Find First and Last Position of Element in Sorted Array (in Python)

Problem

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).

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

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Solution

• Approach 1 : Linear Scan

Checking every index for target exhausts the search space, so it must work.

But its time complexity is \(O(n)\).

• Approach 2 : Binary Search

Because the array is sorted, we can use binary search to locate the left and rightmost indices.

class Solution:
 
     def searchRange(self, nums, target) :
 
        n = len(nums)
        
        if n == 0 or (n==1 and target != nums[0]):
            return [-1, -1]
 
        def search(arr, t):
 
            l, r = 0, len(arr)
 
            while l<r :
 
                mid = (l+r)//2
                num = arr[mid]
 
                if num < t :
                    l = mid+1
                else :
                    r = mid
            
            return l
 
        start = search(nums, target)
 
        if start >= n or target != nums[start] : 
            return [-1, -1]
 
        end = start + search(nums[start:], target+1)-1
 
        return [start, end]

Complexity Analysis

•Time complexity : \(O(logn)\)

•Space complexity : \(O(1)\). All work is done in place, so the overall memory usage is constant.

Source

https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/