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

🟧 Medium

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

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

You must 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 <= 10^5

• -10^9 <= nums[i] <= 10^9

• nums is a non-decreasing array.

• -10^9 <= target <= 10^9

Solution

My Solution

func SearchRange(nums []int, target int) []int {
    l := LowerBound(nums, target)
    if l == len(nums) || nums[l] != target {
        return []int{-1, -1}
    }

    r := UpperBound(nums, target)
    return []int{l, r - 1}
}

func LowerBound(nums []int, target int) int {
    l, r := 0, len(nums)
    for l < r {
        mid := l + (r-l)/2
        if nums[mid] < target {
            l = mid + 1
        } else {
            r = mid
        }
    }

    return l
}

func UpperBound(nums []int, target int) int {
    l, r := 0, len(nums)
    for l < r {
        mid := l + (r-l)/2
        if nums[mid] <= target {
            l = mid + 1
        } else {
            r = mid
        }
    }

    return l
}

Approach

This solution uses two modified binary searches to find the range:

1. LowerBound (First Position):

   • Uses binary search to find the first occurrence

   • When nums[mid] == target, continues searching left half

   • Returns leftmost position where target could be inserted

2. UpperBound (Last Position):

   • Uses binary search to find position after last occurrence

   • When nums[mid] == target, continues searching right half

   • Returns position after the last occurrence

3. Main Function:

   • If LowerBound doesn't find target, return [-1, -1]

   • Otherwise, return [LowerBound, UpperBound-1]

Complexity Analysis

Time Complexity: O(log n)

• Performs two binary searches

• Each binary search takes O(log n) time

• Total time is O(log n) + O(log n) = O(log n)

Space Complexity: O(1)

• Only uses a constant amount of extra space:

  • Two pointers (left, right)

  • Mid index

  • Result array of size 2

Why it works

• LowerBound finds first position by treating equal elements as "too big"

• UpperBound finds last position by treating equal elements as "too small"

• Using [0, len] range instead of [0, len-1] handles edge cases elegantly

• The difference between < and <= in the two functions is crucial:

  • LowerBound: moves right when element is too small

  • UpperBound: moves right when element is too small OR equal

result

Leetcode: link