Solution for SPOJ AGGRCOW - c++

Question

I have the folowing problem:
Farmer John has built a new long barn, with N (2 <= N <= 100,000)
stalls. The stalls are located along a straight line at positions
x1,...,xN (0 <= xi <= 1,000,000,000).
His C (2 <= C <= N) cows don't like this barn layout and become
aggressive towards each other once put into a stall. To prevent the
cows from hurting each other, FJ wants to assign the cows to the
stalls, such that the minimum distance between any two of them is as
large as possible. What is the largest minimum distance?
Input
t – the number of test cases, then t test cases follows.
* Line 1: Two space-separated integers: N and C
* Lines 2..N+1: Line i+1 contains an integer stall location, xi
Output
For each test case output one integer: the largest minimum distance.
Example
Input:
1
5 3
1
2
8
4
9
Output:
3 Output details:
FJ can put his 3 cows in the stalls at positions 1, 4 and 8,
resulting in a minimum distance of 3. Submit solution!
My approach was to pick some pair which has a certain gap and check if there are enough elements in the array to satisfy the need of all the cows.
To find these elements,I used binary search.
When I find an element , I reset my left to mid so I can continue based on the number of cows left.
My code:
#include <iostream>
int bsearch(int arr[],int l,int r,int gap , int n,int c){
int stat = 0;
for (int i = 1;i <= c; ++i) {
while(l <= r) {
int mid = (l+r)/2;
int x = n+(i*gap);
if (arr[mid] > x && arr[mid-1] < x) {
l = mid;
++stat;
break;
}
if(arr[mid] < x) {
l = mid + 1;
}
if (arr[mid] > x) {
r = mid - 1;
}
}
}
if (stat == c) {
return 0;
}
else {
return -1;
}
}
int calc(int arr[],int n , int c) {
int max = 0;
for (int i = 0; i < n; ++i) {
for (int j = i+1;j < n; ++j) {
int gap = arr[j] - arr[i];
if (gap > max) {
if (bsearch(arr,j,n-1,gap,arr[j],c) == 0) {
max = gap;
}
}
}
}
return max;
}
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int t , n ,c;
cin >> t;
for(int i = 0 ; i < t; ++i) {
cin >> n >> c;
int arr[n];
for (int z = 0 ; z < n; ++z) {
cin >> arr[z];
}
sort(arr,arr+n);
//Output;
int ans = calc(arr,n,c);
cout << ans;
}
return 0;
}
Problem Page:
https://www.spoj.com/problems/AGGRCOW/