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

Difficulty: Medium | Pattern: Binary Search (Left & Right Bounds) | Company tags: Facebook, Amazon, Google, LinkedIn

Problem Statement​

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]

Approach: Two Binary Searches — O(log n)​

Run binary search twice: once to find the leftmost occurrence, once for the rightmost.

def searchRange(nums: list[int], target: int) -> list[int]:
    def find_left():
        left, right = 0, len(nums) - 1
        idx = -1
        while left <= right:
            mid = (left + right) // 2
            if nums[mid] == target:
                idx = mid
                right = mid - 1  # keep searching left
            elif nums[mid] < target:
                left = mid + 1
            else:
                right = mid - 1
        return idx
    
    def find_right():
        left, right = 0, len(nums) - 1
        idx = -1
        while left <= right:
            mid = (left + right) // 2
            if nums[mid] == target:
                idx = mid
                left = mid + 1   # keep searching right
            elif nums[mid] < target:
                left = mid + 1
            else:
                right = mid - 1
        return idx
    
    return [find_left(), find_right()]

Using bisect Module​

import bisect

def searchRange(nums: list[int], target: int) -> list[int]:
    left = bisect.bisect_left(nums, target)
    if left == len(nums) or nums[left] != target:
        return [-1, -1]
    right = bisect.bisect_right(nums, target) - 1
    return [left, right]

Dry Run​

nums = [5,7,7,8,8,10], target = 8

Left search:

  • left=0, right=5, mid=2: nums[2]=7 lt 8 → left=3
  • left=3, right=5, mid=4: nums[4]=8 → idx=4, right=3
  • left=3, right=3, mid=3: nums[3]=8 → idx=3, right=2
  • left=3, right=2: stop → return 3

Right search:

  • left=0, right=5, mid=2: nums[2]=7 lt 8 → left=3
  • left=3, right=5, mid=4: nums[4]=8 → idx=4, left=5
  • left=5, right=5, mid=5: nums[5]=10 gt 8 → right=4
  • left=5, right=4: stop → return 4

[3, 4] ✓

Complexity​

  • Time: O(log n) — two binary searches
  • Space: O(1)