Binary Search for a range - c++

Question

I was making binary search program to find the number of elements between the Left and Right values in a range .
I code it :
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
vector<int> arr(20);
int search(int value,int low,int high)
{
if (high <= low)
return low;
int mid = (low + high) / 2;
if (arr[mid] > value)
return search(value,low,mid-1);
else
return search(value,mid+1,high);
}
int main(){
int n;
cin>>n;
//ENTER SORTED ARRAY
for(int i=0;i<n;i++){
cin>>arr[i];
}
int left;
cin>>left;
//RIGHT IS GREATER OR EQUAL TO LEFT
int right;
cin>>right;
cout<<search(right,0,n-1)-search(left,0,n-1)+1<<"\n";
}
It's giving right answer for some ranges.
But for some its giving wrong like If N=6 and array be [1 3 5 8 10 13] and say the range be [5,9] then it's giving 1 as the answer but it should be 2 as 5 and 8 both are in the range.

Answer 1

try this
int search(int value,int low,int high)
{
if (high <= low)
return low;
int mid = (low + high) / 2;
if(arr[mid]==value){ // add this line it would be work for you
return mid;
}
if (arr[mid] > value)
return search(value,low,mid-1);
else
return search(value,mid+1,high);
}
and make correction in main()
cout<<search(right,0,n-1)-search(left,0,n-1)<<"\n";

Answer 2

int search(int value,int low,int high)
{
if (high <= low + 1)
return low;
int mid = (low + high) / 2;
if (arr[mid] > value)
return search(value,low,mid);
else
return search(value,mid,high);
}
And in your main function
cout<<search(right+1,0,n-1)-search(left,0,n-1)<<"\n";

Answer 3

One problem is that when arr[mid] == value, you just ignore it and recurse to the right.
You'll need to either include mid in your right range, or return mid if arr[mid] == value.
I also see duplicate values (if these are possible) being a problem - when recursing to find the left-most position, you need to find the first duplicate value, when recursing to find the right-most position, you need to find the last duplicate value, so a single function without a flag to indicate which one we're doing isn't going to work. To illustrate the problem:
If the range is [5,5] and the input is [1,2,5,5,5,6,8], the same recursive call finding the position of 5 will always return the position of the same 5, where-as you need to return index 2 for the left range and index 4 for the right, as to get 3 as your output.

Answer 4

There is no check that arr[mid] can be == value. In your example, first iteration for left == 5 gives mid == ( 0 + (6-1) )/2 = 5/2 = 2 and arr[2] is exactly 5. We should stop, but your code goes to the branch search(5, 3, 5);

Answer 5

The logic of your program seems wrong, if you want to find the number of elements in the arr that are in the range of [left,right], try this:
int i;
int count = 0;
for(i = 0; i < n; i++) {
if (arr[i] >= left && arr[i] <= right)
count++;
}
If you insist on using binary search try this :
static int search(int value,int low,int high)
{
if (high <= low)
return low;
int mid = (low + high) / 2;
if (arr[mid] == value)
return mid;
int idx;
if (arr[mid] > value)
idx = search(value,low,mid-1);
else
idx = search(value,mid+1,high);
if (value == arr[idx]) {
return idx;
}
else {
if(value > arr[idx])
return mid +1;
else
return mid;
}
}