adamgarcia4

class Solution:
    # return max gold sum by starting at (sr, sc)
    def getMaxGoldFromCell(self, grid, sr, sc):
        originalValue = grid[sr][sc]
        bestByVisitingNeighbors = 0
        
        grid[sr][sc] = 0
        # explore neighbors
        for dx, dy in ((-1, 0), (1, 0), (0, -1), (0, 1)):
            newR = sr + dy

adamgarcia4

class Heap:
    def __init__(self, arr = None):
        self.arr = arr or []
    def __bool__(self):
        return bool(self.arr)
    def __getitem__(self, key):
        return self.arr[key]
    def bubbleup(self, idx):
        # index at root node. Cannot bubble up any further
        if idx == 0:

adamgarcia4

from collections import defaultdict
class Range:
    def __init__(self):
        self.start = None
        self.end = None
        self.char = None
    def __repr__(self):
        return '(' + self.char + ':' + str(self.start) + '->' + str(self.end) + ')'
class Solution:
    def partitionLabels(self, S: str) -> List[int]:

adamgarcia4

class Solution:
    def getMax(self, arr, k, idx, cache):
        # I am partitioning 1 element
        if idx == len(arr) - 1:
            return arr[idx]
        # I need to get max sum after partitioning
        maxSum = 0
        
        # [1,2,9,30]
        # I need to try partitioning from 1 -> K elements

adamgarcia4

from collections import defaultdict
import heapq
class Solution:
    def maxProbability(self, n: int, edges: List[List[int]], succProb: List[float], start: int, end: int) -> float:
        '''
        Input
            n - number of nodes
            edges - (u,v) for an undirected edge.
            succProb - edge[i]'s weight
            start - start node

adamgarcia4

from collections import deque
class MaxMonoQueue:
    # Invariant: all preceeding numbers that are < newVal need to get out!
    def __init__(self):
        self.queue = deque()
    def enqueue(self, val, idx):
        # maintaining the invariant
        while self.queue and self.queue[-1][0] <= val:
            self.queue.pop()
        self.queue.append((val, idx))

adamgarcia4

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def generateTrees(self, N, treeList):
        # for all trees:
        #     treeList[N - 1].append(tree)

adamgarcia4

class Solution:
    def backtrack(self, n, k, partial):
        # Base Case
        # If I have generated a string of the correct length
        # do not proceed any longer.
        if len(partial) == n:
            self.counter += 1
            
            # If I have found the 'k'th happy string, Return this to the main caller.
            if self.counter == k:

adamgarcia4

class Solution:
    # This function updates the lowLink of nodeNum.
    def dfs(self, at, parent, bridges):
        # Initialize current node
        self.visited[at] = True
        self.lowLink[at] = self.rankNum
        self.ids[at] = self.rankNum
        
        self.rankNum += 1
        

adamgarcia4

{
	"licenseKey": "552258f2-1f24-43f9-914a-de3c4ff21b88",
	"devtools_port": 9090,
	"runtime": {
		"arguments": "",
		"version": "18.87.55.19",
		"fallbackVersion": "",
		"futureVersion": ""
	},
	"platform": {