the below idea/Intuition:

Can we make both array similar? 

The answer is Yes as we are reducing each number by its length so eventually all numbers in both the array can be 1. The important thing here is to notice that every number greater than 1 always decreases by performing the operation. Thus, if a number appears in only one of the arrays, you will have to do one of the followings two things:

  • Decrease some greater number to make it equal to this one.
  • Decrease this number.

So, the proposed solution is the following. Consider the largest element in each array. If they are equal, remove both. If not, apply above operation to the larger of them and continue until the arrays are empty. This can be done efficiently using a max heap (priority queue).

Follow the below steps to solve the problem:

  • Declare priority queue (max heap) ‘First’ and push all the elements of A[] on First.
  • Declare priority queue (max heap) ‘Second’ and push all the elements of B[] on Second.
  • Initialize the counter variable to 0.
  • Start a while loop till any of the queues become empty and check.
    • If (First.top() > Second.top()) then pop the top element from First and push the number of digits in that element back to First and increment the counter by 1.
    • If (Second.top() > First.top()) then pop the top element from Second and push the number of digits in that element back to Second and increment the counter by 1.
    • If ( first.top() = second.top() ) then pop the top elements from both heaps.
  • After the iteration is over, If the counter is less than or equal to K, print “Equal“, otherwise, print “Unequal“

Below is the implementation of the above approach.

C++

// Code for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function calculating number of digit
// of the given element
int len(int x)
{
    string temp = to_string(x);
    return temp.size();
}
 
bool canbothEqual(int N, int K, vector<int>& A,
                  vector<int>& B)
{
 
    // Declaring first priority
    // queue MAX HEAP
    priority_queue<int> First;
    for (int i = 0; i < N; i++) {
 
        // Pushing all the elements of
        // array A to the queue
        First.push(A[i]);
    }
 
    // Declaring second priority queue MAX HEAP
    priority_queue<int> Second;
    for (int i = 0; i < N; i++) {
 
        // Pushing all the elements of
        // array B to the queue
        Second.push(B[i]);
    }
 
    // Initialising counter variable to 0
    int counter = 0;
    while (!First.empty() && !Second.empty()) {
        int num1 = First.top();
        int num2 = Second.top();
 
        // If top element of first queue is
        // greater than top element of
        // second queue.
        if (num1 > num2) {
            First.push(len(num1));
            First.pop();
            counter++;
        }
 
        // If top element of second
        // queue is greater than top
        // element of first queue.
        else if (num2 > num1) {
            Second.push(len(num2));
            Second.pop();
            counter++;
        }
 
        // If top element of both the
        // queue is same
        else {
            First.pop();
            Second.pop();
        }
    }
 
    // If counter is smaller than or equal
    // to K then returning 1
    if (K >= counter)
        return true;
 
    // Returning 0
    return false;
}
 
// Driver function
int main()
{
    int N = 3;
    int K = 3;
    vector<int> A = { 2, 9, 3 };
    vector<int> B = { 1, 100, 9 };
 
    // Function Call
    if (canbothEqual(N, K, A, B))
        cout << "Equal";
    else
        cout << "Unequal";
    return 0;
}

Java

// Java code for the above approach
import java.util.*;
 
class GFG {
 
  // Function calculating number of digit
  // of the given element
  static int len(int x)
  {
    String temp = String.valueOf(x);
    return temp.length();
  }
 
  static boolean canbothEqual(int N, int K, int[] A,
                              int[] B)
  {
 
    // Declaring first priority
    // queue MAX HEAP
    PriorityQueue<Integer> First = new PriorityQueue<>(
      (x, y) -> Integer.compare(y, x));
    for (int i = 0; i < N; i++) {
 
      // Pushing all the elements of
      // array A to the queue
      First.add(A[i]);
    }
 
    // Declaring second priority queue MAX HEAP
    PriorityQueue<Integer> Second = new PriorityQueue<>(
      (x, y) -> Integer.compare(y, x));
    for (int i = 0; i < N; i++) {
 
      // Pushing all the elements of
      // array B to the queue
      Second.add(B[i]);
    }
 
    // Initialising counter variable to 0
    int counter = 0;
    while (First.size() != 0 && Second.size() != 0) {
      int num1 = First.peek();
      int num2 = Second.peek();
 
      // If top element of first queue is
      // greater than top element of
      // second queue.
      if (num1 > num2) {
        First.add(len(num1));
        First.poll();
        counter++;
      }
 
      // If top element of second
      // queue is greater than top
      // element of first queue.
      else if (num2 > num1) {
        Second.add(len(num2));
        Second.poll();
        counter++;
      }
 
      // If top element of both the
      // queue is same
      else {
        First.poll();
        Second.poll();
      }
    }
 
    // If counter is smaller than or equal
    // to K then returning 1
    if (K >= counter)
      return true;
 
    // Returning 0
    return false;
  }
 
  // Driver function
 
  public static void main(String[] args)
  {
    int N = 3;
    int K = 3;
    int[] A = { 2, 9, 3 };
    int[] B = { 1, 100, 9 };
 
    // Function Call
    if (canbothEqual(N, K, A, B))
      System.out.println("Equal");
    else
      System.out.println("Unequal");
  }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python code for the above approach
from queue import PriorityQueue
 
def canbothEqual(N, K, A, B):
   
    # Declaring first priority
    # queue MAX HEAP
    First = PriorityQueue()
    for i in range(N):
       
        # Pushing all the elements of
        # array A to the queue
        First.put(A[i])
 
    # Declaring second priority queue MAX HEAP
    Second = PriorityQueue()
    for i in range(N):
        # Pushing all the elements of
        # array B to the queue
        Second.put(B[i])
 
    # Initialising counter variable to 0
    counter = 0
    while (First.empty() == False and Second.empty() == False):
        num1 = First.get()
        num2 = Second.get()
 
        # If top element of first queue is
        # greater than top element of
        # second queue.
        if (num1 > num2):
            First.put(len(str(num1)))
            nn = First.get()
            counter += 1
 
        # If top element of second
        # queue is greater than top
        # element of first queue.
        elif (num2 > num1):
            Second.put(len(str(num2)))
            cc = Second.get()
            counter += 1
             
        # If top element of both the
        # queue is same
        else:
            cc = First.get()
            nn = Second.get()
 
    # If counter is smaller than or equal
    # to K then returning 1
    if (K >= counter):
        return True
       
    # Returning 0
    return False
 
# Driver function
if __name__ == "__main__":
    N = 3
    K = 3
    A = [2, 9, 3]
    B = [1, 100, 9]
 
    # Function Call
    if (canbothEqual(N, K, A, B)):
        print("Equal")
    else:
        print("Unequal")
 
# This code is contributed by Rohit Pradhan

C#

// C# code for the above approach
using System;
using System.Collections.Generic;
 
class Program {
 
  // Function calculating number of digit
  // of the given element
  static int len(int x)
  {
    string temp = x.ToString();
    return temp.Length;
  }
 
  static bool canbothEqual(int N, int K, int[] A, int[] B)
  {
    // Declaring first priority
    // queue MAX HEAP
    List<int> First = new List<int>();
 
    for (int i = 0; i < N; i++) {
      // Pushing all the elements of
      // array A to the queue
      First.Add(A[i]);
    }
 
    // Declaring second priority queue MAX HEAP
    List<int> Second = new List<int>();
 
    for (int i = 0; i < N; i++) {
      // Pushing all the elements of
      // array B to the queue
      Second.Add(B[i]);
    }
 
    // Initialising counter variable to 0
    int counter = 0;
 
    while (First.Count != 0 && Second.Count != 0) {
      First.Sort((a, b) => b.CompareTo(a));
      Second.Sort((a, b) => b.CompareTo(a));
 
      int num1 = First[0];
      int num2 = Second[0];
 
      // If top element of first queue is
      // greater than top element of
      // second queue.
      if (num1 > num2) {
        First.Add(len(num1));
        First.RemoveAt(0);
        counter++;
      }
 
      // If top element of second
      // queue is greater than top
      // element of first queue.
      else if (num2 > num1) {
        Second.Add(len(num2));
        Second.RemoveAt(0);
        counter++;
      }
 
      // If top element of both the
      // queue is same
      else {
        First.RemoveAt(0);
        Second.RemoveAt(0);
      }
    }
 
    // If counter is smaller than or equal
    // to K then returning 1
    if (K >= counter)
      return true;
 
    // Returning 0
    return false;
  }
 
  // Driver function
  static void Main(string[] args)
  {
    int N = 3;
    int K = 3;
    int[] A = { 2, 9, 3 };
    int[] B = { 1, 100, 9 };
 
    // Function Call
    if (canbothEqual(N, K, A, B))
      Console.WriteLine("Equal");
    else
      Console.WriteLine("Unequal");
  }
}
 
// This code is contributed by Tapesh(tapeshdua420)

Javascript

<script>
 
// Priority Queue Implementation in Javascript
function PriorityQueue () {
    this.collection = [];
    this.printCollection = function() {
      console.log(this.collection);
    };
    this.enqueue=function(item){
        if(this.collection.length===0) {
        this.collection.push(item);
      }
      else if(this.collection.length===1){
        if(item[1]>=this.collection[0][1]){
          this.collection.push(item);
        }
        else if(item[1]<this.collection[0][1]){
          this.collection.unshift(item);
        }
      }
     else if(this.collection.length>1){
        let check=false;
        for(let i=0;i<this.collection.length;i++){
            if(this.collection[i][1]>item[1]){
              if(i===0){
                check=true;
                this.collection.unshift(item);
                break;
              }
              else {
                check=true;
                let sliced=this.collection.slice(0,i);
                let slicedend=this.collection.slice(i,this.collection.length);
                sliced.push(item);
                this.collection=sliced.concat(slicedend);
                break;
              }
            }
        }
        if(!check){
                this.collection.push(item);
        }
    };
 
 
    this.dequeue = function(){
        return this.collection.shift()[0];
    };
    this.size = function(){
        return this.collection.length;
    };
    this.isEmpty=function(){
        return this.collection.length===0?true:false;
    };
    this.front=function(){
        return this.collection[0];
    };
    }
}
 
// Code for above approach
 
// Function calculating number of digit
// of the given element
function len(x)
{
    let temp = String(x);
    return temp.length;
}
 
function canbothEqual(N, K, A, B)
{
    // Declaring first priority
    // queue MAX HEAP
    let First = new PriorityQueue();
    // priority_queue<int> First;
    for (let i = 0; i < N; i++) {
 
        // Pushing all the elements of
        // array A to the queue
        First.enqueue(A[i]);
    }
 
    // Declaring second priority queue MAX HEAP
    let Second = new PriorityQueue();
    // priority_queue<int> Second;
    for (let i = 0; i < N; i++) {
 
        // Pushing all the elements of
        // array B to the queue
        Second.enqueue(B[i]);
    }
 
    // Initialising counter variable to 0
    let counter = 0;
    while (!First.isEmpty() && !Second.isEmpty()) {
        let num1 = First.front();
        let num2 = Second.front();
 
        // If top element of first queue is
        // greater than top element of
        // second queue.
        if (num1 > num2) {
            First.enqueue(len(num1));
            First.dequeue();
            counter++;
        }
 
        // If top element of second
        // queue is greater than top
        // element of first queue.
        else if (num2 > num1) {
            Second.enqueue(len(num2));
            Second.dequeue();
            counter++;
        }
 
        // If top element of both the
        // queue is same
        else {
            First.dequeue();
            Second.dequeue();
        }
    }
 
    // If counter is smaller than or equal
    // to K then returning 1
    if (K >= counter)
        return true;
 
    // Returning 0
    return false;
}
 
// Driver function
N = 3;
K = 3;
let A = [ 2, 9, 3 ];
let B = [ 1, 100, 9 ];
 
// Function Call
if (canbothEqual(N, K, A, B))
    console.log("Equal");
else
    console.log("Unequal");
 
// This code is contributed by akashish__
 
</script>

Output

Equal

Time Complexity: O(N * log(N))
Auxiliary Space: O(N) 


 


Reffered: https://www.geeksforgeeks.org

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Category:
 Coding
Sub Category:
 Tutorial
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