Given two integers N and K, the task is to find the element at the Kth position if all odd numbers from 1 to N are written down in increasing order followed by all the even numbers from 1 to N in increasing order.
Examples:
Input: N = 10, K = 3
Output: 5
The required sequence is 1, 3, 5, 7, 9, 2, 4, 6, 8 and 10.
Input: N = 7, K = 7
Output: 6
Approach: It is known that the Nth even number is given by 2 * K and the Nth odd number is given by 2 * K – 1. But since the even numbers are written after (N + 1) / 2 odd numbers here. Therefore, Kth even number is given by 2 * (K – (N + 1) / 2) and the odd numbers will remain the same as 2 * K – 1
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 number// from the required sequenceint kthNum(int n, int k){ // Count of odd integers // in the sequence int a = (n + 1) / 2; // kth number is even if (k > a) return (2 * (k - a)); // It is odd return (2 * k - 1);}// Driver codeint main(){ int n = 7, k = 7; cout << kthNum(n, k); return 0;} |
Java
// Java implementation of the approachclass GFG {// Function to return the kth number// from the required sequencestatic int kthNum(int n, int k){ // Count of odd integers // in the sequence int a = (n + 1) / 2; // kth number is even if (k > a) return (2 * (k - a)); // It is odd return (2 * k - 1);}// Driver codepublic static void main(String []args) { int n = 7, k = 7; System.out.println(kthNum(n, k));}}// This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach # Function to return the kth number # from the required sequence def kthNum(n, k) : # Count of odd integers # in the sequence a = (n + 1) // 2; # kth number is even if (k > a) : return (2 * (k - a)); # It is odd return (2 * k - 1); # Driver code if __name__ == "__main__" : n = 7; k = 7; print(kthNum(n, k)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System; class GFG {// Function to return the kth number// from the required sequencestatic int kthNum(int n, int k){ // Count of odd integers // in the sequence int a = (n + 1) / 2; // kth number is even if (k > a) return (2 * (k - a)); // It is odd return (2 * k - 1);}// Driver codepublic static void Main(String []args) { int n = 7, k = 7; Console.WriteLine(kthNum(n, k));}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript implementation of the approach// Function to return the kth number// from the required sequencefunction kthNum(n, k){ // Count of odd integers // in the sequence var a = (n + 1) / 2; // kth number is even if (k > a) return (2 * (k - a)); // It is odd return (2 * k - 1);}// Driver codevar n = 7, k = 7;document.write(kthNum(n, k));</script> |
Output:
6
Time Complexity: O(1)
Auxiliary Space: O(1)
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