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Value of k-th index of a series formed by append and insert MEX in middle
  • Difficulty Level : Hard
  • Last Updated : 14 Apr, 2021

Given two integers, n and k. Initially we have a sequence consisting of a single number 1. We need to consider series formed after n steps. In each step, we append the sequence to itself and insert the MEX(minimum excluded)(>0) value of the sequence in the middle. Perform (n-1) steps. Finally, find the value of the k-th index of the resulting sequence.
Example: 
 

Input : n = 3, k = 2.
Output: Initially, we have {1}, we have to perform 2 steps.
        1st step :  {1, 2, 1}, since MEX of {1} is 2.
        2nd step:   {1, 2, 1, 3, 1, 2, 1}, 
                     since MEX of {1, 2, 1} is 3.
        Value of 2nd Index = 2.

Input : n = 4, k = 8.
Output: Perform 3 steps.
        After second step, we have  {1, 2, 1, 3, 1, 2, 1}
        3rd step: {1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 
                               1, 2, 1}, since MEX = 4.
        Value of 8th index = 4.

A simple solution is to generate the series using given steps and store in an array. Finally, we return k-th element of the array.
An efficient solution is to use Binary Search. Observe that the middle element of our resulting sequence is n itself. Length of the sequence is 2n – 1, because lengths will be like (1, 3, 7, 15….2n -1). We use Binary search to solve this problem. As we know that the middle element of a sequence is the number of the step(from 1) performed on it. 
In fact, every element in the sequence is a middle element at one or other step. 
We start searching the element from step n and compare the middle index with ‘k’ if found we return n, else we decrease n by 1 and update our range of our search. 
we repeat this step until the index is reached. 
 

CPP




// CPP program to fin k-th element after append
// and insert middle operations
#include <bits/stdc++.h>
using namespace std;
void findElement(int n, int k)
{
    int ans = n; // Middle element of the sequence
    int left = 1;
 
    // length of the resulting sequence.
    int right = pow(2, n) - 1;
    while (1) {
        int mid = (left + right) / 2;
        if (k == mid) {
            cout << ans << endl;
            break;
        }
 
        // Updating the middle element of next sequence
        ans--;
 
        // Moving to the left side of the middle element.
        if (k < mid)
            right = mid - 1;
         
        // Moving to the right side of the middle element.
        else
            left = mid + 1;        
    }
}
 
// Driver code
int main()
{
    int n = 4, k = 8;
    findElement(n, k);
    return 0;
}

Java




// Java program to fin k-th element after append
// and insert middle operations
 
class GFG
{
 
    static void findElement(int n, int k)
    {
        // Middle element of the sequence
        int ans = n;
        int left = 1;
 
        // length of the resulting sequence.
        int right = (int) (Math.pow(2, n) - 1);
        while (true)
        {
            int mid = (left + right) / 2;
            if (k == mid)
            {
                System.out.println(ans);
                break;
            }
 
            // Updating the middle element
            // of next sequence
            ans--;
 
            // Moving to the left side of
            // the middle element.
            if (k < mid)
            {
                right = mid - 1;
            }
             
            // Moving to the right side of
            // the middle element.
            else
            {
                left = mid + 1;
            }
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 4, k = 8;
        findElement(n, k);
    }
 
}
 
// This code has been contributed by 29AjayKumar

Python3




# Python3 code to find k-th element after append
# and insert middle operations
import math
 
def findElement(n , k):
    ans = n      # Middle element of the sequence
    left = 1
     
    # length of the resulting sequence.
    right = math.pow(2, n) - 1
    while 1:
        mid = int((left + right) / 2)
        if k == mid:
            print(ans)
            break
         
        # Updating the middle element of next sequence
        ans-=1
         
        # Moving to the left side of the middle element.
        if k < mid:
            right = mid - 1
         
        # Moving to the right side of the middle element.
        else:
            left = mid + 1
 
# Driver code
n = 4
k = 8
findElement(n, k)
 
# This code is contributed by "Sharad_Bhardwaj".

C#




// C# program to fin k-th element after append
// and insert middle operations
using System;
 
class GFG
{
 
    static void findElement(int n, int k)
    {
        // Middle element of the sequence
        int ans = n;
        int left = 1;
 
        // length of the resulting sequence.
        int right = (int) (Math.Pow(2, n) - 1);
        while (true)
        {
            int mid = (left + right) / 2;
            if (k == mid)
            {
                Console.WriteLine(ans);
                break;
            }
 
            // Updating the middle element
            // of next sequence
            ans--;
 
            // Moving to the left side of
            // the middle element.
            if (k < mid)
            {
                right = mid - 1;
            }
             
            // Moving to the right side of
            // the middle element.
            else
            {
                left = mid + 1;
            }
        }
    }
 
    // Driver code
    public static void Main()
    {
        int n = 4, k = 8;
        findElement(n, k);
    }
 
}
 
/* This code contributed by PrinciRaj1992 */

PHP




<?php
// PHP program to fin k-th element
// after append and insert middle
// operations
 
function findElement($n, $k)
{
     
    // Middle element of the sequence
    $ans = $n;
    $left = 1;
 
    // length of the resulting sequence.
    $right = pow(2, $n) - 1;
    while (1)
    {
        $mid = ($left + $right) / 2;
        if ($k ==$mid)
        {
            echo $ans ,"\n";
            break;
        }
 
        // Updating the middle element
        // of next sequence
        $ans--;
 
        // Moving to the left side of
        // the middle element.
        if ($k < $id)
            $right = $mid - 1;
         
        // Moving to the right side
        // of the middle element.
        else
            $left = $mid + 1;    
    }
}
 
    // Driver code
    $n = 4;
    $k = 8;
    findElement($n, $k);
 
// This code is contributed by aj_36
?>

Javascript




<script>
 
// JavaScript program to fin k-th element after append
// and insert middle operations
 
   function findElement(n, k)
    {
        // Middle element of the sequence
        let ans = n;
        let left = 1;
   
        // length of the resulting sequence.
        let right = (Math.pow(2, n) - 1);
        while (true)
        {
            let mid = (left + right) / 2;
            if (k == mid)
            {
                document.write(ans);
                break;
            }
   
            // Updating the middle element
            // of next sequence
            ans--;
   
            // Moving to the left side of
            // the middle element.
            if (k < mid)
            {
                right = mid - 1;
            }
               
            // Moving to the right side of
            // the middle element.
            else
            {
                left = mid + 1;
            }
        }
    }
 
// Driver code            
        let n = 4, k = 8;
        findElement(n, k);
          
         // This code is contributed by code_hunt.
</script>

Output: 
 

4

Time Complexity:O(log n)
 




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