diff --git a/Problem1.py b/Problem1.py
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+++ b/Problem1.py
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+# https://leetcode.com/problems/delete-and-earn/description/
+
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Recursive approach (exhaustive)
+
+class Solution:
+    def deleteAndEarn(self, nums: List[int]) -> int:
+
+        if len(nums) == 1:
+            return nums[0]
+
+        maxNum = 0
+
+        for num in nums:
+            maxNum = max(maxNum, num)
+        
+        arr = [0] * (maxNum+1)
+
+        for num in nums:
+            arr[num] += num
+
+        return self.helper(arr, 0, 0)
+
+    def helper(self, arr, i, earnings):
+
+        if i >= len(arr): return earnings
+
+        case0 = self.helper(arr, i+1, earnings)
+
+        case1 = self.helper(arr, i+2, earnings + arr[i])
+
+        return max(case0, case1)
+    
+# DP solution
+
+class Solution:
+    def deleteAndEarn(self, nums: List[int]) -> int:
+        n = len(nums)
+
+        maxVal = 0
+        for n in nums:
+            maxVal = max(maxVal, n)
+
+        arr = [0] * (maxVal+1)
+        dp = [0] * (maxVal+1)
+
+        m = len(arr)
+
+        for num in nums:
+            arr[num] += num
+
+        dp[0] = arr[0]
+        dp[1] = max(arr[0], arr[1])
+        
+        for i in range(2, maxVal+1):
+            dp[i] = max(dp[i-1], arr[i]+ dp[i-2])
+
+        return dp[m-1]
+    
+# DP using prev and curr
+
+class Solution:
+    def deleteAndEarn(self, nums: List[int]) -> int:
+        n = len(nums)
+
+        maxVal = 0
+        for n in nums:
+            maxVal = max(maxVal, n)
+
+        arr = [0] * (maxVal+1)
+        dp = [0] * (maxVal+1)
+
+        m = len(arr)
+
+        for num in nums:
+            arr[num] += num
+
+        prev = arr[0]
+        curr = max(arr[0], arr[1])
+        
+        for i in range(2, maxVal+1):
+            temp = curr
+            curr = max(curr, arr[i]+ prev)
+            prev = temp
+
+        return curr
+    
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diff --git a/Problem2.py b/Problem2.py
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+# https://leetcode.com/problems/unique-paths/description/
+
+# exhaustive approach
+# TC: 2^(m+n)
+# SC: O(m+n)
+class Solution:
+    count = 0
+    def uniquePaths(self, m: int, n: int) -> int:
+        self.helper(0, 0, m, n)
+        return self.count
+
+    def helper(self, i, j, m, n):
+
+        if i == m-1 and j == n-1:
+            self.count += 1
+            return
+
+        if i == m or j == n:
+            return
+
+        # move right
+        self.helper(i, j+1, m, n)
+
+        # move down
+        self.helper(i+1, j, m, n)
+
+
+# int based recursion
+class Solution:
+    count = 0
+    def uniquePaths(self, m: int, n: int) -> int:
+        return self.helper(0, 0, m, n)
+
+    def helper(self, i, j, m, n):
+
+        if i == m-1 and j == n-1:
+            return 1
+
+        if i == m or j == n:
+            return 0
+
+        # move right
+        case0 = self.helper(i, j+1, m, n)
+
+        # move down
+        case1 = self.helper(i+1, j, m, n)
+
+        return case0 + case1
+    
+# DP solution with memoization using 2d matrix
+# TC: O(m*n)
+# SC: O(m*n)
+class Solution:
+     
+    def uniquePaths(self, m: int, n: int) -> int:
+        self.memo = [[0] * n for _ in range(m)]
+        return self.helper(0, 0, m, n)
+
+    def helper(self, i, j, m, n):
+
+        if i == m-1 and j == n-1:
+            return 1
+
+        if i == m or j == n:
+            return 0
+
+        if self.memo[i][j] != 0: return self.memo[i][j]
+
+        # move right
+        case0 = self.helper(i, j+1, m, n)
+
+        # move down
+        case1 = self.helper(i+1, j, m, n)
+
+        self.memo[i][j] = case0 + case1
+
+        return case0 + case1
+    
+# DP with tabulation (bottom-up)
+# TC: O(m*n)
+# SC: O(m*n)
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+
+        dp = [[0] * n for _ in range(m)]
+
+        for i in range(m-1, -1, -1):
+            for j in range(n-1, -1, -1):
+                if i == m-1 or j == n-1:
+                    dp[i][j] = 1
+                else:
+                    dp[i][j] = dp[i+1][j] + dp[i][j+1]
+
+        return dp[0][0]
+    
+# DP with tabulation (top-down)
+# TC: O(m*n)
+# SC: O(m*n)
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+
+        dp = [[0] * n for _ in range(m)]
+
+        for i in range(m):
+            for j in range(n):
+                if i == 0 or j == 0:
+                    dp[i][j] = 1
+                else:
+                    dp[i][j] = dp[i-1][j] + dp[i][j-1]
+
+        return dp[m-1][n-1]
+    
+# DP with 1D array
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+
+        dp = [0] * n 
+
+        for i in range(m):
+            for j in range(n):
+                if i == 0 or j == 0:
+                    dp[j] = 1
+                else:
+                    dp[j] = dp[j] + dp[j-1]
+
+        return dp[n-1]
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