# Day-17 Binary Search

##### Background

This problem statement is a part of LeetCode's learn card binary search. Under the sub-heading background.

##### Problem Statement

Given a sorted (in ascending order) integer array nums of n elements and a target value, write a function to search target in nums. If target exists, then return its index, otherwise, return -1.

##### Example 1
``````Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4
``````
##### Example 2
``````Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1
``````

Note:

1. You may assume that all elements in nums are unique n will be in the range [1, 10000].
2. The value of each element in nums will be in the range [-9999, 9999].
##### Solution Approach
1. Keep dividing the array into halves.
2. left = 0, right = len(array)-1. Mid would be left+right/2.
3. Then determine where the element to be searched lies.
``````class Solution:
def search(self, nums: List[int], target: int) -> int:
left = 0
right = len(nums)-1
while left <= right:
mid = abs(int((left+right)/2))
if target < nums[mid]:
right = mid - 1
elif target > nums[mid]:
left = mid + 1
elif nums[mid] == target:
return mid
return -1
``````

#### Learnings

1. Time complexity is O(logn).
2. Take care of the edge conditions.