diff --git a/n-queens.py b/n-queens.py
new file mode 100644
index 00000000..7a6c93df
--- /dev/null
+++ b/n-queens.py
@@ -0,0 +1,74 @@
+''' Time Complexity : O(n * n!) ; 
+    Space Complexity : O(n^2 + n)  
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+
+    Approach - Place the queen if safe , go to next row. if cannot place in next row, backtrack by making 1->0
+'''
+
+class Solution:
+    def solveNQueens(self, n: int) -> List[List[str]]:
+        self.mat=[[False for _ in range(n)] for _ in range(n)]
+        self.result = []
+
+        def helper(i, j, n):
+            #base
+            if i == n:
+                li = []
+                for r in range(n):
+                    s = ""
+                    for c in range(n):
+                        if self.mat[r][c]:
+                            s += "Q"
+                        else:
+                            s += "."
+                    li.append(s)
+                self.result.append(li)
+                return
+
+            for j in range(n):
+                if self.isSafe(i,j,n):
+                    self.mat[i][j] = True
+                    helper(i+1,j,n)
+                    self.mat[i][j] = False
+        
+        helper(0,0,n)
+        return self.result
+
+
+    def isSafe(self,i, j,n):
+        r,c = i, j
+        #up check
+        while r>=0:
+            if self.mat[r][c]:
+                return False
+            r -= 1
+        
+        r,c = i, j
+        #right check diagonal
+        while r>=0 and c<n:
+            if self.mat[r][c]:
+                return False
+            r -= 1
+            c += 1
+
+        r,c = i, j
+        #right left diagonal
+        while r>=0 and c>=0:
+            if self.mat[r][c]:
+                return False
+            r -= 1
+            c -= 1
+        return True
+
+
+
+
+
+
+
+
+
+
+        
\ No newline at end of file
diff --git a/word-search.py b/word-search.py
new file mode 100644
index 00000000..22988db2
--- /dev/null
+++ b/word-search.py
@@ -0,0 +1,41 @@
+''' Time Complexity : O(m * n * 4^L) ; L - len of word
+    Space Complexity : O(L)  
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+   Approach : First do left pass to calculate running product. 
+              Then second iteration from right to left and updating the new product
+'''
+
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        rows, cols = len(board),len(board[0])
+        dirs = [(-1,0),(0,1),(1,0),(0,-1)]
+
+        if rows==1 and cols == 1 and board[0][0] == word:
+            return True
+
+        def dfs(board, word, i, j, s):
+            #base
+            if s == len(word):
+                return True
+            if board[i][j] != word[s]:
+                return False
+            #logic
+            #action 
+            char = board[i][j]
+            board[i][j] = "#"
+            for dir in dirs:
+                r = dir[0] + i
+                c = dir[1] + j
+                if 0<=r<rows and 0<=c<cols:
+                    if dfs(board, word, r, c, s+1):
+                        return True
+            #backtrack
+            board[i][j] = char
+
+        for i in range(rows):
+            for j in range(cols):
+                if dfs(board, word, i, j, 0):
+                    return True
+        return False
\ No newline at end of file