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Find kth smallest number in range [1, n] when all the odd numbers are deleted in C++
In this problem, we are given two integer values n and k. Our task is to find kth smallest number in range [1, n] when all the odd numbers are deleted.
We need to find the kth smallest number in the range [1, n] which contains only even values.
So, from range [1, 5] -> number will be 2, 4.
Let’s take an example to understand the problem,
Input: n = 12, k = 4
Output: 8
Explanation:
Even elements in the range [1, n] : 2, 4, 6, 8, 10, 12
The 4th smallest element is 8.
Solution approach:
The solution is simple as we need to find the kth element from even numbers upto n. This can be easily calculated using the formula,
Element = 2*k.
Program to illustrate the working of our solution,
Example
#include <bits/stdc++.h>
using namespace std;
int main() {
int n = 124, k = 12;
if(n > 2*k){
cout<<"kth smallest number is "<<(2 * k);
}
else
cout<<"kth smallest number cannot be found";
return 0;
}Output
kth smallest number is 24
Updated on: 25-Jan-2021
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