Find number of candidates in the Exam
Given the last rank ‘L’ and the number of candidates at the last rank ‘T’, the task is to find the total number of candidates in the Exam.
Input: L = 5, T = 1
Output: 5
Input: L = 10, T = 2
Output: 11
Approach:
- Suppose L = 5 and T = 2.
- Then there can be many possible rank combinations, like 1, 2, 3, 3, 5, 5.
- So now in this, as you can see, the last rank is 5 and there are 2 students at rank 5,
- Therefore, the total number of candidates is 6.
- This can be understood by a simple formula:
L+T-1
Below is the implementation of the above approach.
C++
#include <iostream>
using namespace std;
int findParticipants( int L, int T)
{
return (L + T - 1);
}
int main()
{
int L = 10, T = 2;
cout << findParticipants(L, T);
return 0;
}
|
Java
class GFG
{
static int findParticipants( int L, int T)
{
return (L + T - 1 );
}
public static void main(String args[])
{
int L = 10 , T = 2 ;
System.out.print(findParticipants(L, T));
}
}
|
Python3
def findParticipants(L, T) :
return (L + T - 1 );
if __name__ = = "__main__" :
L = 10 ; T = 2 ;
print (findParticipants(L, T));
|
C#
using System;
class GFG
{
static int findParticipants( int L, int T)
{
return (L + T - 1);
}
public static void Main()
{
int L = 10, T = 2;
Console.Write(findParticipants(L, T));
}
}
|
Javascript
<script>
function findParticipants(L, T)
{
return (L + T - 1);
}
var L = 10, T = 2;
document.write( findParticipants(L, T));
</script>
|
Time complexity: O(1) as constant operations are being performed
Auxiliary Space: O(1)
Last Updated :
13 Mar, 2023
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