Combination Sum,Combination Sum II, Combination Sum III

Combination Sum

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
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 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]

Solution

Simple DFS

public static List<List<Integer>> combinationSum(int[] candidates, int target) {
    List<Integer> tmp = new ArrayList<Integer>();
    List<List<Integer>> res = new ArrayList<List<Integer>>();
    Arrays.sort(candidates);
    dfs(res, tmp, candidates, target, 0);
    return res;
}

private static void dfs(List<List<Integer>> res, List<Integer> tmp, int[] candidates, int target, int start){
	if(target==0) {
		res.add(new ArrayList<Integer>(tmp));
		return;
	}
	
	if(target<0) return;
	
	for(int i=start;i<candidates.length;i++){
		tmp.add(candidates[i]);
		dfs(res,tmp,candidates,target-candidates[i],i);
		tmp.remove(tmp.size()-1);
	}
}

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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]

Solution

Two small changes:

• pass in i+1 as new start in each recursive call
• keep track of a prev value of the last value in the last tuple, if the next one equals to prev then it is duplicate

public static List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<Integer> tmp = new ArrayList<Integer>();
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        Arrays.sort(candidates);
        dfs2(res, tmp, candidates, target, 0);
        return res;
    }
    
    private static void dfs2(List<List<Integer>> res, List<Integer> tmp, int[] candidates, int target, int start){
         
        if(target==0) {
            res.add(new ArrayList<Integer>(tmp));
            return;
        }
        
        if(target<0) return;
        
        int prev=-1;
        
        for(int i=start;i<candidates.length;i++){
            if(target<candidates[i]) break;
            if(candidates[i]!=prev){
                tmp.add(candidates[i]);
                dfs2(res,tmp,candidates,target-candidates[i],i+1);
                prev=candidates[i];
                tmp.remove(tmp.size()-1);
            }
        }
    }

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Combination Sum III

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
Ensure that numbers within the set are sorted in ascending order.
Example 1:
Input: k = 3, n = 7
Output:
[[1,2,4]]
Example 2:
Input: k = 3, n = 9
Output:
[[1,2,6], [1,3,5], [2,3,4]]

Solution

public static List<List<Integer>> combinationSum3(int k, int n) {
        List<Integer> tmp = new ArrayList<Integer>();
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        dfs(res,tmp,k,n,1);
        return res;
    }
    private static void dfs(List<List<Integer>> res, List<Integer>tmp, int k, int n, int start){
        if(tmp.size()==k && n==0){
            res.add(new ArrayList<Integer>(tmp));
            return;
        }
        if(n<0 || tmp.size()>k) return;
        for(int i=start; i<=9; i++){
            tmp.add(i);
            dfs(res,tmp,k,n-i,i+1);
            tmp.remove(tmp.size()-1);
        }
    }