Better solution

#1

#include
using namespace std;

int main()
{
int n,a[10000];
cin>>n;
int m=0;
for(int i=0;i<n;i++)
{
cin>>a[i];
if(a[i]>a[i-1])
{
m=i;
}
else
{
break;
}
}
cout<<m;
}

this problem doesnt require binary search to solve.

#2

No your solution is worse. Please explain how is it better than the O(logN) that we get from binary search?

#3

i dont have to use binary search as i calculated my answer while taking inputs only

#4
#5

That’s one way to look at it, but when someone asks, they assume that you already have the array. And given the array you have to design the algorithm. So in that sense your algorithm is linear search.

#6

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