Add "Combination Sum" backtracking algorithm.

This commit is contained in:
Oleksii Trekhleb
2018-06-30 10:19:14 +03:00
parent b41cffea0f
commit b33f1d52dc
4 changed files with 151 additions and 0 deletions

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# Combination Sum Problem
Given a **set** of candidate numbers (`candidates`) **(without duplicates)** and
a target number (`target`), find all unique combinations in `candidates` where
the candidate numbers sums to `target`.
The **same** repeated number may be chosen from `candidates` unlimited number
of times.
**Note:**
- All numbers (including `target`) will be positive integers.
- The solution set must not contain duplicate combinations.
## Examples
```
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]
```
```
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
```
## Explanations
Since the problem is to get all the possible results, not the best or the
number of result, thus we dont need to consider DP (dynamic programming),
backtracking approach using recursion is needed to handle it.
Here is an example of decision tree for the situation when `candidates = [2, 3]` and `target = 6`:
```
0
/ \
+2 +3
/ \ \
+2 +3 +3
/ \ / \ \
+2 ✘ ✘ ✘ ✓
/ \
✓ ✘
```
## References
- [LeetCode](https://leetcode.com/problems/combination-sum/description/)

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import combinationSum from '../combinationSum';
describe('combinationSum', () => {
it('should find all combinations with specific sum', () => {
expect(combinationSum([1], 4)).toEqual([
[1, 1, 1, 1],
]);
expect(combinationSum([2, 3, 6, 7], 7)).toEqual([
[2, 2, 3],
[7],
]);
expect(combinationSum([2, 3, 5], 8)).toEqual([
[2, 2, 2, 2],
[2, 3, 3],
[3, 5],
]);
expect(combinationSum([2, 5], 3)).toEqual([]);
expect(combinationSum([], 3)).toEqual([]);
});
});

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/**
* @param {number[]} candidates - candidate numbers we're picking from.
* @param {number} remainingSum - remaining sum after adding candidates to currentCombination.
* @param {number[][]} finalCombinations - resulting list of combinations.
* @param {number[]} currentCombination - currently explored candidates.
* @param {number} startFrom - index of the candidate to start further exploration from.
* @return {number[][]}
*/
function combinationSumRecursive(
candidates,
remainingSum,
finalCombinations = [],
currentCombination = [],
startFrom = 0,
) {
if (remainingSum < 0) {
// By adding another candidate we've gone below zero.
// This would mean that last candidate was not acceptable.
return finalCombinations;
}
if (remainingSum === 0) {
// In case if after adding the previous candidate out remaining sum
// became zero we need to same current combination since it is one
// of the answer we're looking for.
finalCombinations.push(currentCombination.slice());
return finalCombinations;
}
// In case if we haven't reached zero yet let's continue to add all
// possible candidates that are left.
for (let candidateIndex = startFrom; candidateIndex < candidates.length; candidateIndex += 1) {
const currentCandidate = candidates[candidateIndex];
// Let's try to add another candidate.
currentCombination.push(currentCandidate);
// Explore further option with current candidate being added.
combinationSumRecursive(
candidates,
remainingSum - currentCandidate,
finalCombinations,
currentCombination,
candidateIndex,
);
// BACKTRACKING.
// Let's get back, exclude current candidate and try another ones later.
currentCombination.pop();
}
return finalCombinations;
}
/**
* Backtracking algorithm of finding all possible combination for specific sum.
*
* @param {number[]} candidates
* @param {number} target
* @return {number[][]}
*/
export default function combinationSum(candidates, target) {
return combinationSumRecursive(candidates, target);
}