Find The K Sum Of An Array - Leetcode Solution

Problem Statement:

Given an array nums of n integers and an integer target, are there k elements a1, a2, ..., ak in nums such that their sum is equal to target? Find all possible combinations.

Solution:

This is a variation of the classic problem of finding the subset sum problem. Here, we need to find k elements from the array whose sum is equal to the target. One approach to solve this problem is to use backtracking. The basic idea of backtracking is to generate all possible combinations of the given elements and check if their sum is equal to the target. Let's discuss the algorithm step by step:

1. Start with an empty set res and an index i=0.

2. At each step, we have two options: either we choose the current element nums[i] or we don't choose it.

3. If we choose the current element nums[i], we add it to the set res and recursively find k-1 elements whose sum is equal to target-nums[i]. We also increment the index i.

4. If we don't choose the current element nums[i], we just increment the index i and continue the recursion.

5. We stop the recursion when we find k elements whose sum is equal to target, or when we have explored all possible combinations.

6. We return all the valid solutions found during the recursion.

Let's now implement the above algorithm in code:

class Solution { public: vector<vector<int>> kSum(vector<int>& nums, int target, int k) { vector<vector<int>> res; vector<int> cur; dfs(res, cur, nums, target, k, 0); return res; }

private: void dfs(vector<vector<int>>& res, vector<int>& cur, vector<int>& nums, int target, int k, int start) { if(k==0 && target==0) { // found a valid combination res.push_back(cur); return; } if(k<=0 || target<=0) return; // no valid solution for(int i=start; i<nums.size(); i++) { cur.push_back(nums[i]); // choose the current element dfs(res, cur, nums, target-nums[i], k-1, i+1); // recursive call cur.pop_back(); // remove the current element } } };

In the above code, we define a private function dfs() that takes the following parameters:

1. res: reference to the result vector that stores all valid solutions.

2. cur: reference to the current set that stores the elements of the current combination.

3. nums: reference to the input array.

4. target: the target sum.

5. k: number of elements to choose.

6. start: the index from where we need to start exploring the array.

The base case of the recursion is when we have found k elements whose sum is equal to target. In this case, we add the current set to the result vector res.

We also handle the edge cases where k is negative or target is negative or zero. In such cases, we return early without exploring the array further.

In the for loop, we explore all possible choices starting from the index start. We choose the current element nums[i], add it to the set cur, and recursively call dfs() with k-1 elements and target-nums[i] as the new parameters. We also increment the index i to avoid choosing the same element again.

After the recursive call, we remove the current element nums[i] from the set cur to backtrack and try the other choices.

Finally, we return the result vector res containing all valid solutions.

Time and Space Complexity:

The time complexity of the above algorithm is O(n^k) where n is the size of the input array and k is the number of elements we need to choose. This is because we explore all possible combinations of the k elements from the array.

The space complexity of the algorithm is O(k) in the worst case, where k is the number of elements we need to choose. This is because we store the current set of elements in the recursion stack. The space complexity is also O(n^k) because we can have at most n^k valid combinations.

Conclusion:

In this problem, we used the backtracking algorithm to find all possible combinations of k elements from the array whose sum is equal to the target. We implemented the algorithm by defining a recursive function dfs() that explores all possible choices and stores the valid solutions in a result vector res. The time complexity of the algorithm is O(n^k) and the space complexity is O(k) in the worst case.

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