680. The Pair Whose Product Lands Nearest

Given n positive integers and a target, choose two of them by position and make their PRODUCT as close to the target as possible.

If two achievable products are equally close, print the SMALLER one.

1 2 3 4, target 7 -> 6 from 2 and 3 2 3 5, target 10 -> 10 exactly 1 2 3 4 5, target 11 -> 10

## The same sweep, with multiplication

Sort, converge, and keep the best product seen. The product rises as i advances and falls as j retreats, because all the values are positive:

p < target -> i += 1 raise the product p > target -> j -= 1 lower it p == target -> stop, nothing is closer

This is problem 615 with the sum replaced by a product and the target moved off zero.

## Why positivity is doing the work

With positive values, replacing a[i] by a larger value strictly increases the product, and replacing a[j] by a smaller one strictly decreases it. That monotonicity is what tells the sweep which pointer to move.

Allow a zero or a negative and it breaks: multiplying by a larger value can decrease the product. The constraint says positive for that reason, and problem 675 needs the same promise for the same reason.

## The tie-break

The statement takes the SMALLER product on a tie, which matches problem 607 and is the opposite of problem 615. Compare abs(p - target) first, then break equality with p < best.

Three problems in this topic ask for a nearest value and they do not all agree, deliberately. Reading the rule each time is the habit; remembering one and reusing it is the bug.

## Overflow

Two values near 10^9 multiply to 10^18, which fits in a signed 64 bit integer and not in a double. JavaScript needs big integers, as in problem 675.

## Boundaries

Two values give one product, whatever it is. All values equal give that value squared.

## Cost

Time O(n log n) for the sort. Space O(1) beyond the input.

Constraints - `2 ≤ n ≤ 200000` - `1 ≤ each value ≤ 10^9` - `1 ≤ target ≤ 10^18`