Table of contents
Given problem
Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array.
Below is an test case for this problem.
Example 1:
Input: [1,3,5,6], 5
Output: 2
Example 2:
Input: [1,3,5,6], 2
Output: 1
Example 3:
Input: [1,3,5,6], 7
Output: 4
Example 4:
Input: [1,3,5,6], 0
Output: 0
Using Binary Search algorithm
Our array is sorted, so we can apply Binary Search algorithm to deal with it.
Below is the source code that using Binary Search algorithm.

Using first invariant
public int searchInsert(int[] nums, int target) { int left = 0; int right = nums.length  1; while (left <= right) { int mid = left + (right  left) / 2; if (nums[mid] == target) { return mid; } else if (nums[mid] < target) { left = mid + 1; } else { right = mid  1; } } return left; }

Using third invariant
public static int searchInsert(int[] nums, int target) { int blue = 1; int red = nums.length; while (blue + 1 != red) { int mid = blue + (red  blue) / 2; if (nums[mid] == target) { return mid; } if (nums[mid] < target) { blue = mid; } else { red = mid; } } return red; }
The complexity of this way:
 Time complexity: O(log(n))
 Space complexity: O(1)
Wrapping up
 Apply Binary Search algorithm when we have sorted array.
References: