Combination Sum – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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.
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]]
Example 3:
Input: candidates = [2], target = 1 Output: []
Constraints:
1 <= candidates.length <= 30
1 <= candidates[i] <= 200
- All elements of
candidates
are distinct. 1 <= target <= 500
C++ Combination Sum LeetCode Solution
class Solution {
public:
void Sum(vector<int>& candidates, int target, vector<vector<int> >& res, vector<int>& r, int i)
{
if(target == 0)
{
// if we get exact answer
res.push_back(r);
return;
}
while(i < candidates.size() && target - candidates[i] >= 0)
{
// Till every element in the array starting
// from i which can contribute to the target
r.push_back(candidates[i]);// add them to vector
// recur for next numbers
Sum(candidates,target - candidates[i],res,r,i);
++i;
// Remove number from vector (backtracking)
r.pop_back();
}
}
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
sort(candidates.begin(),candidates.end()); // sort candidates array
// remove duplicates
candidates.erase(unique(candidates.begin(),candidates.end()),candidates.end());
vector<int> r;
vector<vector<int> > res;
Sum(candidates,target,res,r,0);
return res;
}
};
Java Combination Sum LeetCode Solution
The main idea reminds an approach for solving the coins/knapsack problem – to store the result for all i < target and create the solution from them. For that, for each t from 1 to our target we try every candidate which is less or equal to t in ascending order. For each candidate “c” we run through all combinations for target t-c starting with the value greater or equal than c to avoid duplicates and store only ordered combinations.
public class Solution {
public List<List<Integer>> combinationSum(int[] cands, int t) {
Arrays.sort(cands); // sort candidates to try them in asc order
List<List<List<Integer>>> dp = new ArrayList<>();
for (int i = 1; i <= t; i++) { // run through all targets from 1 to t
List<List<Integer>> newList = new ArrayList(); // combs for curr i
// run through all candidates <= i
for (int j = 0; j < cands.length && cands[j] <= i; j++) {
// special case when curr target is equal to curr candidate
if (i == cands[j]) newList.add(Arrays.asList(cands[j]));
// if current candidate is less than the target use prev results
else for (List<Integer> l : dp.get(i-cands[j]-1)) {
if (cands[j] <= l.get(0)) {
List cl = new ArrayList<>();
cl.add(cands[j]); cl.addAll(l);
newList.add(cl);
}
}
}
dp.add(newList);
}
return dp.get(t-1);
}
}
Python 3 Combination Sum LeetCode Solution
class Solution(object):
def combinationSum(self, candidates, target):
ret = []
self.dfs(candidates, target, [], ret)
return ret
def dfs(self, nums, target, path, ret):
if target < 0:
return
if target == 0:
ret.append(path)
return
for i in range(len(nums)):
self.dfs(nums[i:], target-nums[i], path+[nums[i]], ret)
Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
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Eulerian Circuit-3
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Strongly Connected Componen-t2
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