0145EasyArrayBinary Search

Survey Marker Lookup

Tracked in this browser only

Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A survey crew filed one elevation per marker along a ridge trail, and the office sorted the file from the deepest reading up to the highest. That sorted file reaches you as nums; no two markers share an elevation, and readings below sea level show up as negatives.

An engineer names an elevation target and wants the position of the marker holding it, counting positions from 0. If the file has no marker at that elevation, report -1.

The file can be enormous and the office runs this query constantly, so your routine must halve the range it is still considering on each step rather than reading through the markers one by one.

Examples

Example 1

Input: nums = [-8, -3, 0, 4, 11], target = 11
Output: 4

The highest marker in the file carries elevation 11 and sits at position 4.

Example 2

Input: nums = [-8, -3, 0, 4, 11], target = 5
Output: -1

Elevation 5 falls between two filed markers, so no position holds it.

Example 3

Input: nums = [-6000, -25, -6], target = -6
Output: 2

Readings under sea level are filed the same way, and this one closes out the file at position 2.

Constraints

1 <= nums.length <= 10^4
-10^4 < nums[i], target < 10^4
No elevation is filed twice.
nums runs from the lowest reading to the highest.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def search(nums: list[int], target: int) -> int:

Java

public int search(int[] nums, int target)

The intended approach

Time O(log n)Space O(1)

Hold a window of positions that could still contain the elevation. Read the middle marker of that window: matching it ends the search, a lower reading means the answer can only sit above it, and a higher reading means it can only sit below. Each comparison throws away half of the remaining window.

Pythonentry point: search

def search(nums, target):
    low = 0
    high = len(nums) - 1
    while low <= high:
        mid = (low + high) // 2
        reading = nums[mid]
        if reading == target:
            return mid
        if reading < target:
            low = mid + 1
        else:
            high = mid - 1
    return -1

The obvious version

Read the file from front to back and report the first position whose elevation matches. Linear, and it never relies on the file being sorted.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_marker_position(nums, target):
    for i in range(len(nums)):
        if nums[i] == target:
            return i
    return -1

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in