[LeetCode#39]Combination Sum

Problem:

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]

Analysis:

This problem is a typical DFS problem, it is simple, but it is very very useful. It actually shares same skills we have used in two sum problem: 
1. How to take advantage of a sorted array to aovid duplicates?
2. How to guarantee the results in no-descending order?

They all could solve through the skill we have used in three sum problem: sort the array first. But there is also an obvious difference between this problem with two sum problem. 
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3 sum problem: there could be duplicates in the array, we should use some skills to avoid the elements with value appear was used on the same index many times. 
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combinaionSum problem: no duplicates in the candidates set, and we allow to use each element many times. (This free us from checking iff nums[i] == nums[i-1]). 
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Basic idea:
step 1: sort the candidates in no-descending order.
Arrays.sort(candidates);
step 2: get a number from the sorted array, then continue the search along the elements after it(inclusive, since each element could be used multi times)
Note: the "inclusive" would allow the same element to be used at other indexes, but not repeat on the same index. 
A tree structure is good start for understanding!
for (int i = start; i < candidates.length; i++) {
    item.add(candidates[i]);
    searchPath(candidates, i, target-candidates[i], item, ret);
    item.remove(item.size()-1);
}

Good programming habit:
Don't do the complex checking when forking!!! Stop the wrong froking branches at base case!!!
if (target == 0)
    ret.add(new ArrayList<Integer> (item));
if (target < 0)
    return;
    
    
Two repeatives in such problem:
1. the element with the same value were used on the same index many times.
    1.1 iff there are duplicates in array (nums[i] != num[i-1])
    1.2 iff there are not duplicates in array (easy, but should allow it at next index)
    
2. mix the order of the same result.
[2 3] [3 2]
The problem was caused by following logic:
for (int i = 0; i < candidates.length; i++) {
    ...
    searchPath(candidates, i, target-candidates[i], item, ret);
}
It must cause duplicates!


The great way to solve above prolem is using sorting candidates array!
And use the fix:
for (int i = start; i < candidates.length; i++) {
    ...
}
start from "start", rather than the elements before start.

Solution:

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        if (candidates == null)
            throw new IllegalArgumentException("candidates is null");
        List<List<Integer>> ret= new ArrayList<List<Integer>> ();
        if (candidates.length == 0 || target < 0)
            return ret;
        ArrayList<Integer> item = new ArrayList<Integer> ();
        Arrays.sort(candidates);
        searchPath(candidates, 0, target, item, ret);
        return ret;
    }
    
    private void searchPath(int[] candidates, int start, int target, List<Integer> item, List<List<Integer>> ret) {
        if (target == 0)
            ret.add(new ArrayList<Integer> (item));
        if (target < 0)
            return;
        for (int i = start; i < candidates.length; i++) {
            item.add(candidates[i]);
            searchPath(candidates, i, target-candidates[i], item, ret);
            item.remove(item.size()-1);
        }
    }
}