Sorted anchors and two pointers

Balance Three Calibration Readings

Original & variant

Original examples and wording. The question below states the exact input format and output contract.

01. The question

Choose three different positions whose signed readings sum to the requested calibration total. Return unique value triples, each increasing, in lexicographic order. Variant: the target is configurable, not always zero. Equal readings may be used only when enough separate positions exist.

Before revealing the idea, predict the result and describe what information you would need to track.

02. Worked example

Input[-2, 0, 1, 1, 3, -1]; total = 2

Output[[-2,1,3],[-1,0,3],[0,1,1]]

Sort to [-2,-1,0,1,1,3].
Anchor -2 pairs 1 with 3; repeated 1 must not duplicate that triple.
Later anchors find [-1,0,3] and [0,1,1].

Try your code

Your Python solution

Define solve(data, parameter="") and return your result. Your code runs locally, not on a server.

Edit your test cases

Custom test cases (JSON)

Use 1-10 objects with data and an optional string parameter. Expected outputs are computed from the lesson. An optional expected field must be correct.

Input contract & reference code

Example data: [-2,0,1,1,3,-1]. Parameter: "2". Numeric parameters arrive as strings; convert them with int(parameter) when needed.

The example and generated cases use the question’s exact variant. This checks behavior on a small test batch—not performance on every possible input.

Python runtime notices

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Printed output

Five-second execution limit per batch. Standard-library Python only; additional packages and network access are unavailable. Stop ends execution without blocking the lesson.

03. The idea

Try sketching your own approach first. What would you track, and why would it work?

Reveal the idea

Sort a copy and skip repeated anchor values.
Move two inner pointers according to whether the sum is too small or too large.
After a match, move both pointers past repeated endpoint values.

04. Watch it happen

Use Step to move slowly, or Play to follow the whole trace. Change a preset or load your own input.

Input & settings

TrySpeedPage style

Red = current item / blue = active window or cells / green = completed work. A value docks at its actual destination; the original input stays unchanged. Positions start at 0.

Working state / what changes

Ready to begin

Enable JavaScript to step through the animation. The worked example and solution below remain readable without it.

Python / why it changes

The line behind the step

def solve(data, parameter=''):
    target = int(parameter)
    values = sorted(data)
    triples = []
    for anchor in range(len(values) - 2):
        if anchor and values[anchor] == values[anchor - 1]:
            continue
        left, right = anchor + 1, len(values) - 1
        while left < right:
            total = values[anchor] + values[left] + values[right]
            if total == target:
                triples.append([values[anchor], values[left], values[right]])
                left += 1
                right -= 1
                while left < right and values[left] == values[left - 1]:
                    left += 1
                while left < right and values[right] == values[right + 1]:
                    right -= 1
            elif total < target:
                left += 1
            else:
                right -= 1
    return triples

How to use this walkthrough

Hover or focus a control for its label. Play forms the input and follows values into the working state. Step shows one change; Back and the slider restore exact snapshots. Reduced motion uses instant updates. Saved snapshots and sorted heap displays are teaching overhead, not algorithm complexity.

05. Python solution

def solve(data, parameter=''):
    target = int(parameter)
    values = sorted(data)
    triples = []
    for anchor in range(len(values) - 2):
        if anchor and values[anchor] == values[anchor - 1]:
            continue
        left, right = anchor + 1, len(values) - 1
        while left < right:
            total = values[anchor] + values[left] + values[right]
            if total == target:
                triples.append([values[anchor], values[left], values[right]])
                left += 1
                right -= 1
                while left < right and values[left] == values[left - 1]:
                    left += 1
                while left < right and values[right] == values[right + 1]:
                    right -= 1
            elif total < target:
                left += 1
            else:
                right -= 1
    return triples

06. Cost & edge cases

Time: O(n²)
Space: O(n) for the sorted copy, plus output

Fewer than three readings returns [].
[2,2,2,2] with target 6 returns [[2,2,2]].
Use 0-16 readings from -30 to 30; target -90 to 90.

Keep going

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