Find the Kth position element of the given sequence - GeeksforGeeks

07 Mar, 2022
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 – 1Below is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// 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 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 implementation of the approach # Function to return the kth number# from the required sequencedef 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 codeif __name__ == "__main__" :     n = 7; k = 7;     print(kthNum(n, k)); # This code is contributed by AnkitRai01
// 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
<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>
6
 
Time Complexity: O(1)
Auxiliary Space: O(1) 
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