77. Combinations (https://leetcode.com/problems/combinations/description/): Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.

77. Combinations (#1 Backtracking +DFS).java

// Backtracking + DFS solution
// Time: O(2 ^ n), 29ms
// Space: O(n) for there will be only n recursion calls (excluding result), 41.6mb
class Solution {
    List<List<Integer>> ans;
    public List<List<Integer>> combine(int n, int k) {
        ans = new ArrayList<>();
        combineR(n, k, 1, new ArrayList<>());
        return ans;
    }
    
    private void combineR(int n, int k, int index, List<Integer> list) {
        // Base case
        if(list.size() == k) {
            ans.add(new ArrayList<>(list));
            return;
        }
        // Recursion
        for(int i = index; i <= n; i++) {
            list.add(i);
            combineR(n, k, i + 1, list);
            list.remove(list.size() - 1);
        }
    }
}

77. Combinations (#2 Recursion).java

// Recursion solution: C(n, k) = C(n - 1, k) + C(n - 1, k - 1)
// Time: O(2 ^ n), 4ms
// Space: O(n), 39.2mb
class Solution {
    public List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> ans;
        
        // Base case 1: k == 0 -> [[]]
        if(k == 0) {
            ans = new ArrayList();
            ans.add(new ArrayList<>());
            return ans;
        }
        
        // Base case 2: n == k -> [[1, 2, .., n]]
        if(n == k) {
            ans = new ArrayList();
            List<Integer> list = new ArrayList<>();
            for(int i = 1; i <= n; i++) list.add(i);
            ans.add(list);
            return ans;
        }
        
        ans = combine(n - 1, k);
        
        for(List<Integer> p : combine(n - 1, k - 1)) {
            p.add(n);
            ans.add(p);
        }
        
        return ans;
    }
}

77. Combinations (#3 DP).java

// DP solution: C(n, k) = C(n - 1, k) + C(n - 1, k - 1)
// Time: O(n * k * C) for n * k table, every cell takes C time to concatenate, where C is the combination, 106ms
// Space: O(n * k ^ 2 * C) for there will be a n * k table, where each cell is at most C arrays with length of O(k)
class Solution {
    public List<List<Integer>> combine(int n, int k) {
        // Define the table
        List<List<Integer>>[][] table = new List[n + 1][k + 1];
        
        // Base case
        for(int j = 0; j <= k; j++) {
            table[0][j] = new ArrayList<>();
            table[0][j].add(new ArrayList<>());
        }
        for(int i = 0; i <= n; i++) {
            table[i][0] = new ArrayList<>();
            table[i][0].add(new ArrayList<>());
        }
        
        // Fill the table
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= i && j <= k; j++) {
                table[i][j] = new ArrayList<>();
                if(table[i - 1][j] != null) {
                    for(List<Integer> list : table[i - 1][j]) {
                        if(list.size() == 0) continue;
                        table[i][j].add(list);
                    }
                }
                for(List<Integer> list : table[i - 1][j - 1]) {
                    List<Integer> newList = new ArrayList<>(list);
                    newList.add(i);
                    table[i][j].add(newList);
                }
            }
        }
        // Return the bottom right cell
        return table[n][k];
    }
}