Find the two that add up
A reconciliation tool looks for the two entries that together explain a difference.
- Return the two positions, counting from zero, smaller position first.
- If several pairs would work, return the one whose second position comes first.
- No pair means an empty list.
- A value may not be paired with itself, but two equal values at different positions are fine.
pairSummingTo(values: list<int>, target: int) → list<int>
Java needs a compiler and Drill does not host one yet, so this page is the reference rather than an exercise: the problem, worked examples, and the solution in full. To type it out, the same problem runs in Python.
Where you start
List<Integer> pairSummingTo(List<Integer> values, int target) {
}Worked examples
| Call | Result |
|---|---|
pairSummingTo(Main.<Integer>ls(2, 7, 11, 15), 9) | Main.<Integer>ls(0, 1) |
pairSummingTo(Main.<Integer>ls(3, 2, 4), 6) | Main.<Integer>ls(1, 2) |
pairSummingTo(Main.<Integer>ls(3, 3), 6) | Main.<Integer>ls(0, 1) |
pairSummingTo(Main.<Integer>ls(1, 2), 99) | Main.<Integer>ls() |
Hint
Walk once, and for each value ask whether the number that would complete it has already gone by.
Reference solution in Java
List<Integer> pairSummingTo(List<Integer> values, int target) {
Map<Integer, Integer> seen = new HashMap<>();
for (int j = 0; j < values.size(); j++) {
int v = values.get(j), need = target - v;
if (seen.containsKey(need)) return new ArrayList<>(Arrays.asList(seen.get(need), j));
if (!seen.containsKey(v)) seen.put(v, j);
}
return new ArrayList<>();
}