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Backtracking 3 Done #1052

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65 changes: 65 additions & 0 deletions nqueens.py
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Original file line number Diff line number Diff line change
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# Time Complexity : O(N!) where N is the size of the board
# Space Complexity : O(N^2) where N is the size of the board
# Did this code successfully run on Leetcode : Yes
# Any problem you faced while coding this : No

# Your code here along with comments explaining your approach:
# I am makeing a boolean grid to keep track of the queens placed on the board.
# I am using backtracking to place the queens on the board row by row.
# For each row, I am iterating through each column and checking if it is safe to place a queen there.
# If it is safe, I place the queen and call the backtrack function recursively for the next row.
# If I reach the end of the board, I add the current configuration to the result list.
# If it is not safe, I backtrack and try the next column.
# Finally, I return the result list.


from typing import List
class Solution:
def solveNQueens(self, n: int) -> List[List[str]]:
grid = [[False]*n for _ in range(n)]
result = []
def backtrack(grid,r):
if r == n:
li = []
for i in range(n):
sb = []
for j in range(n):
if grid[i][j]:
sb.append("Q")
else:
sb.append(".")
li.append("".join(sb))
result.append(li)
return

for c in range(n):
if isSafe(grid,r,c):
grid[r][c] = True
backtrack(grid,r+1)
grid[r][c] = False
def isSafe(grid,r,c):
i = r
j = c
while i >=0:
if grid[i][j]:
return False
i -= 1
i = r
j = c
while i >= 0 and j >= 0:
if grid[i][j]:
return False
i -= 1
j -= 1
i = r
j = c
while i >= 0 and j < n :
if grid[i][j] :
return False
i -= 1
j += 1
return True
backtrack(grid,0)
return result


44 changes: 44 additions & 0 deletions word.py
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# Time Complexity : O(M*N*4^L) where M is the length of the word, N is the size of the board and L is the length of the word
# Space Complexity : O(L) where L is the length of the word
# Did this code successfully run on Leetcode : Yes
# Any problem you faced while coding this : No

# Your code here along with comments explaining your approach:
# I am using DFS to traverse the board starting from each cell.
# I am keeping track of the current index in the word and the current cell in the board.
# If the current index is equal to the length of the word, I return True.
# If the current cell is out of bounds or the character in the cell is not equal to the character in the word at the current index, I return False.
# I am marking the current cell as visited by changing its value to "#".
# I am then recursively calling the DFS function for the 4 adjacent cells (up, down, left, right) and incrementing the index in the word.
# If any of the recursive calls return True, I return True.
# After exploring all 4 directions, I backtrack by restoring the original value of the cell.
# Finally, if no starting cell leads to a successful match, I return False.


from typing import List
class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
m = len(board)
n = len(board[0])
dirs = [(1,0),(0,1),(0,-1),(-1,0)]
def dfs(i,j,idx):
if idx == len(word):
return True
if i < 0 or j < 0 or i == m or j == n or board[i][j] == "#":
return False
if board[i][j] != word[idx]:
return False
board[i][j] = "#"
for dr,dc in dirs:
nr = dr + i
nc = dc + j
if dfs(nr,nc,idx+1):
return True
board[i][j] = word[idx]
for i in range(m):
for j in range(n):
if dfs(i,j,0):
return True
return False