Search Insert Position

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.

Here are few examples.

[1,3,5,6], 5 → 2

[1,3,5,6], 2 → 1

[1,3,5,6], 7 → 4

[1,3,5,6], 0 → 0

Solution1: 

Run time complexity is O(n), constant space.

public class Solution {
    public int searchInsert(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] >= target) {
                return i;
            }
        }
        
        return nums.length;
    }
}

Solution2: binary search
Run time complexity is O(logn), constant space.

public class Solution {
    public int searchInsert(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        int left = 0;
        int right = nums.length - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return left;
    }
}