CODING CHALLENGE · N°26

Merge Intervals

Medium SortingArraysInterview Classic

Calendar apps, memory allocators, and half of all interview loops run on this: sort intervals by start, then sweep once, growing or closing the current merged block. The classic that punishes off-by-one thinking at the boundaries.

The problem

Given intervals, a list of [start, end] pairs (start ≤ end, any order), merge all overlapping intervals and return the result sorted by start. Two intervals overlap if one starts before or exactly when the other ends — touching intervals like [1,4] and [4,5] merge into [1,5].

EXAMPLE 1

Input intervals = [[1,3],[2,6],[8,10],[15,18]]

Output [[1,6],[8,10],[15,18]]

[1,3] and [2,6] overlap

EXAMPLE 2

Input intervals = [[1,4],[4,5]]

Output [[1,5]]

touching endpoints merge

EXAMPLE 3

Input intervals = [[1,10],[2,3]]

Output [[1,10]]

nested intervals disappear into the outer one

CONSTRAINTS

Input may be unsorted — sort by start first.
Touching counts as overlapping: next.start ≤ current.end merges.
Merging takes max(end) — nested intervals must not shrink the block.

SOLVE IT YOURSELF

Your turn — write it

Edit the stub, hit Run (or ⌘/Ctrl + Enter), and watch the hidden tests. Stuck? the hints are right above and Reveal solution is one click away.

YOUR TASK

Implement merge_intervals(intervals) → the minimal list of non-overlapping intervals covering the input, sorted by start.

HINTS — 4 IDEAS

Sort by start. After that, only the CURRENT merged block can ever absorb the next interval.
If next.start ≤ current.end, extend: current.end = max(current.end, next.end).
Otherwise close the block, append it, and start a new one.
The nested case [[1,10],[2,3]] is why the max() matters.

CPython · WebAssembly

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