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

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

Description

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?

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]

Example 3:

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

Constraints:

• 0 <= nums.length <= 105

• -109 <= nums[i] <= 109

• nums is a non-decreasing array.

• -109 <= target <= 109

Tags

Array, Binary Search

Solution

Based on the binary search template, we need to re-design the case of nums[mid] == target, because we do not want to return the found index immediately when it might not be either the starting or the ending position. In the case of equality of the both, if we need to find the starting position of target, we assign mid to r. Otherwise we assign mid to l. After the loop, If we are looking for the starting position, we check the nums[l] first and then nums[r], otherwise the process is reversed. Return -1 when target does not present in the array.

Complexity

• Time complexity: $O(log(n))$

• Space complexity: $O(1)$

Code

func searchRange(nums []int, target int) []int {
	if len(nums) == 0 {
		return []int{-1, -1}
	}
	return []int{binarySearch(nums, target, true), binarySearch(nums, target, false)}
}

func binarySearch(nums []int, target int, first bool) int {
	l, r := 0, len(nums)-1
	for l+1 < r {
		mid := l + (r-l)/2
		if nums[mid] > target {
			r = mid
		} else if nums[mid] < target {
			l = mid
		} else {
			if first {
				r = mid
			} else {
				l = mid
			}
		}
	}
	if !first {
		l, r = r, l
	}
	if nums[l] == target {
		return l
	}
	if nums[r] == target {
		return r
	}
	return -1
}