Search Insert Position and Search for a Range

Both problems require binary search. There are two ways to write the loop:

 while (start + 1 < end) {
            int mid = (start + end) / 2;
            if (A[mid] == target)
                return mid;
            else if (A[mid] < target) 
                start = mid;
            else
                end = mid;
        }

Or:

 while (start < end) {
            int mid = (start + end) / 2;
            if (A[mid] == target)
                return mid;
            else if (A[mid] < target) 
                start = mid + 1;
            else
                end = mid - 1;
        }

The first one exits the loop when there are two elements left and the second one exits when there is only one element. I always prefer the second one because it requires less comparison after exiting the loop.

However, for these problem, it is better to use the first one, because both problems require us to search for a range.

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

If there are two elements left when we exit the loop, either the target is at the start, the end, or we need to insert it in between. If we exit the loop with only one element, and if the target doesn't equal to the element at that position, then it's hard for us to determine where to add the target.

public int searchInsert(int[] A, int target) {
        if (A == null)
            throw new NullPointerException("Null array!");
        if (A.length == 0)
            return 0;
        if (target <= A[0])
            return 0;
        if(target > A[A.length - 1])
            return A.length;
        int start = 0;
        int end = A.length - 1;
        while (start + 1 < end) {
            int mid = (start + end) / 2;
            if (A[mid] == target)
                return mid;
            else if (A[mid] < target) 
                start = mid;
            else
                end = mid;
        }
        if (A[start] == target)
            return start;
        else if (A[end] == target)
            return end;
        return start + 1;
    }

Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

For this problem, my approach is to do binary search twice, first time to search for the start position of the range, the second to search for the end position. Since we are looking for a range, when A[mid] == target, we can not simply return the mid, to search for the start point, we search the first half of the array, and to search for the end point, we search the second half of the array.

public int[] searchRange(int[] A, int target) {
        if (A == null)
            throw new NullPointerException("Null array!");
        int[] bound = new int[2];
        bound[0] = -1;
        bound[1] = -1;
        if (A.length == 0 || target < A[0] || target > A[A.length - 1])
            return bound;
        int start = 0;
        int end = A.length - 1;
        while (start + 1< end) {
            int mid = (start + end) / 2;
            if (A[mid] == target) {
                end = mid;
            }
            else if (A[mid] < target)
                start = mid;
            else
                end = mid;
        }
        if (A[start] == target)
            bound[0] = start;
        else if (A[end] == target)
            bound[0] = end;
        else
            return bound;
        start = 0;
        end = A.length - 1;
        while (start + 1 < end) {
            int mid = (start + end) / 2;
            if (A[mid] == target)
                start = mid;
            else if (A[mid] < target)
                start = mid;
            else
                end = mid;
        }
        if (A[end] == target)
            bound[1] = end;
        else if (A[start] == target)
            bound[1] = start;
        else 
            bound[0] = -1;
        return bound;
        
    }