34. Find First and Last Position of Element in Sorted Array

Given an array of integers nums sorted in ascending order, 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).

Binary search, twice

The O(log n ) run time combined with the sorted array tell us to use binary search. Maybe twice.

The first binary search to find the first occurrence of the target

The normal binary search will stop after nums[mid] == target. EG: [0,3,3,3,3], t = 3. We will stop at index 2. To keep the binary search going, we will search again in [0:mid – 1]. If we didn’t find any occurrence in the range, that means the previous found target is the first occurrence.

The second bs finds the last occurrence of the target

Same as above but search repeatedly in [mid + 1: end]

Recursive code

    def searchRange(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        We will call modified binary search twice. Once search leftmost and once search rightmost

        """
        first = self.find_first(nums, target, 0, len(nums) - 1)
        last = self.find_last(nums, target, 0, len(nums) - 1)
        return (first,last)


    def find_first(self, nums, target, start, end):
        """

        """
        if not nums: return -1
        if nums[start] == target:
            return start
        if start == end:
            return -1

        mid = start + (end - start) // 2


        if nums[mid] >= target:
            return self.find_first(nums, target, start, mid)
        else:
            return self.find_first(nums, target, mid + 1, end)


    def find_last(self, nums, target, start, end):

        if not nums: return -1
        if nums[end] == target:
            return end
        if start == end:       
            return -1

        mid = end - (end - start) // 2


        if nums[mid] <= target:
            return self.find_last(nums, target, mid, end)

        else:
            return self.find_last(nums, target, start, mid - 1)

Take out return statement when nums[mid] == target

We change it from basic binary search by taking out the return statement when we found nums[mid] == target. This allow us to keep searching.

Find first

mid = start + (end – start) // 2

With this formula for mid, mid will be at the center when the number of numbers is odd. And mid will be at the left of the middle when we have a even number of numbers. This is because we calculate middle using start and add the floor of half of the length.

This helps terminate the code when we don’t have any target in the range.

  1. find_first(nums, target, mid + 1, end) will terminate as usual binary search since mid + 1 will cross end
  2. find_first(nums, target, start, mid) If we use mid = end – (end – start) // 2, then it won’t terminate if we recursively call this function. An example is [2,3,4] target = 0. We will call : find_first(nums, 0, 0, 2) find_first(nums, 0, 0, 1) find_first(nums, 0, 0, 1) for many times

This problem will be solved if we let mid = start + (end – start) // 2.

If found a target, maintain it in the range by searching again in [start, mid] instead of [start, mid – 1]

This allow us to have at least 1 occurrence of target to “fall back” on.

Return start if nums[start] == target

After many recursive calls, our effective range will have 4 conditions

  1. [target, target] nums[start] is definitely the first target. So we return it
  2. [not target, target] Let’s assume they have index k and k + 1. mid = k and we will call find_first(nums, target, k, k) and it nums[k] == target. We indeed return the first occurrence of the target
  3. [target, not target] nums[start] == target, so return start
  4. [not target, not target] It will return -1

So in all 4 situation, if nums[start] == target, and we return, this will lead to correct behavior.

Find last

It is almost identical to find first. But we need to change the recursive call conditions and the calculation of mid.

mid = end – ( end – start) // 2

So we can let let mid squeeze to the back in the case of even length range. This will allow use to terminate.

Iterative code

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