C++ Program to get number at position k after positioning n natural numbers

Suppose we have two numbers n and k. We are determined to rearrange natural numbers. But there are too many natural numbers, so we have decided to start with the first n. Pick the following sequence of numbers: firstly, all odd integers from 1 to n (in ascending order), then all even integers from 1 to n (also in ascending order). We have to find which number will stand at the position number k.

Problem Category

Various problems in programming can be solved through different techniques. To solve a problem, we have to devise an algorithm first, and to do that we have to study the particular problem in detail. A recursive approach can be used if there is a recurring appearance of the same problem over and over again; alternatively, we can use iterative structures also. Control statements such as if-else and switch cases can be used to control the flow of logic in the program. Efficient usage of variables and data structures provides an easier solution and a lightweight, low-memory-requiring program. We have to look at the existing programming techniques, such as Divide-and-conquer, Greedy Programming, Dynamic Programming, and find out if they can be used. This problem can be solved by some basic logic or a brute-force approach. Follow the following contents to understand the approach better.

So, if the input of our problem is like n = 10; k = 3, then the output will be 5.


To solve this, we will follow these steps −

n := (n + 1) / 2
return (k * 2 - 1) if (k <= n), otherwise (2 * (k - n))


Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
int solve(int n, int k){
   n = (n + 1) / 2;
   return (k <= n ? k * 2 - 1 : 2 * (k - n));
int main(){
   int n = 10;
   int k = 3;
   cout << solve(n, k) << endl;


10, 3