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>
C# 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
public List<int> PairSummingTo(List<int> values, int target) {
}Worked examples
| Call | Result |
|---|---|
PairSummingTo(new List<int> { 2, 7, 11, 15 }, 9) | new List<int> { 0, 1 } |
PairSummingTo(new List<int> { 3, 2, 4 }, 6) | new List<int> { 1, 2 } |
PairSummingTo(new List<int> { 3, 3 }, 6) | new List<int> { 0, 1 } |
PairSummingTo(new List<int> { 1, 2 }, 99) | new List<int> { } |
Hint
Walk once, and for each value ask whether the number that would complete it has already gone by.
Reference solution in C#
public List<int> PairSummingTo(List<int> values, int target) {
var seen = new Dictionary<int, int>();
for (int j = 0; j < values.Count; j++) {
int v = values[j], need = target - v;
if (seen.ContainsKey(need)) return new List<int> { seen[need], j };
if (!seen.ContainsKey(v)) seen[v] = j;
}
return new List<int>();
}