class Solution(object):
    def floodFill(self, image, sr, sc, newColor):
        R, C = len(image), len(image[0])
        color = image[sr][sc]
        if color == newColor: return image
        def dfs(r, c):
            if image[r][c] == color:
                image[r][c] = newColor
                if r >= 1: dfs(r-1, c)
                if r+1 < R: dfs(r+1, c)
                if c >= 1: dfs(r, c-1)
                if c+1 < C: dfs(r, c+1)

        dfs(sr, sc)
        return image

def max_n(lst, n = 1):
  return sorted(lst, reverse = True)[:n]

max_n([1, 2, 3]) # [3]
max_n([1, 2, 3], 2) # [3, 2]

# Python program for implementation of Bubble Sort

def bubbleSort(arr):
	n = len(arr)

	# Traverse through all array elements
	for i in range(n-1):
	# range(n) also work but outer loop will repeat one time more than needed.

		# Last i elements are already in place
		for j in range(0, n-i-1):

			# traverse the array from 0 to n-i-1
			# Swap if the element found is greater
			# than the next element
			if arr[j] > arr[j+1] :
				arr[j], arr[j+1] = arr[j+1], arr[j]

# Driver code to test above
arr = [64, 34, 25, 12, 22, 11, 90]

bubbleSort(arr)

print ("Sorted array is:")
for i in range(len(arr)):
	print ("%d" %arr[i]),

def key_of_min(d):
  return min(d, key = d.get)

key_of_min({'a':4, 'b':0, 'c':13}) # b

#84 48 13 20 61 20 33 97 34 45 6 63 71 66 24 57 92 74 6 25 51 86 48 15 64 55 77 30 56 53 37 99 9 59 57 61 30 97 50 63 59 62 39 32 34 4 96 51 8 86 10 62 16 55 81 88 71 25 27 78 79 88 92 50 16 8 67 82 67 37 84 3 33 4 78 98 39 64 98 94 24 82 45 3 53 74 96 9 10 94 13 79 15 27 56 66 32 81 77
# xor a list of integers to find the lonely integer
res = a[0]
    for i in range(1,len(a)):
        res = res ^ a[i]

weigh = lambda a,b: sum(b)-sum(a)
FindCoin = lambda A: 0 if (n := len(A)) == 1 else (m := n//3) * (w := 1 + weigh(A[:m], A[2*m:])) + FindCoin(A[m*w:m*(w+1)])
print(FindCoin([1,1,1,1,1,1,1,2,1]))