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Reduce the given Array of [1, N] by rotating left or right based on given conditions

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  • Last Updated : 12 Aug, 2021

Given a sorted array arr[] of the first N Natural Numbers and an integer X, the task is to print the last remaining element after performing the below operations (N – 1) times:

Examples:

Input: N = 5, arr[] = {1, 2, 3, 4, 5}, X = 1
Output: 3
Explanation:
The operations are performed as follows:

  1. Rotating the array right by 1 unit modifies the array to {5, 1, 2, 3, 4}. Deleting the array element modifies the array to {5, 1, 2, 3}.
  2. Rotating the array right by 1 unit modifies the array to {3, 5, 1, 2}. Deleting the array element modifies the array to {3, 5, 1}.
  3. Rotating the array right by 1 unit modifies the array to {1, 3, 5}. Deleting the array element modifies the array to {1, 3}.
  4. Rotating the array right by 1 unit modifies the array to {3, 1}. Deleting the array element modifies the array to {3}.

Therefore, the last remaining element is 3.

Input: N = 5, arr[] = {1, 2, 3, 4, 5}, X = 2
Output: 3

Naive Approach: The simplest approach to solve the given problem is to push all the numbers of the range [1, N] in a deque and perform the given operations (N – 1) times according to the given value of X and then print the last remaining element.

Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: The given problem can be optimized based on the following observations: 

  1. If the value of X is 1, then the last left element will be the difference between the smallest power of 2 greater N and N.
  2. If the value of X is 2, then the last left element will be 2*(N – D) + 1, where D represents the largest power of 2 less than or equal to N.

Follow the steps below to solve the problem:

  • Store the power of 2 greater than N in a variable, say nextPower.
  • If the value of X is 1, then print the value (nextPower – N) as the result.
  • Otherwise, print the value 2*(N – nextPower / 2) + 1 as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the last element
// left after performing N - 1 queries
// of type X
int rotate(int arr[], int N, int X)
{
    // Stores the next power of 2
    long long int nextPower = 1;
 
    // Iterate until nextPower is at most N
    while (nextPower <= N)
        nextPower *= 2;
 
    // If X is equal to 1
    if (X == 1)
        return nextPower - N;
 
    // Stores the power of 2 less than or
    // equal to N
    long long int prevPower = nextPower / 2;
 
    // Return the final value
    return 2 * (N - prevPower) + 1;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int X = 1;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << rotate(arr, N, X);
 
    return 0;
}


Java




// Java program for the above approach
import java.io.*;
 
class GFG
{
   
  // Function to find the last element
// left after performing N - 1 queries
// of type X
static int rotate(int arr[], int N, int X)
{
   
    // Stores the next power of 2
    long nextPower = 1;
 
    // Iterate until nextPower is at most N
    while (nextPower <= N)
        nextPower *= 2;
 
    // If X is equal to 1
    if (X == 1)
        return (int) nextPower - N;
 
    // Stores the power of 2 less than or
    // equal to N
    long prevPower = nextPower / 2;
 
    // Return the final value
    return 2 * (N - (int)prevPower) + 1;
}
 
// Driver Code
    public static void main (String[] args) {
        int arr[] = { 1, 2, 3, 4, 5 };
    int X = 1;
    int N =arr.length;
 
   System.out.println(rotate(arr, N, X));
 
    }
}
 
    // This code is contributed by Potta Lokesh


Python3




# Python3 program for the above approach
 
# Function to find the last element
# left after performing N - 1 queries
# of type X
def rotate(arr, N, X):
    # Stores the next power of 2
    nextPower = 1
 
    # Iterate until nextPower is at most N
    while (nextPower <= N):
        nextPower *= 2
 
    # If X is equal to 1
    if (X == 1):
        ans = nextPower - N
        return ans
 
    # Stores the power of 2 less than or
    # equal to N
    prevPower = nextPower // 2
 
    # Return the final value
    return 2 * (N - prevPower) + 1
 
 
# Driver Code
arr = [1, 2, 3, 4, 5]
X = 1
N = len(arr)
print(rotate(arr, N, X))
 
# This code is contributed by amreshkumar3.


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the last element
// left after performing N - 1 queries
// of type X
static int rotate(int []arr, int N, int X)
{
     
    // Stores the next power of 2
    int nextPower = 1;
 
    // Iterate until nextPower is at most N
    while (nextPower <= N)
        nextPower *= 2;
 
    // If X is equal to 1
    if (X == 1)
        return nextPower - N;
 
    // Stores the power of 2 less than or
    // equal to N
    int prevPower = nextPower / 2;
 
    // Return the final value
    return 2 * (N - prevPower) + 1;
}
 
// Driver Code
public static void Main()
{
    int []arr = { 1, 2, 3, 4, 5 };
    int X = 1;
    int N = arr.Length;
 
    Console.Write(rotate(arr, N, X));
}
}
 
// This code is contributed by SURENDRA_GANGWAR


Javascript




<script>
       // JavaScript program for the above approach
 
       // Function to find the last element
       // left after performing N - 1 queries
       // of type X
       function rotate(arr, N, X) {
           // Stores the next power of 2
           let nextPower = 1;
 
           // Iterate until nextPower is at most N
           while (nextPower <= N)
               nextPower *= 2;
 
           // If X is equal to 1
           if (X == 1)
               return nextPower - N;
 
           // Stores the power of 2 less than or
           // equal to N
           let prevPower = nextPower / 2;
 
           // Return the final value
           return 2 * (N - prevPower) + 1;
       }
 
       // Driver Code
       let arr = [1, 2, 3, 4, 5];
       let X = 1;
       let N = arr.length;
 
       document.write(rotate(arr, N, X));
 
   // This code is contributed by Potta Lokesh
 
   </script>


Output

3

Time Complexity: O(log N)
Auxiliary Space: O(1)


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