Binary search test case 1 and 3

#1

#include
#include

using namespace std;

int main()
{
int key,u,l,m,n,a[10000];
cin >> n;
for(int i=0;i<n;i++)
cin >> a[i];
cin >> key;
sort(a,a+n);
l=0;
u=n-1;
while(l<=u)
{
m=(u+l)/2;
if(key==a[m])
{
cout << m << endl;
break;
}
else if(key > a[m])
l=m+1;
else if(key < a[m])
u=m-1;
}
return 0;
}

#2

Hi, your code doesnt consider the case when the number is not found in the array -1 has to be printed

also define an array of size n and not a[10000] it might not cover all cases

also since the i/p array is always in increasing order , u are not required to call the sort function

try doing all the changes by ur own
in case u fail
refer to the below code

#3

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