Binary Search

The basic pattern of a binary search problem is to find a target element in an orderly list.

The model problem is Leetcode 704

Here are two points that would make mistakes:

• The boundary conditions for the loop
• The redefinition of search interval boundaris

To the boundary conditions for the loop

while (left < right)

or

while (left <= right)

we need to know when conditions can make the left and right boundary values equal and when not.

To the value of search interval boundaris

right = mid;

or

right = mid + 1;

we need to know how to redefine the boundary values of left and right.

Loop Invariant

The loop invariant is a property (or expression) that always be true within every iteration of the loop. For the binary search problem, the loop invariant here is the search interval, aka. the value of boundaries left and right.

Here we discuss this search interval should be left-closed, right-open or left-closed, right-closed.

Left-closed Right-closed Interval

Since this is a closed interval, both the left and right boundary values of the interval should be valid (i.e., legitimate values). There for, at that time, the left pointer will point to 0, and the right pointer will point to nums.size()-1.

int left = 0, right = nums.size() - 1;

For the loop condition, since both the left and right boundaris can be included in a closed interval, it is reasonable for the left and right boundaries to be equal. Therefore, the condition for the while loop should be left <= right.

When updating the boundary values in this case, we should notice that the right boundary is closed, thus the new right boudary is $mid - 1$ when nums[mid] > target and left boundary is $mid + 1$ when nums[mid] < target.

Left-closed Right-open Interval

Similar to the left-closed right-closed interval, the boundary values are updated in the same way. However, in this case, since the right boundary is open, the condition of the while loop should be left < right here. And when updating boundary values, the right one should be $ mid$ rather than $mid + 1$ because the boundary value is included by the interval.