Find Maximum Possible Smallest Time Gap Between Two Clock Readings in C++

Suppose we have an array D with N elements. Consider in a code festival there are N+1 participants including Amal. Amal checked and found that the time gap between the local times in his city and the i-th person's city was D[i] hours. The time gap between two cities: For any two cities A and B, if the local time in city B is d o'clock at the moment when the local time in city A is 0 o'clock, then the time gap between these two cities is minimum of d and 24−d hours.

Here, we are using 24-hour notation. Then, for each pair of two people chosen from the N+1 people, he wrote out the time gap between their cities. Let the smallest time gap among them be s hours. We have to find the maximum possible value of s.

So, if the input is like D = [7, 12, 8], then the output will be 4, because the time gap between the second and third persons' cities is 4 hours.

Steps

To solve this, we will follow these steps −

n := size of D
sort the array D
Define an array t
insert 0 at the end of t
for initialize i := 0, when i < n, update (increase i by 1), do:
   if i mod 2 is same as 0, then:
      insert D[i] at the end of t
   Otherwise
      insert 24 - D[i] at the end of t
sort the array t
ans := inf
for initialize i := 1, when i < size of t, update (increase i by 1), do:
   ans := minimum of ans and t[i] - t[i - 1]
return ans

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
int solve(vector<int> D) {
   int n = D.size();
   sort(D.begin(), D.end());
   vector<int> t;
   t.push_back(0);
   for (int i = 0; i < n; i++){
      if (i % 2 == 0)
         t.push_back(D[i]);
      else
         t.push_back(24 - D[i]);
   }
   sort(t.begin(), t.end());
   int ans = 1e9;
   for (int i = 1; i < t.size(); i++){
      ans = min(ans, t[i] - t[i - 1]);
   }
   return ans;
}
int main(){
   vector<int> D = { 7, 12, 8 };
   cout << solve(D) << endl;
}

Input

{ 7, 12, 8 }

Output

4