What if you could find any hidden word in a puzzle without missing a single twist or turn?
Why Word Search in Grid Using Backtracking in DSA Go?
Imagine you have a big crossword puzzle on paper, and you want to find if a certain word is hidden inside it. You try to look for the word by scanning each row and column manually, checking every letter one by one.
This manual search is slow and tiring. You might miss the word if it bends or twists in the puzzle. Also, checking every possible path by hand is confusing and error-prone.
Backtracking helps by trying each letter step-by-step and going back if the path doesn't lead to the word. It explores all possible directions automatically, making sure no possibility is missed.
for i := 0; i < rows; i++ { for j := 0; j < cols; j++ { if grid[i][j] == word[0] { // Manually check neighbors } } }
func exist(grid [][]byte, word string) bool {
// Use backtracking to explore all paths
// Return true if word found
return false
}This lets you quickly and correctly find words hidden in any direction inside a grid, even if they twist or turn.
Finding words in a word search puzzle game on your phone, where words can be horizontal, vertical, or diagonal.
Manual search is slow and error-prone for complex grids.
Backtracking tries all paths and steps back when stuck.
It ensures no possible word path is missed in the grid.
Practice
Solution
Step 1: Understand backtracking in word search
Backtracking tries all possible paths in the grid to find the target word by exploring neighbors recursively.Step 2: Eliminate other options
Sorting, counting, or replacing characters are unrelated to the search process using backtracking.Final Answer:
To explore all possible paths to find the word -> Option CQuick Check:
Backtracking = Explore paths [OK]
- Thinking backtracking sorts or modifies the grid
- Confusing backtracking with simple iteration
- Assuming backtracking counts words directly
Solution
Step 1: Understand grid indexing in Go
Grid indices start at 0 and go up to length-1, so valid indices satisfy 0 <= row < len(grid) and 0 <= col < len(grid[0]).Step 2: Check each option
if row >= 0 && row < len(grid) && col >= 0 && col < len(grid[0]) correctly uses >= 0 and < length. Options A, C, D use incorrect bounds like > 0 or <= length which are off by one.Final Answer:
if row >= 0 && row < len(grid) && col >= 0 && col < len(grid[0]) -> Option DQuick Check:
Index check = if row >= 0 && row < len(grid) && col >= 0 && col < len(grid[0]) [OK]
- Using > 0 instead of >= 0 excludes first row/column
- Using <= length allows out-of-range index
- Confusing row and column lengths
grid := [][]byte{{'A','B','C','E'},{'S','F','C','S'},{'A','D','E','E'}}
word := "ABCCED"
found := exist(grid, word)
fmt.Println(found)
func exist(board [][]byte, word string) bool {
for i := 0; i < len(board); i++ {
for j := 0; j < len(board[0]); j++ {
if backtrack(board, word, i, j, 0) {
return true
}
}
}
return false
}
func backtrack(board [][]byte, word string, row, col, index int) bool {
if index == len(word) {
return true
}
if row < 0 || row >= len(board) || col < 0 || col >= len(board[0]) || board[row][col] != word[index] {
return false
}
temp := board[row][col]
board[row][col] = '#'
found := backtrack(board, word, row+1, col, index+1) ||
backtrack(board, word, row-1, col, index+1) ||
backtrack(board, word, row, col+1, index+1) ||
backtrack(board, word, row, col-1, index+1)
board[row][col] = temp
return found
}Solution
Step 1: Understand the grid and word
The grid contains the letters to form the word "ABCCED" starting at (0,0) moving right and down.Step 2: Trace backtracking calls
Backtracking explores neighbors and marks visited cells with '#'. The word "ABCCED" can be found following the path A->B->C->C->E->D.Final Answer:
true -> Option AQuick Check:
Word found = true [OK]
- Not marking visited cells causing revisits
- Incorrect boundary checks causing false negatives
- Assuming word is not found without tracing path
func backtrack(board [][]byte, word string, row, col, index int) bool {
if index == len(word) {
return true
}
if row < 0 || row >= len(board) || col < 0 || col >= len(board[0]) || board[row][col] != word[index] {
return false
}
// Missing marking visited cell
found := backtrack(board, word, row+1, col, index+1) ||
backtrack(board, word, row-1, col, index+1) ||
backtrack(board, word, row, col+1, index+1) ||
backtrack(board, word, row, col-1, index+1)
return found
}Solution
Step 1: Analyze the backtrack function
The function checks boundaries and matches characters but does not mark visited cells, so it can revisit the same cell multiple times.Step 2: Understand the impact of missing visited marking
Without marking visited cells, the search may loop or fail to find the word correctly because it revisits cells already used in the current path.Final Answer:
The function does not mark visited cells, causing revisits -> Option AQuick Check:
Missing visited marking = The function does not mark visited cells, causing revisits [OK]
- Ignoring visited marking causes infinite loops
- Misunderstanding base case logic
- Assuming boundary checks fix revisits
Solution
Step 1: Understand the problem of multiple words
Searching each word separately is inefficient because it repeats similar searches multiple times.Step 2: Use Trie with backtracking
Building a Trie (prefix tree) of all words allows pruning paths early if no word matches, improving efficiency significantly.Step 3: Eliminate incorrect options
Sorting or replacing letters does not help search. Running backtracking separately is slow.Final Answer:
Build a Trie of words and use backtracking with pruning to find all words -> Option BQuick Check:
Trie + backtracking = efficient multi-word search [OK]
- Searching each word independently wastes time
- Ignoring Trie data structure benefits
- Trying to sort grid which doesn't help search
Ready to prove it?
Pass the test and this pattern moves unknown
