Find First and Last Position of Element in Sorted Array

Binary Search

Problem

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

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

Follow up: Could you write an algorithm with O(log n) runtime complexity?

For example:

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

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

Input: nums = [], target = 0
Output: [-1,-1]

Thought Process

• To find the first occurence/position of target, if num[mid] is greater than the target, then we have to move down. Along with this, if num[mid] is equal to the target, we have to move down because this potentially is not the first position of this element.

• To find the last occurence/position of target, if num[mid] is less than the target then we have to keep moving up. Along with this we have to keep moving up if num[mid] is equal to target because it potentially may not be the last occurence/position of this element.

Solution

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        first = self.first(nums,target)
        last = self.last(nums, target)
          
        return [first,last]
        
    def first(self, nums, target):
        low, high = 0, len(nums)-1
        res = -1
        
        while low <= high:
            mid = low + (high-low)//2
            if nums[mid] >= target:
                high = mid-1
            elif nums[mid] < target:
                low = mid+1
                
            if nums[mid] == target:
                res = mid
        return res
    
    def last(self, nums, target):
        low, high = 0, len(nums)-1
        res = -1
        
        while low <= high:
            mid = low + (high-low)//2
            if nums[mid] <= target:
                low = mid+1
            elif nums[mid] > target:
                high = mid-1
            
            if nums[mid] == target:
                res = mid
            
        return res

Time Complexity

• Time: O(log n)

• Space: O(1)