adding answer algo

Author: ivanpenaloza

src/my_project/interviews/amazon_high_frequency_23/round_2/jump_game_ii.py

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+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(largeStack)
+        print(smallStack)
+
+        print(dp)
+        
+        return dp[0]  # Minimum cost from start

src/my_project/interviews/amazon_high_frequency_23/round_2/merge_intervals.py

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+from typing import List, Union, Collection, Mapping, Optional
+from abc import ABC, abstractmethod
+
+class Solution:
+    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
+
+        list_answer = list()
+
+        for i in sorted(intervals, key=lambda i: i[0]):
+
+            if list_answer and i[0] <= list_answer[-1][-1]:
+                list_answer[-1][-1] = max(i[1], list_answer[-1][-1])
+            else: 
+                list_answer.append(i)
+
+        return list_answer