From 4f216cc059dedda050e5ed22e184fd40c0ec9031 Mon Sep 17 00:00:00 2001
From: ivan <ivan1016017@gmail.com>
Date: Wed, 29 Jul 2026 05:15:59 -0600
Subject: [PATCH] adding updates

---
 .../round_7/jump_game_vii.py                  | 61 +++++++++++++++++++
 .../round_1/26_rotting_oranges.py             |  5 +-
 2 files changed, 64 insertions(+), 2 deletions(-)
 create mode 100644 src/my_project/interviews/amazon_high_frequency_23/round_7/jump_game_vii.py

diff --git a/src/my_project/interviews/amazon_high_frequency_23/round_7/jump_game_vii.py b/src/my_project/interviews/amazon_high_frequency_23/round_7/jump_game_vii.py
new file mode 100644
index 00000000..483e38c5
--- /dev/null
+++ b/src/my_project/interviews/amazon_high_frequency_23/round_7/jump_game_vii.py
@@ -0,0 +1,61 @@
+from typing import List, Union, Collection, Mapping, Optional
+
+
+class Solution:
+    def minCost(self, nums: List[int], costs: List[int]) -> int:
+        """
+        Example: nums = [3, 1, 2, 4], costs = [1, 2, 3, 4]
+        
+        From index 0 (value=3):
+          - Next smaller: index 1 (value=1)
+          - Next larger: index 3 (value=4)
+        
+        DP builds cost backwards:
+        - dp[3] = 0 (at end, no cost)
+        - dp[2] = dp[3] + costs[3] = 0 + 4 = 4 (can jump to 4)
+        - dp[1] = dp[2] + costs[2] = 4 + 3 = 7 (can jump to 2)
+        - dp[0] = min(dp[1]+costs[1], dp[3]+costs[3]) = min(7+2, 0+4) = 4
+        """
+        
+        smallStack = []  # Monotonic increasing (next smaller element)
+        largeStack = []  # Monotonic decreasing (next larger element)
+        dp = [0] * len(nums)  # dp[i] = min cost from i to end
+        
+        # Process backwards (right to left)
+        for i in range(len(nums) - 1, -1, -1):
+            
+            # Maintain monotonic increasing stack (for next smaller)
+            # Remove elements >= current (they can't be "next smaller")
+            while smallStack and nums[smallStack[-1]] >= nums[i]:
+                smallStack.pop()
+            
+            # Maintain monotonic decreasing stack (for next larger)
+            # Remove elements < current (they can't be "next larger")
+            while largeStack and nums[largeStack[-1]] < nums[i]:
+                largeStack.pop()
+            
+            # Calculate minimum cost for this position
+            nxtCost = []
+            
+            if largeStack:
+                lid = largeStack[-1]  # Next larger index
+                nxtCost.append(dp[lid] + costs[lid])
+            
+            if smallStack:
+                sid = smallStack[-1]  # Next smaller index
+                nxtCost.append(dp[sid] + costs[sid])
+            
+            # Min cost from current position to end
+            dp[i] = min(nxtCost) if nxtCost else 0
+            
+            # Add current index to both stacks
+            largeStack.append(i)
+            smallStack.append(i)
+
+        print(smallStack)
+        print(largeStack)
+        
+
+        print(dp)
+        
+        return dp[0]  # Minimum cost from start
\ No newline at end of file
diff --git a/src/my_project/interviews/google_top_exercises/round_1/26_rotting_oranges.py b/src/my_project/interviews/google_top_exercises/round_1/26_rotting_oranges.py
index f5213e85..e8ac4cff 100644
--- a/src/my_project/interviews/google_top_exercises/round_1/26_rotting_oranges.py
+++ b/src/my_project/interviews/google_top_exercises/round_1/26_rotting_oranges.py
@@ -15,7 +15,7 @@ def orangesRotting(self, grid: List[List[int]]) -> int:
             for c in range(n):
                 if grid[r][c] == 1:
                     freshy += 1
-                if grid[r][c] == 2:
+                elif grid[r][c] == 2:
                     q.append((r, c))
 
         # Run multi-source BFS
@@ -35,4 +35,5 @@ def orangesRotting(self, grid: List[List[int]]) -> int:
             if rotten_oranges > 0:
                 time += 1
 
-        return time if freshy == 0 else -1
\ No newline at end of file
+        return time if freshy == 0 else -1
+        
\ No newline at end of file