Description

Given a set of candidate numbers (C) (without duplicates) 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. 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] ]

Hint

Train of Thought

A general approach to backtracking questions in Java (Subsets, Permutations, Combination Sum, Palindrome Partitioning) This structure might apply to many other backtracking questions, but here I am just going to demonstrate Subsets, Permutations, and Combination Sum.

Subsets : https://leetcode.com/problems/subsets/

    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> list = new ArrayList<>();
        Arrays.sort(nums);
        backtrack(list, new ArrayList<>(), nums, 0);
        return list;
    }

    private void backtrack(List<List<Integer>> list , List<Integer> tempList, int [] nums, int start){
        list.add(new ArrayList<>(tempList));
        for(int i = start; i < nums.length; i++){
            tempList.add(nums[i]);
            backtrack(list, tempList, nums, i + 1);
            tempList.remove(tempList.size() - 1);
        }
    }

Subsets II (contains duplicates) : https://leetcode.com/problems/subsets-ii/

    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> list = new ArrayList<>();
        Arrays.sort(nums);
        backtrack(list, new ArrayList<>(), nums, 0);
        return list;
    }

    private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int start){
        list.add(new ArrayList<>(tempList));
        for(int i = start; i < nums.length; i++){
            if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates
            tempList.add(nums[i]);
            backtrack(list, tempList, nums, i + 1);
            tempList.remove(tempList.size() - 1);
        }
    }     

Permutations : https://leetcode.com/problems/permutations/

public List<List<Integer>> permute(int[] nums) {
   List<List<Integer>> list = new ArrayList<>();
   // Arrays.sort(nums); // not necessary
   backtrack(list, new ArrayList<>(), nums);
   return list;
}

private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums){
   if(tempList.size() == nums.length){
      list.add(new ArrayList<>(tempList));
   } else{
      for(int i = 0; i < nums.length; i++){ 
         if(tempList.contains(nums[i])) continue; // element already exists, skip
         tempList.add(nums[i]);
         backtrack(list, tempList, nums);
         tempList.remove(tempList.size() - 1);
      }
   }
} 

Permutations II (contains duplicates) : https://leetcode.com/problems/permutations-ii/

public List> permuteUnique(int[] nums) { List> list = new ArrayList<>(); Arrays.sort(nums); backtrack(list, new ArrayList<>(), nums, new boolean[nums.length]); return list; }

private void backtrack(List> list, List tempList, int [] nums, boolean [] used){ if(tempList.size() == nums.length){ list.add(new ArrayList<>(tempList)); } else{ for(int i = 0; i < nums.length; i++){ if(used[i] || i > 0 && nums[i] == nums[i-1] && !used[i - 1]) continue; used[i] = true; tempList.add(nums[i]); backtrack(list, tempList, nums, used); used[i] = false; tempList.remove(tempList.size() - 1); } } } Combination Sum : https://leetcode.com/problems/combination-sum/

public List> combinationSum(int[] nums, int target) { List> list = new ArrayList<>(); Arrays.sort(nums); backtrack(list, new ArrayList<>(), nums, target, 0); return list; }

private void backtrack(List> list, List tempList, int [] nums, int remain, int start){ if(remain < 0) return; else if(remain == 0) list.add(new ArrayList<>(tempList)); else{ for(int i = start; i < nums.length; i++){ tempList.add(nums[i]); backtrack(list, tempList, nums, remain - nums[i], i); // not i + 1 because we can reuse same elements tempList.remove(tempList.size() - 1); } } } Combination Sum II (can't reuse same element) : https://leetcode.com/problems/combination-sum-ii/

public List> combinationSum2(int[] nums, int target) { List> list = new ArrayList<>(); Arrays.sort(nums); backtrack(list, new ArrayList<>(), nums, target, 0); return list;

}

private void backtrack(List> list, List tempList, int [] nums, int remain, int start){ if(remain < 0) return; else if(remain == 0) list.add(new ArrayList<>(tempList)); else{ for(int i = start; i < nums.length; i++){ if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates tempList.add(nums[i]); backtrack(list, tempList, nums, remain - nums[i], i + 1); tempList.remove(tempList.size() - 1); } } } Palindrome Partitioning : https://leetcode.com/problems/palindrome-partitioning/

public List> partition(String s) { List> list = new ArrayList<>(); backtrack(list, new ArrayList<>(), s, 0); return list; }

public void backtrack(List> list, List tempList, String s, int start){ if(start == s.length()) list.add(new ArrayList<>(tempList)); else{ for(int i = start; i < s.length(); i++){ if(isPalindrome(s, start, i)){ tempList.add(s.substring(start, i + 1)); backtrack(list, tempList, s, i + 1); tempList.remove(tempList.size() - 1); } } } }

public boolean isPalindrome(String s, int low, int high){ while(low < high) if(s.charAt(low++) != s.charAt(high--)) return false; return true; }

Code

Complexity