The two readings that add up
A reconciliation tool has a sorted column of amounts and a difference to explain. It looks for the two amounts that together account for it.
- The amounts arrive sorted ascending.
- Return the two amounts, smaller first.
- If several pairs work, return the one with the smallest first amount.
- If no pair adds up, return an empty list.
- An amount cannot pair with itself — the two have to be at different positions.
pairSummingTo(amounts: 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> amounts, int target) {
}Worked examples
| Call | Result |
|---|---|
pairSummingTo(Main.<Integer>ls(1, 2, 4, 7, 11), 9) | Main.<Integer>ls(2, 7) |
pairSummingTo(Main.<Integer>ls(1, 2, 3, 4), 5) | Main.<Integer>ls(1, 4) |
pairSummingTo(Main.<Integer>ls(1, 2, 3), 100) | Main.<Integer>ls() |
pairSummingTo(Main.<Integer>ls(), 3) | Main.<Integer>ls() |
Hint
Sorted input means you can start at both ends. If the two ends add up to too much, the right end is too big; if too little, the left end is too small.
Reference solution in Java
List<Integer> pairSummingTo(List<Integer> amounts, int target) {
int i = 0, j = amounts.size() - 1;
while (i < j) {
int sum = amounts.get(i) + amounts.get(j);
if (sum == target) return new ArrayList<>(List.of(amounts.get(i), amounts.get(j)));
if (sum < target) i++;
else j--;
}
return new ArrayList<>();
}