Question

Formatted question description: https://leetcode.ca/all/40.html

40. Combination Sum II

Given a collection of candidate numbers (C) and a target number (T),
find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used **once** in the combination.

Note:
    All numbers (including target) will be positive integers.
    Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
    The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8,

A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

Algorithm

Based on Combination_Sum_I, add sorting to move duplicated numbers together and skip.

Code

Java

  • import java.util.ArrayList;
    import java.util.Arrays;
    import java.util.List;
    
    
    public class Combination_Sum_II {
    
        class Solution {
    
            List<List<Integer>> result = new ArrayList<>();
    
            public List<List<Integer>> combinationSum2(int[] candidates, int target) {
                if (candidates == null || candidates.length == 0) {
                    return result;
                }
    
                // nlogn
                Arrays.sort(candidates);
    
    
                dfs(candidates, target, 0, new ArrayList<Integer>());
    
                return result;
            }
    
            private void dfs(int[] candidates, int target, int start, ArrayList<Integer> onePath) {
    
                if (target < 0) { // target<0, assumption is all positive
                    return;
                }
    
                if (target == 0) {
                    result.add(new ArrayList<>(onePath));
                    // continue, and possible 1,-1 are next ones
                }
    
                for (int i = start; i < candidates.length; i++) {
                    if (i > start && candidates[i] == candidates[i - 1]) continue;
                    onePath.add(candidates[i]);
                    dfs(candidates, target - candidates[i], i+1, onePath); // i+1, to use only once
                    onePath.remove(onePath.size() - 1);
                }
            }
        }
    
        // https://leetcode.com/problems/combination-sum-ii/solution/
        // official solution got "Time Limit Exceeded" error, funny but still a good idea solving this problem
        class Solution_lc {
            public List<List<Integer>> combinationSum2(int[] candidates, int target) {
                // container to hold the final combinations
                List<List<Integer>> results = new ArrayList<>();
                LinkedList<Integer> comb = new LinkedList<>();
    
                HashMap<Integer, Integer> counter = new HashMap<>();
                for (int candidate : candidates) {
                    if (counter.containsKey(candidate))
                        counter.put(candidate, counter.get(candidate) + 1);
                    else
                        counter.put(candidate, 1);
                }
    
                // convert the counter table to a list of (num, count) tuples
                List<int[]> counterList = new ArrayList<>();
                counter.forEach((key, value) -> {
                    counterList.add(new int[]{key, value});
                });
    
                backtrack(comb, target, 0, counterList, results);
                return results;
            }
    
            private void backtrack(LinkedList<Integer> comb,
                                   int remain, int curr,
                                   List<int[]> counter,
                                   List<List<Integer>> results) {
    
                if (remain == 0) {
                    // make a deep copy of the current combination.
                    results.add(new ArrayList<Integer>(comb));
                    return;
                }
    
                for (int nextCurr = curr; nextCurr < counter.size(); ++nextCurr) {
                    int[] entry = counter.get(nextCurr);
                    Integer candidate = entry[0], freq = entry[1];
    
                    if (freq <= 0)
                        continue;
    
                    // add a new element to the current combination
                    comb.addLast(candidate);
                    counter.set(nextCurr, new int[]{candidate, freq - 1});
    
                    // continue the exploration with the updated combination
                    backtrack(comb, remain - candidate, nextCurr, counter, results);
    
                    // backtrack the changes, so that we can try another candidate
                    counter.set(nextCurr, new int[]{candidate, freq});
                    comb.removeLast();
                }
            }
        }
    }
    
  • // OJ: https://leetcode.com/problems/combination-sum-ii/
    // Time: O(N^2)
    // Space: O(N)
    class Solution {
        vector<vector<int>> ans;
        void dfs(vector<int> &A, int target, int start, vector<int> &tmp) {
            if (!target) {
                ans.push_back(tmp);
                return;
            }
            for (int i = start; i < A.size() && target >= A[i]; ++i) {
                if (i != start && A[i] == A[i - 1]) continue;
                tmp.push_back(A[i]);
                dfs(A, target - A[i], i + 1, tmp);
                tmp.pop_back();
            }
        }
    public:
        vector<vector<int>> combinationSum2(vector<int>& A, int target) {
            sort(A.begin(), A.end());
            vector<int> tmp;
            dfs(A, target, 0, tmp);
            return ans;
        }
    };
    
  • class Solution(object):
      def combinationSum2(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
    
        def dfs(nums, target, start, visited, path, res):
          if target == 0:
            res.append(path + [])
            return
    
          for i in range(start, len(nums)):
            if i > start and nums[i] == nums[i - 1]:
              continue
            if target - nums[i] < 0:
              return 0
            if i not in visited:
              visited.add(i)
              path.append(nums[i])
              dfs(nums, target - nums[i], i + 1, visited, path, res)
              path.pop()
              visited.discard(i)
    
        candidates.sort()
        res = []
        visited = set([])
        dfs(candidates, target, 0, visited, [], res)
        return res
    
    

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