A general approach to backtrack problem


  • 1
    W

    The answer for this problem based on the template provided by @issac3, and is able to provide detailed answer if the problem changed.

    class Solution {
    public:
        int countArrangement(int N) {
            vector<int> tempList;
            vector<vector<int>> result;
            buildBeautiful(result, tempList, 1, N);
            return result.size();
        }
        
        void buildBeautiful(vector<vector<int>> &result, vector<int> tempList, int start, int end){
            if(start > end ){
                result.push_back(tempList);
                return;
            }
            for(int i=1; i<=end; i++){
                if(((tempList.size()+1)%i) != 0 && (i%(tempList.size()+1)) != 0)    continue;
                else if(find(tempList.begin(), tempList.end(), i) != tempList.end())    continue;
                tempList.push_back(i);
                buildBeautiful(result, tempList, start+1, end);
                tempList.pop_back();
            }
        }
    };
    

    @issac3 said in 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<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> list = new ArrayList<>();
        Arrays.sort(nums);
        backtrack(list, new ArrayList<>(), nums, new boolean[nums.length]);
        return list;
    }
    
    private void backtrack(List<List<Integer>> list, List<Integer> 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<List<Integer>> combinationSum(int[] nums, int target) {
        List<List<Integer>> list = new ArrayList<>();
        Arrays.sort(nums);
        backtrack(list, new ArrayList<>(), nums, target, 0);
        return list;
    }
    
    private void backtrack(List<List<Integer>> list, List<Integer> 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<List<Integer>> combinationSum2(int[] nums, int target) {
        List<List<Integer>> list = new ArrayList<>();
        Arrays.sort(nums);
        backtrack(list, new ArrayList<>(), nums, target, 0);
        return list;
        
    }
    
    private void backtrack(List<List<Integer>> list, List<Integer> 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<List<String>> partition(String s) {
       List<List<String>> list = new ArrayList<>();
       backtrack(list, new ArrayList<>(), s, 0);
       return list;
    }
    
    public void backtrack(List<List<String>> list, List<String> 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;
    } 
    

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