How can i find sum of three element in a given away - dart

Question

Is there a method we use to reach the desired number in an array given in dart language.. I can do this for binary ones, but I can't do it for a code that finds the sum of 3 or more elements
For example
Input: candidates = [10,1,2,7,6,1,5], target = 8
Output:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
this is the my code i have done until now
void main() {
var candidates = [10, 1, 2, 7, 6, 1, 5], target = 8;
var answer = [];
for (int i = 0; i < candidates.length; i++) {
for (int j = 0; j < candidates.length; j++) {
if (candidates[i] + candidates[j] == target && i != j && i < j) {
answer.add([candidates[i], candidates[j]]);
}
}
}
}

Answer 1

I am sure this can be done more efficient but since the solution is for some Leetcode assignment, I don't really want to spend too much time on optimizations.
I have tried added some comments in the code which explains my way of doing it:
void main() {
getSumLists([10, 1, 2, 7, 6, 1, 5], 8).forEach(print);
// [5, 1, 2]
// [1, 6, 1]
// [1, 7]
// [6, 2]
getSumLists([2, 5, 2, 1, 2], 5).forEach(print);
// [2, 1, 2]
// [5]
}
Iterable<List<int>> getSumLists(
List<int> candidates,
int target, {
List<int>? tempAnswer,
int sum = 0,
}) sync* {
// We cannot use default value in parameter since that makes list const
final tempAnswerNullChecked = tempAnswer ?? [];
if (sum == target) {
// We got a result we can return.
// OPTIMIZATION: If you know the returned list from each found result is not
// being used between found results, you can remove the `.toList()` part.
yield tempAnswerNullChecked.toList();
} else if (sum > target) {
// No need to search further in this branch since we are over the target
return;
}
// Make a copy so we don't destroy the input list but also so it works even
// if provided list as input is non-growing / non-modifiable
final newCandidates = candidates.toList();
while (newCandidates.isNotEmpty) {
// We take numbers from the end of the list since that is more efficient.
final number = newCandidates.removeLast();
// Recursive call where we return all results we are going to find given
// the new parameters
yield* getSumLists(
newCandidates,
target,
tempAnswer: tempAnswerNullChecked..add(number),
sum: sum + number,
);
// Instead of creating a new tempAnswerNullChecked, we just reuse it and
// make sure we remove any value we are temporary adding
tempAnswerNullChecked.removeLast();
// Ensure we don't get duplicate combinations. So if we have checked the
// number `1` we remove all `1` so we don't try the second `1`.
newCandidates.removeWhere((element) => element == number);
}
}