# Recursion, Backtracking and Subset Problems

These pattern of questions include subset formation, backtracking and recursive calls. Almost all of the questions are leetcode medium level questions. Solving these following problems will help people gain a better understanding of

1. Backtracking,
2. Recursion calls,
3. Subset pattern of questions.

## Find all the subsets

Problem on Leetcode $$\to$$

Given an integer array nums of unique elements, return all possible subsets (the power set). The solution set must not contain duplicate subsets. Return the solution in any order.

### Approach

With basic recursive approach, we first include an element and don't include an element. This is how we can generate all the unique subsets from a given set. We push the subset to the answer only at the last step of the recursion tree when the index reaches to the last element.

### Code

class Solution {
private:

void recurse(vector<int>& v, int index) {
static vector<int> b;

if (index > v.size() - 1) {

// Push the subset only at the last step of the recursion tree

vector<int> c(b); // create a copy of b and add it into answer

return;
}

// with the element
b.push_back(v[index]);
recurse(v, index + 1);

// without the current element
b.pop_back();
recurse(v, index + 1);
}

public:
vector<vector<int>> subsets(vector<int>& nums) {
recurse(nums, 0);
}
};


## Subsets II

Subsets II is a little bit different, given an integer array nums that may contain duplicates, return all possible subsets (the power set).

The solution set must not contain duplicate subsets. Return the solution in any order.

### Approach

• The approach should be similar to the subset approach, now in order to avoid duplicates in the power set what we can do is the following
• We can first sort the input array
• then skip duplicates while traversing through the array in the recursive calls by instead of skipping 1 index, skip multiple indexes until we can't find a new value.
• This specific modification helps us avoid duplicates
while(index < nums.size() - 1 and nums[index] == nums[index + 1]){
index++;
}

• Now we can avoid duplicates in the power set.

### Code

class Solution {
private:

void recurse(vector<int>& nums, int index){

static vector<int> b;

if (index >= nums.size()) {
// at the end of tree
vector<int> c(b);
return;
}

b.push_back(nums[index]);
recurse(nums, index + 1);

b.pop_back();
while(index < nums.size() - 1 and nums[index] == nums[index + 1]){
index++;
}
recurse(nums, index + 1);
}

public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
std::sort(nums.begin(), nums.end());
recurse(nums, 0);
}
};


## Combination Sum

Problem on Leetcode $$\to$$

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.

It is guaranteed that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

### Examples

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

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

Input: candidates = , target = 1
Output: []


### Constraints:

• $$1 \leq$$ candidates.length $$\leq 30$$
• $$1 \leq$$ <= candidates[i] $$\leq 200$$
• All elements of candidates are distinct.
• $$1 \leq$$ target $$\leq 500$$

### Approach

• We'll create a recursive subroutine called recurse which will recursively find all the sub-sequences that sums up to target
• Recall a similar problem in the intro section where we solved Find all the sub-sequences that sums up to K. But here the problem is little different, here one element can be added multiple times.
• So we create a modification in the first recursive call: recurse(candidates, target, sum, index);. In this case we are again calling with the same index so that we can check if multiple times we can add the same element.
• For the case where we are not considering a particular element [as] we are not considering a particular element, we'll simply call the recursive call recurse(candidates, target, sum, index + 1; with $$\text{Index} + 1$$.

### Code

class Solution {
public:

void recurse(vector<int> &candidates, int target, int sum, int index) {

static vector<int> b; // shared data structure

if (index > candidates.size() - 1) return;

if (sum < target) {

// check with the current index
b.push_back(candidates[index]);
sum = sum + candidates[index];
recurse(candidates, target, sum, index); // can take the same index multiple times

// remove the current index and check again
b.pop_back();
sum = sum - candidates[index];
recurse(candidates, target, sum, index+1); // index+1 bcz we are no longer interested with that index
}

if (sum == target) {
vector<int> bk(b);
}

}

vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
recurse(candidates, target, 0, 0);
}
};


## Combination Sum III

Find the problem on Leetcode $$\to$$

### Problem Statement

This is a slight modification of the previous problem, find all valid combinations of k numbers that sum up to n such that the following conditions are true:

• Only numbers 1 through 9 are used.
• Each number is used at most once.

Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.

### Approach

Similar to the last problem we'll continue with a recursive approach and use a shared data structure b to keep track of the combinations. The code will be the almost same as the previous problem but we'll include the answer if and only if the size of the answer is equal to k according to the question specifications.

For candidates the question says they are $$1 \to 9$$, but we'll include $$10$$ because the following test case will fail if we don't include $$10$$. Figure out why on your own. (Hint: draw the recursion tree)

TEST CASE FAILURE:
target = 45
k = 9

ACTUAL RESULT SHOULD BE RETURENED:
[1,2,3,4,5,6,7,8,9]

Failure: fails to return the result if the candidates are {1 to 9}


### Code

class Solution {
private:

void build(int k, int target, int sum, int index, vector<int>& candidates) {

// shared data structure across recursion
static vector<int> b;

if (index > candidates.size() - 1) return;

if (sum < target) {
b.push_back(candidates[index]);
build(k, target, sum+candidates[index], index+1, candidates);

b.pop_back();
build(k, target, sum, index+1, candidates);
}

if (sum == target and b.size() == k) {
vector<int> lvl(b);
}
}

public:
vector<vector<int>> combinationSum3(int k, int n) {
vector<int> candidates = {1,2,3,4,5,6,7,8,9,10};
build(k, n, 0, 0, candidates);