Kth array element after M replacements of array elements by XOR of adjacent pairs

Difficulty Level : Easy
Last Updated : 11 Sep, 2020

Given an array arr[] of size N and two integers M and K, the task is to find the array element at the K^th index after performing following M operations on the given array.

In a single operation, a new array is formed whose elements have the Bitwise XOR values of the adjacent elements of the current array.

If the number of operations M or the value of K after M operations is invalid then print -1.

Examples:

Input: arr[] = {1, 4, 5, 6, 7}, M = 1, K = 2
Output: 3
Explanation:
Since M = 1, therefore, the operation has to be performed only once on the array.
The array is modified to {1^4, 4^5, 5^6, 6^7} = {5, 1, 3, 1}.
The value of the element at index K = 2 in the updated array is 3.

Input: arr[] = {3, 2, 6}, M = 2, K = 1
Output: -1
Explanation:
Since M = 2, operation has be performed twice.
On performing 1st operation on the array {3, 2, 6} it is modified as {3^2, 2^6} = {1, 4}.
On performing 2nd operation on the array {1, 4}, it is modified as {1^4} = {5}.
After performing 2 operations, the size of the array is reduced to 1. This implies that the index K = 1 doesn’t exist after 2 operations.
Hence, the given input is invalid. So the output is -1.

Approach: The idea is to generate the array after every operation and check-in each iteration whether it is possible to do the operation further or not. If it is not possible, then return “-1”. Below are the steps:

Check if the given number of operations can be performed on the array or not.
After every iteration, the length of the array is reduced by 1 which means if M is greater than or equal to N, then it is not possible to perform the operation. Hence, print “-1”.
If the value of M is valid then check if the given index K is valid or not.
After M operations, (N – M) elements are left in the array. So, if the index value is greater than or equal to the value of (N – M), then print “-1”.
If both M and K values are valid, then iterate a loop M times and do the following:
- In each iteration, create a temporary array temp[] which stores the values obtained by performing operation on the adjacent values of the original array.
- Traverse the array arr[] and calculate the Bitwise XOR values of adjacent elements and save these calculated values in the temporary array temp[].
- Update the current array arr[] with temp[].
After the M iterations, return the value of arr[K], which is the required output.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Method that returns the
// corresponding output by 
// taking the given inputs.
int xor_operations(int N, int arr[],  
                   int M, int K)
{
     
    // If this condition is satisfied,
    // value of M is invalid
    if (M < 0 or M >= N)
        return -1;
     
    // Check if index K is valid
    if (K < 0 or K >= N - M)
        return -1;
     
    // Loop to perform M operations
    for(int p = 0; p < M; p++)
    {
         
        // Creating a temporary list
        vector<int>temp;
     
        // Traversing the array
        for(int i = 0; i < N; i++)
        {
             
            // Calculate XOR values 
            // of adjacent elements
            int value = arr[i] ^ arr[i + 1];
             
            // Adding this value to 
            // the temporary list
            temp.push_back(value);
             
            // Update the original array
            arr[i] = temp[i];
        }
    }
     
    // Getting value at index K
    int ans = arr[K];
     
    return ans;
}
 
// Driver Code
int main()
{
    // Number of elements
    int N = 5;
     
    // Given array arr[]
    int arr[] = {1, 4, 5, 6, 7};
    int M = 1, K = 2;
     
    // Function call
    cout << xor_operations(N, arr, M, K);
    return 0;
}
 
// This code is contributed by gauravrajput1

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Method that returns the
// corresponding output by 
// taking the given inputs.
static int xor_operations(int N, int arr[], 
                          int M, int K)
{
     
    // If this condition is satisfied,
    // value of M is invalid
    if (M < 0 || M >= N)
        return -1;
     
    // Check if index K is valid
    if (K < 0 || K >= N - M)
        return -1;
     
    // Loop to perform M operations
    for(int p = 0; p < M; p++)
    {
         
        // Creating a temporary list
        Vector<Integer>temp = new Vector<Integer>();
     
        // Traversing the array
        for(int i = 0; i < N - 1; i++)
        {
             
            // Calculate XOR values 
            // of adjacent elements
            int value = arr[i] ^ arr[i + 1];
             
            // Adding this value to 
            // the temporary list
            temp.add(value);
             
            // Update the original array
            arr[i] = temp.get(i);
        }
    }
     
    // Getting value at index K
    int ans = arr[K];
     
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Number of elements
    int N = 5;
     
    // Given array arr[]
    int arr[] = { 1, 4, 5, 6, 7 };
    int M = 1, K = 2;
     
    // Function call
    System.out.print(xor_operations(N, arr, M, K));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Method that returns the
# corresponding output by 
# taking the given inputs.
def xor_operations(N, arr, M, K):
     
    # If this condition is satisfied,
    # value of M is invalid
    if M < 0 or M >= N:
        return -1
     
    # Check if index K is valid
    if K < 0 or K >= N-M:
        return -1
     
 
    # Loop to perform M operations
    for _ in range(M):
     
        # Creating a temporary list
        temp = []
     
        # Traversing the array
        for i in range(len(arr)-1):
             
            # Calculate XOR values 
            # of adjacent elements
            value = arr[i]^arr[i + 1]
             
            # Adding this value to 
            # the temporary list
            temp.append(value)
         
        # Update the original array
        arr = temp[:]
     
    # Getting value at index K
    ans = arr[K]
    return ans
 
# Driver Code
 
# Number of elements
N = 5
 
# Given array arr[]
arr = [1, 4, 5, 6, 7]
M = 1
K = 2
 
# Function Call
print(xor_operations(N, arr, M, K))

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Method that returns the
// corresponding output by 
// taking the given inputs.
static int xor_operations(int N, int []arr, 
                          int M, int K)
{
  // If this condition is satisfied,
  // value of M is invalid
  if (M < 0 || M >= N)
    return -1;
 
  // Check if index K is valid
  if (K < 0 || K >= N - M)
    return -1;
 
  // Loop to perform M operations
  for(int p = 0; p < M; p++)
  {
    // Creating a temporary list
    List<int>temp = new List<int>();
 
    // Traversing the array
    for(int i = 0; i < N - 1; i++)
    {
      // Calculate XOR values 
      // of adjacent elements
      int value = arr[i] ^ arr[i + 1];
 
      // Adding this value to 
      // the temporary list
      temp.Add(value);
 
      // Update the original array
      arr[i] = temp[i];
    }
  }
 
  // Getting value at index K
  int ans = arr[K];
 
  return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
  // Number of elements
  int N = 5;
 
  // Given array []arr
  int []arr = {1, 4, 5, 6, 7};
  int M = 1, K = 2;
 
  // Function call
  Console.Write(xor_operations(N, arr, 
                               M, K));
}
}
 
// This code is contributed by Rajput-Ji

Output:

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

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