Leetcode Combination Sum II problem solution

YASH PAL,

In this Leetcode Combination Sum II problem solution we have given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sum to target. Each number in candidates may only be used once in the combination.

Leetcode Combination Sum II problem solution

Problem solution in Python.

def combinationSum2(self, nums, target):
        def bt(tlist, start, target):
            s = sum(tlist)
            if s == target:
                res.append(tlist[::])
                return
            if s > target:
                return

            for i in range(start, len(nums)):
                if i > start and nums[i] == nums[i - 1]:
                    continue
                tlist.append(nums[i])
                bt(tlist, i + 1, target)
                tlist.pop()
        nums.sort()
        res = []
        bt([], 0, target)
        return res

Problem solution in Java.

class Solution {
    HashSet<List<Integer>> set = new HashSet<>();

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        List<Integer> tempList = new ArrayList<>();
        backtrack(tempList,candidates,target,0);
        List<List<Integer>> res = new ArrayList<>();
        for(List<Integer> ls:set){
            res.add(ls);
        }
        return res;
    }
    
    public void backtrack(List<Integer> tempList,int[] candidates,int needed_target,int start){
        if(needed_target==0){
            set.add(new ArrayList<>(tempList));
            return;
        }
        for(int i=start;i<candidates.length;i++){
            if(candidates[i]>needed_target) continue;
            tempList.add(candidates[i]);
            backtrack(tempList,candidates,needed_target-candidates[i],i+1);
            tempList.remove(tempList.size()-1);
        }
    }
}

Problem solution in C++.

#define pb push_back
class Solution {
public:
    vector<vector<int>> res;
    void getlist(vector<int>& ar,int index,int target , vector<int>& s){
        int sum = accumulate(s.begin(),s.end(),0);
        if(sum == target){
            res.pb(s);
            return;
        }
        if(sum > target || ar.size() <= index){
            return;
        }
        getlist(ar,index+1,target,s);
        s.pb(ar[index]);
        getlist(ar,index+1,target,s);
        s.pop_back();
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<int> s;
        getlist(candidates,0,target,s);
        vector<vector<int>> ans;
        set<vector<int>> st;
        for(auto x : res){
            vector<int> temp = x;
            sort(temp.begin(),temp.end());
            st.insert(temp);
        }
        for(auto x : st){
            ans.pb(x);
        }
        return ans;
    }
};

Problem solution in C.

void dfs(int* nums, int* colSizes, int numsSize, int target, 
         int** sols, int* sols_size, int* sol, int sol_size, int pointer) {
  if(target < 0) return;
  else if(target == 0) {
    sols[*sols_size] = calloc(sol_size, sizeof(int));
    for(int i = 0; i < sol_size; i++) sols[*sols_size][i] = sol[i];
    colSizes[*sols_size] = sol_size;
    (*sols_size)++;
  }
  else{
    for(int i = pointer; i < numsSize; i++) {
      if(nums[i] > target) return;
      if(i != pointer && nums[i] == nums[i - 1]) continue;
      sol[sol_size] = nums[i];
      dfs(nums, colSizes, numsSize, target - nums[i], 
          sols, sols_size, sol, sol_size + 1, i + 1);
    }
  }
}

int comp(const void* v1, const void* v2) {
  return (*(int*) v1) - (*(int*) v2);
}

int** combinationSum2(int* nums, int numsSize, int target, int** colSizes, int* returnSize) {
  int** sols = calloc(500, sizeof(int*));
  *colSizes = calloc(500, sizeof(int));
  int* sol = calloc(500, sizeof(int));
  qsort((void*) nums, numsSize, sizeof(int), comp);
  dfs(nums, *colSizes, numsSize, target, sols, returnSize, sol, 0, 0);
  return sols;
}

coding problems