The first overtake

 Problem:-

In a car race, there are N cars 1,2 ... N where the ith car is moving with a speed of (N - i + 1). All the cars are moving in the same direction.

The current position of the cars is Xi. Determine the time at which the first overtake takes place.

Note: Overtake is defined as the place when a car crosses another car that is ahead of it.

Input format

• Each test contains multiple test cases.
• The first line contains the number of test cases T.
• The first line of each test case contains two space-separated integers N and M representing the number of cars and the length of the circular path.
• The second line of each test case contains N integers where the ith integer is Xi representing the position of the ith car on the track.

Output format

For each test case, print a single line containing one integer representing the time at which the first overtake takes place.

Constraints

$$\begin{gathered} 1\le T\le {10}^4 \\ 2\le N\le {10}^5 \\ 1\le M\le {10}^9 \\ 0\le X_i\le {10}^9 \\ {X}_i<X_{i+1}\text{ for }(1\le i\le N-1) \\ \text{It is guaranteed that the sum of N over all test cases does not exceed }2\cdot {10}^5 \end{gathered}$$

Sample Input

1
2
1 2

Sample Output

1

Time Limit: 2

Memory Limit: 256

Source Limit:

Explanation

The first overtake takes place at time t = 1.

Code:-

#include<stdio.h>

int main()

{

    int t;

    scanf("%d",&t);

    long int n;

    while(t--)

    {

        scanf("%ld",&n);

        long long int a[n],i,m=0,min=10000000007,t,j,flag=0;

        for(i=0;i<n;i++)

        {

            scanf("%lld",&a[i]);

            if(flag==0)

         {

             m=a[i];

             flag=1;

            }

            else

            {

                t=abs(a[i]-m);

                if(min>t)

                min=t;

            }

            m=a[i];

        }

    /*  for(i=1;i<n;i++)

        {

            for(j=i+1;j<n;j++)

            {

                t=abs(a[i]-a[j]);

                if(min>t)

                min=t;

            }

        }*/

        printf("%lld\n",min);

    }

    return 0;

}