There are n spectators in the stadium, labeled from 1 to n.
At time 1, the first spectator stands.
At time 2, the second spectator stands.
…
At time k, the k-th spectator stands.
At time k + 1, the (k + 1)-th spectator stands and the first spectator sits.
At time k + 2, the (k + 2)-th spectator stands and the second spectator sits.
…
At time n, the n-th spectator stands and the (n – k)-th spectator sits.
At time n + 1, the (n + 1 – k)-th spectator sits.
…
At time n + k, the n-th spectator sits.
Find the number of spectators standing in the stadium at a time t.
Examples :
Input : 10 5 3 Output : 3 Explanation : n = 10, k = 5, t = 3 At time 1, 1st spectator stands. At time 2, 2nd spectator stands. At time 3, 3rd spectator stands. Thus, the result is 3 as there are total of 3 spectators standing. Input :10 5 7 Output : 5 Explanation : n = 10, k = 5, t = 7 At time 1, 1st spectator stands. At time 2, 2nd spectator stands. At time 3, 3rd spectator stands. At time 4, 4th spectator stands. At time 5, 5th spectator stands. At time 6, 6th spectator stands and 1st spectator sits [(n-k) = 6 - 5 = 1 as mentioned above]. At time 7, 7th spectator stands and 2nd spectator sits. So, now there are total of 5 spectators standing. Thus, result is 5.
Approach :
Now we can observe certain conditions in this problem :
NOTE : There can only be maximum of k spectators standing in the stadium.
1) If the time is less than k value then that means spectators are still standing. So the result would be t.
2) In time between n and k(inclusive), the spectators standing are k. [Observation]
3) After n time, the spectators starts to sit one by one. So we calculate the spectators sitting by ‘t – n’. Then subtract k by this value. This produces the result.
Below is the implementation of the above approach :
C++
// CPP program to find number of spectators // standing at a time#include <bits/stdc++.h>using namespace std;void result(long long n, long long k, long long t){ // If the time is less than k // then we can print directly t time. if (t <= k) cout << t; // If the time is n then k spectators // are standing. else if (t <= n) cout << k; // Otherwise we calculate the spectators // standing. else { long long temp = t - n; temp = k - temp; cout << temp; }}// Driver codeint main(){ // Stores the value of n, k and t // t is time // n & k is the number of spectators long long n, k, t; n = 10; k = 5; t = 12; result(n, k, t); return 0;} |
Java
// Java program to find number of spectators // standing at a timeclass GFG { static void result(long n, long k,long t) { // If the time is less than k // then we can print directly t time. if (t <= k) System.out.println(t); // If the time is n then k spectators // are standing. else if (t <= n) System.out.println(k); // Otherwise we calculate the // spectators standing. else { long temp = t - n; temp = k - temp; System.out.println(temp); } } // Driver code public static void main(String args[]) { // Stores the value of n, k and t // t is time // n & k is the number of spectators long n, k, t; n = 10; k = 5; t = 12; result(n, k, t); }}/*This code is contributed by Nikita Tiwari.*/ |
Python3
# Python program to find number of spectators # standing at a timedef result(n, k, t) : # If the time is less than k # then we can print directly t time. if (t <= k) : print(t ) # If the time is n then k spectators # are standing. elif (t <= n) : print( k) # Otherwise we calculate the # spectators standing. else : temp = t - n temp = k - temp print (temp) # Driver code# Stores the value of n, k and t# t is time# n & k is the number of spectatorsn = 10k = 5t = 12result(n, k, t)# This code is contributed by Nikita Tiwari. |
C#
// C# program to find number of // spectators standing at a timeusing System;class GFG { static void result(long n, long k,long t) { // If the time is less than k // then we can print directly t time. if (t <= k) Console.WriteLine(t); // If the time is n then k spectators // are standing. else if (t <= n) Console.WriteLine(k); // Otherwise we calculate the // spectators standing. else { long temp = t - n; temp = k - temp; Console.WriteLine(temp); } } // Driver code public static void Main() { // Stores the value of n, k and t // t is time // n & k is the number of spectators long n, k, t; n = 10; k = 5; t = 12; result(n, k, t); }} //This code is contributed by Anant Agarwal. |
PHP
<?php// PHP program to find number of // spectators standing at a timefunction result($n, $k, $t){ // If the time is less // than k then we can // print directly t time. if ($t <= $k) echo t; // If the time is n then // k spectators are standing. else if ($t <= $n) echo k; // Otherwise we calculate // the spectators standing. else { $temp = $t - $n; $temp = $k - $temp; echo $temp; }}// Driver code// Stores the value of n, k and t// t is time// n & k is the number of spectators$n = 10;$k = 5;$t = 12;result($n, $k, $t);// This code is contributed by Sam007?> |
Javascript
<script>// Javascript program to find number of // spectators standing at a timefunction result(n, k, t){ // If the time is less than k // then we can print directly t time. if (t <= k) document.write(t); // If the time is n then k spectators // are standing. else if (t <= n) document.write(k); // Otherwise we calculate the // spectators standing. else { let temp = t - n; temp = k - temp; document.write(temp); }}// Driver code// Stores the value of n, k and t// t is time// n & k is the number of spectatorslet n, k, t;n = 10;k = 5;t = 12;result(n, k, t);// This code is contributed by susmitakundugoaldanga </script> |
Output :
3
Time Complexity : O(1)
Space Complexity : O(1)
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