C++ Program to find maximum possible smallest time gap between two pair of clock readings

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 
Example
Let us see the following implementation to get better understanding −
#include 
using namespace std;
int solve(vector D) {
   int n = D.size();
   sort(D.begin(), D.end());
   vector t;
   t.push_back(0);
   for (int i = 0; i  D = { 7, 12, 8 };
   cout 
Input
{ 7, 12, 8 }

Output
4