We’re working through binary search problems. This one has an initial solution with many lines duplicated. We further work the solution to use a function and a boolean variable to make the code more clean via the DRY principle.

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Initial Solution

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        
        start, end = 0, len(nums) - 1
        result = [-1, -1]
        while start <= end:
            mid = start + (end - start) // 2
            left = nums[mid - 1] if mid > 0 else float('-inf')
            if nums[mid] == target:
                if left != target:
                    result[0] = mid
                    break
                end = mid - 1
            elif nums[mid] < target:
                start = mid + 1
            else:
                end = mid - 1
        start, end = 0, len(nums) - 1
        while start <= end:
            mid = start + (end - start) // 2
            right = nums[mid + 1] if mid < len(nums) - 1 else float('inf')
            if nums[mid] == target:
                if right != target:
                    result[1] = mid
                    break
                start = mid + 1
            elif nums[mid] < target:
                start = mid + 1
            else:
                end = mid - 1
        return result

Time complexity is O(logn) and space complexity is O(1).