Problem

You are given an N by N matrix of random letters and a dictionary of words. Find the maximum number of words that can be packed on the board from the given dictionary.

A word is considered to be able to be packed on the board if:

  • It can be found in the dictionary
  • It can be constructed from untaken letters by other words found so far on the board
  • The letters are adjacent to each other (vertically and horizontally, not diagonally).
  • Each tile can be visited only once by any word.

For example, given the following dictionary:

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{ 'eat', 'rain', 'in', 'rat' }

and matrix:

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[['e', 'a', 'n'],
 ['t', 't', 'i'],
 ['a', 'r', 'a']]

Your function should return 3, since we can make the words ’eat’, ‘in’, and ‘rat’ without them touching each other. We could have alternatively made ’eat’ and ‘rain’, but that would be incorrect since that’s only 2 words.

Examples

Example 1

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Input: 
board = [['e', 'a', 'n'],
         ['t', 't', 'i'],
         ['a', 'r', 'a']]
dictionary = ['eat', 'rain', 'in', 'rat']
Output: 3
Explanation:
We can form 'eat' (e->a->t), 'in' (i->n), and 'rat' (r->a->t) without overlap.
Path for 'eat': (0,0)->(0,1)->(1,0)
Path for 'in': (1,2)->(0,2)  
Path for 'rat': (2,1)->(2,2)->(1,1)
Total: 3 words can be packed simultaneously.

Example 2

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Input:
board = [['a', 'b'],
         ['c', 'd']]
dictionary = ['ab', 'cd', 'ac']
Output: 2
Explanation:
We can form either 'ab' and 'cd' or just 'ac'.
Best choice: 'ab' (a->b) and 'cd' (c->d) = 2 words
Alternative: 'ac' (a->c) = 1 word
Maximum is 2 words.

Example 3

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Input:
board = [['c', 'a', 't'],
         ['a', 'r', 's'],
         ['t', 's', 'e']]
dictionary = ['cat', 'car', 'rat', 'tar', 'art', 'set']
Output: 3
Explanation:
One optimal packing:
'cat': (0,0)->(0,1)->(0,2)
'art': (1,0)->(1,1)->(2,0)  
'set': (2,2)->(2,1)->(2,0) - but conflicts with 'art'
Better: 'cat', 'art', 'set' by using different paths.

Example 4

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Input:
board = [['a']]
dictionary = ['a', 'aa']
Output: 1
Explanation:
Single cell board can only form word 'a'.
Word 'aa' requires 2 'a's but we only have 1 cell.

Example 5

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Input:
board = [['x', 'y'],
         ['z', 'w']]
dictionary = ['hello', 'world']
Output: 0
Explanation:
No words from dictionary can be formed with available letters on the board.

Solution

Intuition

This problem requires finding the maximum number of non-overlapping words that can be formed on the board. We can use backtracking to try all possible combinations of words. For each word in the dictionary, we attempt to find it on the board using DFS, then recursively try to find more words on the remaining board. We backtrack by unmarking used cells and trying different combinations.

Approach

  1. For each word in the dictionary, try to find all possible paths on the board
  2. Use DFS to search for each word starting from every cell
  3. When a word is found, mark the cells as used and recursively search for more words
  4. Use backtracking: after exploring a path, unmark the cells and try other combinations
  5. Keep track of the maximum number of words found across all combinations
  6. Return the maximum count achieved

Code

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class Solution {
private:
    int maxWords = 0;
    vector<vector<int>> dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    
    bool dfs(vector<vector<char>>& board, const string& word, int idx, int r, int c, vector<vector<bool>>& visited, vector<pair<int, int>>& path) {
        if (idx == word.length()) return true;
        if (r < 0 || r >= board.size() || c < 0 || c >= board[0].size() || visited[r][c] || board[r][c] != word[idx]) {
            return false;
        }
        
        visited[r][c] = true;
        path.push_back({r, c});
        
        for (auto& dir : dirs) {
            if (dfs(board, word, idx + 1, r + dir[0], c + dir[1], visited, path)) {
                return true;
            }
        }
        
        visited[r][c] = false;
        path.pop_back();
        return false;
    }
    
    vector<vector<pair<int, int>>> findWordPaths(vector<vector<char>>& board, const string& word) {
        vector<vector<pair<int, int>>> allPaths;
        vector<vector<bool>> visited(board.size(), vector<bool>(board[0].size(), false));
        
        for (int i = 0; i < board.size(); i++) {
            for (int j = 0; j < board[0].size(); j++) {
                vector<pair<int, int>> path;
                if (dfs(board, word, 0, i, j, visited, path)) {
                    allPaths.push_back(path);
                    // Reset for finding other paths
                    for (auto& cell : path) {
                        visited[cell.first][cell.second] = false;
                    }
                }
            }
        }
        return allPaths;
    }
    
    void backtrack(vector<vector<char>>& board, vector<string>& dictionary, int idx, vector<vector<bool>>& used, int count) {
        maxWords = max(maxWords, count);
        
        for (int i = idx; i < dictionary.size(); i++) {
            auto paths = findWordPaths(board, dictionary[i]);
            
            for (auto& path : paths) {
                bool canUse = true;
                for (auto& cell : path) {
                    if (used[cell.first][cell.second]) {
                        canUse = false;
                        break;
                    }
                }
                
                if (canUse) {
                    // Mark cells as used
                    for (auto& cell : path) {
                        used[cell.first][cell.second] = true;
                    }
                    
                    backtrack(board, dictionary, i + 1, used, count + 1);
                    
                    // Backtrack
                    for (auto& cell : path) {
                        used[cell.first][cell.second] = false;
                    }
                }
            }
        }
    }
    
public:
    int maxWordPacking(vector<vector<char>>& board, vector<string>& dictionary) {
        if (board.empty() || dictionary.empty()) return 0;
        
        vector<vector<bool>> used(board.size(), vector<bool>(board[0].size(), false));
        maxWords = 0;
        backtrack(board, dictionary, 0, used, 0);
        return maxWords;
    }
};
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func maxWordPacking(board [][]byte, dictionary []string) int {
    if len(board) == 0 || len(dictionary) == 0 {
        return 0
    }
    
    dirs := [][]int{{0, 1}, {0, -1}, {1, 0}, {-1, 0}}
    rows, cols := len(board), len(board[0])
    maxWords := 0
    
    var dfs func(word string, idx, r, c int, visited [][]bool, path *[][]int) bool
    dfs = func(word string, idx, r, c int, visited [][]bool, path *[][]int) bool {
        if idx == len(word) {
            return true
        }
        if r < 0 || r >= rows || c < 0 || c >= cols || visited[r][c] || board[r][c] != word[idx] {
            return false
        }
        
        visited[r][c] = true
        *path = append(*path, []int{r, c})
        
        for _, dir := range dirs {
            if dfs(word, idx+1, r+dir[0], c+dir[1], visited, path) {
                return true
            }
        }
        
        visited[r][c] = false
        *path = (*path)[:len(*path)-1]
        return false
    }
    
    findWordPaths := func(word string) [][][]int {
        var allPaths [][][]int
        visited := make([][]bool, rows)
        for i := range visited {
            visited[i] = make([]bool, cols)
        }
        
        for i := 0; i < rows; i++ {
            for j := 0; j < cols; j++ {
                path := [][]int{}
                if dfs(word, 0, i, j, visited, &path) {
                    allPaths = append(allPaths, append([][][]int{}, path...))
                    // Reset visited
                    for _, cell := range path {
                        visited[cell[0]][cell[1]] = false
                    }
                }
            }
        }
        return allPaths
    }
    
    var backtrack func(idx int, used [][]bool, count int)
    backtrack = func(idx int, used [][]bool, count int) {
        if count > maxWords {
            maxWords = count
        }
        
        for i := idx; i < len(dictionary); i++ {
            paths := findWordPaths(dictionary[i])
            
            for _, path := range paths {
                canUse := true
                for _, cell := range path {
                    if used[cell[0]][cell[1]] {
                        canUse = false
                        break
                    }
                }
                
                if canUse {
                    // Mark cells as used
                    for _, cell := range path {
                        used[cell[0]][cell[1]] = true
                    }
                    
                    backtrack(i+1, used, count+1)
                    
                    // Backtrack
                    for _, cell := range path {
                        used[cell[0]][cell[1]] = false
                    }
                }
            }
        }
    }
    
    used := make([][]bool, rows)
    for i := range used {
        used[i] = make([]bool, cols)
    }
    
    backtrack(0, used, 0)
    return maxWords
}
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class Solution {
    private int maxWords = 0;
    private int[][] dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    
    private boolean dfs(char[][] board, String word, int idx, int r, int c, boolean[][] visited, List<int[]> path) {
        if (idx == word.length()) return true;
        if (r < 0 || r >= board.length || c < 0 || c >= board[0].length || visited[r][c] || board[r][c] != word.charAt(idx)) {
            return false;
        }
        
        visited[r][c] = true;
        path.add(new int[]{r, c});
        
        for (int[] dir : dirs) {
            if (dfs(board, word, idx + 1, r + dir[0], c + dir[1], visited, path)) {
                return true;
            }
        }
        
        visited[r][c] = false;
        path.remove(path.size() - 1);
        return false;
    }
    
    private List<List<int[]>> findWordPaths(char[][] board, String word) {
        List<List<int[]>> allPaths = new ArrayList<>();
        boolean[][] visited = new boolean[board.length][board[0].length];
        
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                List<int[]> path = new ArrayList<>();
                if (dfs(board, word, 0, i, j, visited, path)) {
                    allPaths.add(new ArrayList<>(path));
                    // Reset visited
                    for (int[] cell : path) {
                        visited[cell[0]][cell[1]] = false;
                    }
                }
            }
        }
        return allPaths;
    }
    
    private void backtrack(char[][] board, String[] dictionary, int idx, boolean[][] used, int count) {
        maxWords = Math.max(maxWords, count);
        
        for (int i = idx; i < dictionary.length; i++) {
            List<List<int[]>> paths = findWordPaths(board, dictionary[i]);
            
            for (List<int[]> path : paths) {
                boolean canUse = true;
                for (int[] cell : path) {
                    if (used[cell[0]][cell[1]]) {
                        canUse = false;
                        break;
                    }
                }
                
                if (canUse) {
                    // Mark cells as used
                    for (int[] cell : path) {
                        used[cell[0]][cell[1]] = true;
                    }
                    
                    backtrack(board, dictionary, i + 1, used, count + 1);
                    
                    // Backtrack
                    for (int[] cell : path) {
                        used[cell[0]][cell[1]] = false;
                    }
                }
            }
        }
    }
    
    public int maxWordPacking(char[][] board, String[] dictionary) {
        if (board.length == 0 || dictionary.length == 0) return 0;
        
        boolean[][] used = new boolean[board.length][board[0].length];
        maxWords = 0;
        backtrack(board, dictionary, 0, used, 0);
        return maxWords;
    }
}
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class Solution:
    def max_word_packing(self, board: List[List[str]], dictionary: List[str]) -> int:
        if not board or not dictionary:
            return 0
        
        rows, cols = len(board), len(board[0])
        dirs = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        self.max_words = 0
        
        def dfs(word: str, idx: int, r: int, c: int, visited: List[List[bool]], path: List[Tuple[int, int]]) -> bool:
            if idx == len(word):
                return True
            if r < 0 or r >= rows or c < 0 or c >= cols or visited[r][c] or board[r][c] != word[idx]:
                return False
            
            visited[r][c] = True
            path.append((r, c))
            
            for dr, dc in dirs:
                if dfs(word, idx + 1, r + dr, c + dc, visited, path):
                    return True
            
            visited[r][c] = False
            path.pop()
            return False
        
        def find_word_paths(word: str) -> List[List[Tuple[int, int]]]:
            all_paths = []
            visited = [[False] * cols for _ in range(rows)]
            
            for i in range(rows):
                for j in range(cols):
                    path = []
                    if dfs(word, 0, i, j, visited, path):
                        all_paths.append(path[:])
                        # Reset visited
                        for r, c in path:
                            visited[r][c] = False
            
            return all_paths
        
        def backtrack(idx: int, used: List[List[bool]], count: int) -> None:
            self.max_words = max(self.max_words, count)
            
            for i in range(idx, len(dictionary)):
                paths = find_word_paths(dictionary[i])
                
                for path in paths:
                    can_use = True
                    for r, c in path:
                        if used[r][c]:
                            can_use = False
                            break
                    
                    if can_use:
                        # Mark cells as used
                        for r, c in path:
                            used[r][c] = True
                        
                        backtrack(i + 1, used, count + 1)
                        
                        # Backtrack
                        for r, c in path:
                            used[r][c] = False
        
        used = [[False] * cols for _ in range(rows)]
        backtrack(0, used, 0)
        return self.max_words

Complexity

  • ⏰ Time complexity: O(W * N² * M^L * 4^L), where W is the number of words, N is the board size, M is the average word length, and L is the maximum word length. For each word, we explore all starting positions and all possible paths
  • 🧺 Space complexity: O(N² + L), for the visited array, recursion stack depth, and path storage

Method 2 - Optimized Backtracking with Trie

Intuition

We can optimize the word search by using a Trie (prefix tree) to store the dictionary words. This allows us to prune paths early when no word in the dictionary can be formed from the current prefix. We still use backtracking to try different combinations, but with more efficient word detection.

Approach

  1. Build a Trie from all dictionary words for efficient prefix checking
  2. Use DFS to explore all possible paths on the board
  3. At each step, check if the current path forms a valid prefix in the Trie
  4. If we reach a word end in the Trie, record the word and continue searching
  5. Use backtracking to try different combinations of words
  6. Track the maximum number of words found

Code

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class TrieNode {
public:
    unordered_map<char, TrieNode*> children;
    string word = "";
};

class Solution {
private:
    TrieNode* buildTrie(vector<string>& words) {
        TrieNode* root = new TrieNode();
        for (string& word : words) {
            TrieNode* curr = root;
            for (char c : word) {
                if (curr->children.find(c) == curr->children.end()) {
                    curr->children[c] = new TrieNode();
                }
                curr = curr->children[c];
            }
            curr->word = word;
        }
        return root;
    }
    
    void dfs(vector<vector<char>>& board, int r, int c, TrieNode* node, vector<vector<bool>>& visited, vector<string>& found) {
        if (r < 0 || r >= board.size() || c < 0 || c >= board[0].size() || visited[r][c]) {
            return;
        }
        
        char ch = board[r][c];
        if (node->children.find(ch) == node->children.end()) {
            return;
        }
        
        node = node->children[ch];
        if (!node->word.empty()) {
            found.push_back(node->word);
            node->word = ""; // Avoid duplicates
        }
        
        visited[r][c] = true;
        vector<vector<int>> dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        
        for (auto& dir : dirs) {
            dfs(board, r + dir[0], c + dir[1], node, visited, found);
        }
        
        visited[r][c] = false;
    }
    
public:
    int maxWordPackingTrie(vector<vector<char>>& board, vector<string>& dictionary) {
        if (board.empty() || dictionary.empty()) return 0;
        
        TrieNode* root = buildTrie(dictionary);
        int maxWords = 0;
        
        // Try all possible starting positions
        for (int i = 0; i < board.size(); i++) {
            for (int j = 0; j < board[0].size(); j++) {
                vector<vector<bool>> visited(board.size(), vector<bool>(board[0].size(), false));
                vector<string> found;
                dfs(board, i, j, root, visited, found);
                maxWords = max(maxWords, (int)found.size());
            }
        }
        
        return maxWords;
    }
};
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type TrieNode struct {
    children map[byte]*TrieNode
    word     string
}

func buildTrie(words []string) *TrieNode {
    root := &TrieNode{children: make(map[byte]*TrieNode)}
    
    for _, word := range words {
        curr := root
        for i := 0; i < len(word); i++ {
            ch := word[i]
            if _, exists := curr.children[ch]; !exists {
                curr.children[ch] = &TrieNode{children: make(map[byte]*TrieNode)}
            }
            curr = curr.children[ch]
        }
        curr.word = word
    }
    return root
}

func maxWordPackingTrie(board [][]byte, dictionary []string) int {
    if len(board) == 0 || len(dictionary) == 0 {
        return 0
    }
    
    root := buildTrie(dictionary)
    maxWords := 0
    rows, cols := len(board), len(board[0])
    dirs := [][]int{{0, 1}, {0, -1}, {1, 0}, {-1, 0}}
    
    var dfs func(r, c int, node *TrieNode, visited [][]bool, found *[]string)
    dfs = func(r, c int, node *TrieNode, visited [][]bool, found *[]string) {
        if r < 0 || r >= rows || c < 0 || c >= cols || visited[r][c] {
            return
        }
        
        ch := board[r][c]
        if _, exists := node.children[ch]; !exists {
            return
        }
        
        node = node.children[ch]
        if node.word != "" {
            *found = append(*found, node.word)
            node.word = "" // Avoid duplicates
        }
        
        visited[r][c] = true
        
        for _, dir := range dirs {
            dfs(r+dir[0], c+dir[1], node, visited, found)
        }
        
        visited[r][c] = false
    }
    
    // Try all possible starting positions
    for i := 0; i < rows; i++ {
        for j := 0; j < cols; j++ {
            visited := make([][]bool, rows)
            for k := range visited {
                visited[k] = make([]bool, cols)
            }
            found := []string{}
            dfs(i, j, root, visited, &found)
            if len(found) > maxWords {
                maxWords = len(found)
            }
        }
    }
    
    return maxWords
}
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class TrieNode {
    Map<Character, TrieNode> children = new HashMap<>();
    String word = "";
}

class Solution {
    private TrieNode buildTrie(String[] words) {
        TrieNode root = new TrieNode();
        for (String word : words) {
            TrieNode curr = root;
            for (char c : word.toCharArray()) {
                curr.children.putIfAbsent(c, new TrieNode());
                curr = curr.children.get(c);
            }
            curr.word = word;
        }
        return root;
    }
    
    private void dfs(char[][] board, int r, int c, TrieNode node, boolean[][] visited, List<String> found) {
        if (r < 0 || r >= board.length || c < 0 || c >= board[0].length || visited[r][c]) {
            return;
        }
        
        char ch = board[r][c];
        if (!node.children.containsKey(ch)) {
            return;
        }
        
        node = node.children.get(ch);
        if (!node.word.isEmpty()) {
            found.add(node.word);
            node.word = ""; // Avoid duplicates
        }
        
        visited[r][c] = true;
        int[][] dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        
        for (int[] dir : dirs) {
            dfs(board, r + dir[0], c + dir[1], node, visited, found);
        }
        
        visited[r][c] = false;
    }
    
    public int maxWordPackingTrie(char[][] board, String[] dictionary) {
        if (board.length == 0 || dictionary.length == 0) return 0;
        
        TrieNode root = buildTrie(dictionary);
        int maxWords = 0;
        
        // Try all possible starting positions
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                boolean[][] visited = new boolean[board.length][board[0].length];
                List<String> found = new ArrayList<>();
                dfs(board, i, j, root, visited, found);
                maxWords = Math.max(maxWords, found.size());
            }
        }
        
        return maxWords;
    }
}
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class TrieNode:
    def __init__(self):
        self.children = {}
        self.word = ""

class Solution:
    def build_trie(self, words: List[str]) -> TrieNode:
        root = TrieNode()
        for word in words:
            curr = root
            for ch in word:
                if ch not in curr.children:
                    curr.children[ch] = TrieNode()
                curr = curr.children[ch]
            curr.word = word
        return root
    
    def dfs(self, board: List[List[str]], r: int, c: int, node: TrieNode, visited: List[List[bool]], found: List[str]) -> None:
        if r < 0 or r >= len(board) or c < 0 or c >= len(board[0]) or visited[r][c]:
            return
        
        ch = board[r][c]
        if ch not in node.children:
            return
        
        node = node.children[ch]
        if node.word:
            found.append(node.word)
            node.word = ""  # Avoid duplicates
        
        visited[r][c] = True
        dirs = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        
        for dr, dc in dirs:
            self.dfs(board, r + dr, c + dc, node, visited, found)
        
        visited[r][c] = False
    
    def max_word_packing_trie(self, board: List[List[str]], dictionary: List[str]) -> int:
        if not board or not dictionary:
            return 0
        
        root = self.build_trie(dictionary)
        max_words = 0
        rows, cols = len(board), len(board[0])
        
        # Try all possible starting positions
        for i in range(rows):
            for j in range(cols):
                visited = [[False] * cols for _ in range(rows)]
                found = []
                self.dfs(board, i, j, root, visited, found)
                max_words = max(max_words, len(found))
        
        return max_words

Complexity

  • ⏰ Time complexity: O(N² * 4^L), where N is the board size and L is the maximum word length. The Trie optimization reduces the constant factors significantly
  • 🧺 Space complexity: O(W * L + N²), where W is the number of words and L is the average word length for the Trie, plus space for visited array

Notes

  • Method 1 provides a complete solution with backtracking to find the optimal combination of words
  • Method 2 optimizes word searching using a Trie, but may not handle the “maximum packing” requirement perfectly as it finds all words from a single starting path
  • The problem requires careful consideration of non-overlapping constraint - each cell can only be used once across all words
  • Backtracking is essential for exploring different combinations and finding the global maximum
  • Method 1 is more suitable for the exact problem requirements, while Method 2 shows optimization techniques
  • For larger boards or dictionaries, additional optimizations like pruning based on remaining cells could be beneficial
  • The problem combines graph traversal (DFS) with combinatorial optimization (finding maximum packing)
  • Consider using bit manipulation for visited state if memory optimization is crucial