Given the length of an array of integers N and an integer K. The task is to modify the array in such a way that the array contains first all odd integers from 1 to N in ascending order, then all even integers from 1 to N in ascending order and then print the Kth element in the modified array.
Examples:
Input: N = 8, K = 5
Output: 2
The array will be {1, 3, 5, 7, 2, 4, 6, 8}
and the fifth element is 2.
Input: N = 7, K = 2
Output: 3
Naive approach: A simple approach is to store the odd numbers first, one by one till N, and then storing the even numbers one by one till N, and then printing the kth element.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the kth element// in the modified arrayint getNumber(int n, int k){ int arr[n]; int i = 0; // First odd number int odd = 1; while (odd <= n) { // Insert the odd number arr[i++] = odd; // Next odd number odd += 2; } // First even number int even = 2; while (even <= n) { // Insert the even number arr[i++] = even; // Next even number even += 2; } // Return the kth element return arr[k - 1];}// Driver codeint main(){ int n = 8, k = 5; cout << getNumber(n, k); return 0;} |
Java
// Java implementation of the approachclass GFG {// Function to return the kth element// in the modified arraystatic int getNumber(int n, int k){ int []arr = new int[n]; int i = 0; // First odd number int odd = 1; while (odd <= n) { // Insert the odd number arr[i++] = odd; // Next odd number odd += 2; } // First even number int even = 2; while (even <= n) { // Insert the even number arr[i++] = even; // Next even number even += 2; } // Return the kth element return arr[k - 1];}// Driver codepublic static void main(String[] args){ int n = 8, k = 5; System.out.println(getNumber(n, k));}} // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach# Function to return the kth element# in the modified arraydef getNumber(n, k): arr = [0] * n; i = 0; # First odd number odd = 1; while (odd <= n): # Insert the odd number arr[i] = odd; i += 1; # Next odd number odd += 2; # First even number even = 2; while (even <= n): # Insert the even number arr[i] = even; i += 1; # Next even number even += 2; # Return the kth element return arr[k - 1];# Driver codeif __name__ == '__main__': n = 8; k = 5; print(getNumber(n, k));# This code is contributed by Rajput-Ji |
C#
// C# implementation of the approachusing System; class GFG {// Function to return the kth element// in the modified arraystatic int getNumber(int n, int k){ int []arr = new int[n]; int i = 0; // First odd number int odd = 1; while (odd <= n) { // Insert the odd number arr[i++] = odd; // Next odd number odd += 2; } // First even number int even = 2; while (even <= n) { // Insert the even number arr[i++] = even; // Next even number even += 2; } // Return the kth element return arr[k - 1];}// Driver codepublic static void Main(String[] args){ int n = 8, k = 5; Console.WriteLine(getNumber(n, k));}} // This code is contributed by PrinciRaj1992 |
Javascript
<script>// C++ implementation of the approach// Function to return the kth element// in the modified arrayfunction getNumber(n, k){ var arr = Array(n).fill(n); var i = 0; // First odd number var odd = 1; while (odd <= n) { // Insert the odd number arr[i++] = odd; // Next odd number odd += 2; } // First even number var even = 2; while (even <= n) { // Insert the even number arr[i++] = even; // Next even number even += 2; } // Return the kth element return arr[k - 1];}// Driver code var n = 8, k = 5; document.write(getNumber(n, k));</script> |
Output:
2
Time Complexity: O(n)
Auxiliary Space: O(n), as extra space of size n is used
Efficient approach: Find the index where the first even element will be stored in the generated array. Now if the value of k is less than or equal to the index then the desired number will be k * 2 – 1 else the desired number will be (k – index) * 2
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the kth element// in the modified arrayint getNumber(int n, int k){ int pos; // Finding the index from where the // even numbers will be stored if (n % 2 == 0) { pos = n / 2; } else { pos = (n / 2) + 1; } // Return the kth element if (k <= pos) { return (k * 2 - 1); } else return ((k - pos) * 2);}// Driver codeint main(){ int n = 8, k = 5; cout << getNumber(n, k); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG{ // Function to return the kth element// in the modified arraystatic int getNumber(int n, int k){ int pos; // Finding the index from where the // even numbers will be stored if ((n % 2) == 0) { pos = n / 2; } else { pos = (n / 2) + 1; } // Return the kth element if (k <= pos) { return (k * 2 - 1); } else return ((k - pos) * 2);}// Driver codepublic static void main (String[] args){ int n = 8, k = 5; System.out.println (getNumber(n, k));}}// This code is contributed by @tushil. |
Python3
# Python3 implementation of the approach # Function to return the kth element # in the modified array def getNumber(n, k) : # Finding the index from where the # even numbers will be stored if (n % 2 == 0) : pos = n // 2; else : pos = (n // 2) + 1; # Return the kth element if (k <= pos) : return (k * 2 - 1); else : return ((k - pos) * 2); # Driver code if __name__ == "__main__" : n = 8; k = 5; print(getNumber(n, k)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return the kth element// in the modified arraystatic int getNumber(int n, int k){ int pos; // Finding the index from where the // even numbers will be stored if ((n % 2) == 0) { pos = n / 2; } else { pos = (n / 2) + 1; } // Return the kth element if (k <= pos) { return (k * 2 - 1); } else return ((k - pos) * 2);}// Driver codestatic public void Main (){ int n = 8, k = 5; Console.Write(getNumber(n, k));}}// This code is contributed by @ajit. |
Javascript
<script> // Javascript implementation of the approach // Function to return the kth element // in the modified array function getNumber(n, k) { let pos; // Finding the index from where the // even numbers will be stored if ((n % 2) == 0) { pos = parseInt(n / 2, 10); } else { pos = parseInt(n / 2, 10) + 1; } // Return the kth element if (k <= pos) { return (k * 2 - 1); } else return ((k - pos) * 2); } let n = 8, k = 5; document.write(getNumber(n, k));</script> |
Output:
2
Time Complexity: O(1)
Auxiliary Space: O(1)
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