Combination Sum — Leetcode
Topic: Array
Difficulty: Medium
Question Number: 39
Problem Statement:
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.
Example 1:
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.
Example 2:
Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]
Code:
in Java
class Solution {
List<List<Integer>> combinations;
int[] candidatesArr;
int targetSum;
public List<List<Integer>> combinationSum(int[] candidates, int target) {
combinations = new ArrayList();
targetSum = target;
candidatesArr = candidates;
findCombinations(0, new ArrayList<Integer>(), target);
return combinations;
}
public void findCombinations(int candidate, ArrayList<Integer> current, int total) {
if (total == 0) {
combinations.add(new ArrayList<>(current));
return;
}
if (total < 0) {
return;
}
for (int i = candidate; i < candidatesArr.length; i++) {
current.add(candidatesArr[i]);
findCombinations(i, current, total - candidatesArr[i]);
current.remove(current.size() - 1);
}
}
}
in Python by Vaishnav Jha
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
ans = []
def dfs(i, cur, s):
if s == target: # Base Case 1
ans.append(cur[:]) # Append copy of list
return
if i>= len(candidates) or s > target: # Base Case 2
return
cur.append(candidates[i])
dfs(i,cur,s+candidates[i]) # Recirsive call with the ith candidate
cur.pop() # Backtrack
dfs(i+1, cur,s) # Recirsive call without the ith candidate
dfs(0,[],0)
return ans
# Contributed By - Vaishnav Jha
Thank You
Oshi Raghav