Word Search II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Given an m x n board of characters and a list of strings words, return all words on the board.

Each word must be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

Example 1:

Input: board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]
Output: ["eat","oath"]

Example 2:

Input: board = [["a","b"],["c","d"]], words = ["abcb"]
Output: []

Constraints:

• m == board.length
• n == board[i].length
• 1 <= m, n <= 12
• board[i][j] is a lowercase English letter.
• 1 <= words.length <= 3 * 104
• 1 <= words[i].length <= 10
• words[i] consists of lowercase English letters.
• All the strings of words are unique.

C++ Word Search II LeetCode Solution

class Solution {
public:
  void combination(vector<vector<int>>& result, vector<int> sol, int k, int n) {
    if (sol.size() == k && n == 0) { result.push_back(sol); return ; }
    if (sol.size() < k) {
      for (int i = sol.empty() ? 1 : sol.back() + 1; i <= 9; ++i) {
        if (n - i < 0) break;
        sol.push_back(i);
        combination(result, sol, k, n - i);
        sol.pop_back();
      }
    }
  }

  vector<vector<int>> combinationSum3(int k, int n) {
    vector<vector<int>> result;
    vector<int> sol;
    combination(result, sol, k, n);
    return result;
  }
};

Java Word Search II LeetCode Solution

 public List<List<Integer>> combinationSum3(int k, int n) {
    List<List<Integer>> ans = new ArrayList<>();
    combination(ans, new ArrayList<Integer>(), k, 1, n);
    return ans;
}

private void combination(List<List<Integer>> ans, List<Integer> comb, int k,  int start, int n) {
	if (comb.size() == k && n == 0) {
		List<Integer> li = new ArrayList<Integer>(comb);
		ans.add(li);
		return;
	}
	for (int i = start; i <= 9; i++) {
		comb.add(i);
		combination(ans, comb, k, i+1, n-i);
		comb.remove(comb.size() - 1);
	}
}
 

Python 3 Word Search II LeetCode Solution

class Solution(object):
    def combinationSum3(self, k, n):
        ret = []
        self.dfs(list(range(1, 10)), k, n, [], ret)
        return ret
    
    def dfs(self, nums, k, n, path, ret):
        if k < 0 or n < 0:
            return 
        if k == 0 and n == 0:
            ret.append(path)
        for i in range(len(nums)):
            self.dfs(nums[i+1:], k-1, n-nums[i], path+[nums[i]], ret)

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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