diff --git a/leetcode_740.py b/leetcode_740.py
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+++ b/leetcode_740.py
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+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Feb 22 13:29:02 2026
+
+@author: rishigoswamy
+
+    LeetCode 740: Delete and Earn
+    Link: https://leetcode.com/problems/delete-and-earn/
+    
+    Approach:
+    Convert the problem into House Robber (LeetCode 198).
+    
+    If we pick value x, we earn (x * freq[x]) points, but we lose the ability to pick
+    x-1 and x+1 (since they get deleted). That creates an "adjacent conflict" on the
+    value line: choosing x conflicts with choosing x-1 and x+1.
+    
+    So we build an array points[] where:
+        points[x] = total points we can earn by taking all occurrences of value x
+    
+    Then the problem becomes:
+        "Choose non-adjacent indices in points[] to maximize sum"
+    which is exactly House Robber.
+    
+    // Time Complexity : O(n + maxVal)
+        n = len(nums), maxVal = max(nums)
+        - O(n) to build points array
+        - O(maxVal) for the DP pass
+    // Space Complexity : O(maxVal)
+        points array size is maxVal + 1
+
+"""
+from typing import List 
+
+class Solution:
+    def deleteAndEarn(self, nums: List[int]) -> int:
+        array = [0] * (max(nums)+1)
+
+        for num in nums:
+            array[num] += num
+        
+        print(array)
+
+        del1 = 0
+        del2 = 0
+        for i in range(len(array)):
+
+            temp = max(array[i]+ del1, del2)
+            del1 = del2
+            del2 = temp
+        return del2
+
+        
+
diff --git a/leetcode_931.py b/leetcode_931.py
new file mode 100644
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+++ b/leetcode_931.py
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+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Feb 22 13:32:07 2026
+
+@author: rishigoswamy
+
+    LeetCode 931: Minimum Falling Path Sum
+    Link: https://leetcode.com/problems/minimum-falling-path-sum/
+
+    Approach:
+    Dynamic Programming (row by row).
+
+    Let prev[j] = minimum falling path sum to reach column j in the previous row.
+    For the current row i and column j:
+        curr[j] = matrix[i][j] + min(prev[j], prev[j-1], prev[j+1])
+    (only include neighbors that are in bounds)
+
+    We only need the previous row to compute the current row, so we use O(n) extra space.
+
+    // Time Complexity : O(n^2)
+        n = number of rows = number of columns (square matrix)
+        Each cell computed once with O(1) work.
+    // Space Complexity : O(n)
+        prev and curr arrays of length n.
+
+"""
+
+from typing import List
+
+
+class Solution:
+    def minFallingPathSum(self, matrix: List[List[int]]) -> int:
+        temp = []
+        prev = matrix[0]
+        for i in range(1, len(matrix)):
+            temp = []
+            temp.append(
+                matrix[i][0] + 
+            min(prev[0], prev[1]))
+            for j in range(1, len(matrix) - 1):
+                temp.append(matrix[i][j]+ min(prev[j-1],
+                prev[j],
+                prev[j+1]))
+            temp.append(matrix[i][-1] + min(prev[-1], prev[-2]))
+            prev = temp
+        return min(prev)
+            
+
+
+        
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