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Eda

Posted on • Originally published at rivea0.github.io

# LeetCode Meditations: Combination Sum

Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

The test cases are generated such that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

For example:

Input: candidates = [2, 3, 6, 7], target = 7

Output: [[2, 2, 3], [7]]

Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.


Or:

Input: candidates = [2, 3, 5], target = 8

Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]


One thing to notice here is that we can have duplicate values — for instance, in the second example, [2, 2, 2, 2] is a possible option for a given target of 8.

We can try adding the same item to the current total, until it's equal to the target or is more than the target. If the current total ends up being equal to the target, we'll add the numbers we've gathered so far to our result. Otherwise, we'll backtrack, and try the next item in candidates.

We'll have two important variables to help us here: currentNums which has the current numbers we're looking at, and currentTotal which is the sum of currentNums.

For the first base case where we can add the currentNums to the result, we'll check if the currentTotal equals target:

if (currentTotal === target) {
result.push([...currentNums]);
return;
}


Another case where we need to return is when we've looked at all the items in candidates or when currentTotal has surpassed target:

if (idx >= candidates.length || currentTotal > target) {
return;
}

Note
idx will be the index of the item in candidates that we're adding to currentNums.

The process mentioned above goes like this:

currentNums.push(candidates[idx]);
backtrack(idx, currentNums, currentTotal + candidates[idx]);
currentNums.pop();
backtrack(idx + 1, currentNums, currentTotal);


And, that's all there is to the backtrack function:

function backtrack(idx: number, currentNums: number[], currentTotal: number) {
if (currentTotal === target) {
result.push([...currentNums]);
return;
}

if (idx >= candidates.length || currentTotal > target) {
return;
}

currentNums.push(candidates[idx]);
backtrack(idx, currentNums, currentTotal + candidates[idx]);
currentNums.pop();
backtrack(idx + 1, currentNums, currentTotal);
}


For example, let's say that the candidates array is [2, 3, 5] and the target is 5.

The first item is 2, so we'll add it to itself until it reaches 6 (2 + 2 + 2), the point where the current total is more than the target.

Note
Remember that we're keeping track of the numbers we gather, at this point the currentNums are [2, 2, 2].

Now that we're over the target, we'll pop the last 2 from currentNums, and add the next item in candidates, which is 3. Now, our current total is 2 + 2 + 3, which is again more than the target, so we'll pop 3. We'll go on to 5, and our current total will be 2 + 2 + 5, which is of course more than the target, so we'll pop 5 as well.

At this point, we're left with 2 + 2, but we tried all the items in candidates, so we'll pop the last 2 from currentNums again.

Now, our current total is just 2. So, we go on, and add the next item in candidates, and now, our current total is 2 + 3 which equals our target. We'll add it to our result and return.

We'll try the next item, our current total is now 2 + 5. It is more than the target, so we'll pop the last item, and once again we're only left with 2 as our current total. But, we tried all the items again, so we'll pop this 2 as well.

We tried every possible combination for 2, so now it's time to look at 3.

We'll add 3 to itself until it's more than or equal to the target. Our current total will reach 3 + 3, which is more than the target, so we'll pop the last 3 from currentNums. Now, we'll go on to the next item, our current total will be 3 + 5, which exceeds the target again, so we'll pop 5.
At this point we've tried all the items for 3, so it's now time to pop 3 as well.

We go on to 5, and as our current total is just 5 which is equal to the target, we'll add it to our result. There is no next item we can try with 5, so we'll pop it off.

We don't have any items left to look at, so we're done.
Our result is [[2, 3], [5]].

The final solution in TypeScript looks like this:

function combinationSum(candidates: number[], target: number): number[][] {
let result: number[][] = [];
let nums = [];

function backtrack(idx: number, currentNums: number[], currentTotal: number) {
if (currentTotal === target) {
result.push([...currentNums]);
return;
}

if (idx >= candidates.length || currentTotal > target) {
return;
}

currentNums.push(candidates[idx]);
backtrack(idx, currentNums, currentTotal + candidates[idx]);
currentNums.pop();
backtrack(idx + 1, currentNums, currentTotal);
}

backtrack(0, nums, 0);

return result;
}


For a more detailed explanation of this solution, see NeetCode's video on this problem.

#### Time and space complexity

There are different opinions on the time complexity of this solution. The most likely one is—I think— $2^t$ where $t$ is the target number. The reason is related to the height of the decision tree. For example, if the first item in candidates is 1, the number of calls to backtrack will be, in the worst case, equal to the target.
The space complexity can be $O(t)$ where $t$ is the target, because the recursive call stack can reach a depth of $t$ in the worst case.

This is, in my opinion, a very challenging problem, so it's now time to take a breath. Next, we'll look at Word Search. Until then, happy coding.