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

40 - Combination Sum II

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