646. Weaving Two Words Together

Two words are given. Build a new word by taking characters ALTERNATELY, starting with the first word. When one word runs out, append whatever is left of the other.

abc and pqr -> apbqcr ab and pqrs -> apbqrs abcd and pq -> apbqcd

## One pointer per word

i = 0, j = 0 while i < len(x) or j < len(y): if i < len(x): emit x[i]; i += 1 if j < len(y): emit y[j]; j += 1

The loop continues while EITHER word has characters left, and each emission is guarded separately. That is what makes the leftover tail come out automatically: once one word is exhausted its guard simply stops firing, and the other keeps emitting.

## Why no separate tail loop is needed

The common first attempt runs while i < len(x) and j < len(y) and then appends the remainders afterwards, which needs two extra loops and two more chances to get a bound wrong.

Changing the and to an or and guarding each emission removes both. Whenever a loop is followed by "and then handle the leftovers", it is worth asking whether the condition can carry them instead.

## The order within a round matters

Both emissions happen in the same round, first word then second. Alternating the WHOLE loop body instead, taking one character per iteration and flipping a flag, also works and is longer for no gain.

## Boundaries

Words of equal length interleave exactly. A one character word against a long one emits one character and then the entire rest of the other.

## Cost

Time O(n + m). Space O(n + m) for the output.

Constraints - `1 ≤ length of each word ≤ 100000` - both words contain only lowercase letters