#question1.py
def rose(n) :
  if n==0 : 
    yield []
  else :
    for k in range(0,n) :    
      for l in rose(k) :
        for r in rose(n-1-k) :      
            yield [l]+[r]+[r]
            
def start(n) :
  for x in rose(n) : 
      print(x) #basically I am printing x for each rose(n) file


print("starting program: \n")
start(2) # here is where I call the start function

def hex_to_rgb(hex):
  return tuple(int(hex[i:i+2], 16) for i in (0, 2, 4))

hex_to_rgb('FFA501') # (255, 165, 1)

# Python program for Bitonic Sort. Note that this program
# works only when size of input is a power of 2.

# The parameter dir indicates the sorting direction, ASCENDING
# or DESCENDING; if (a[i] > a[j]) agrees with the direction,
# then a[i] and a[j] are interchanged.*/
def compAndSwap(a, i, j, dire):
	if (dire==1 and a[i] > a[j]) or (dire==0 and a[i] > a[j]):
		a[i],a[j] = a[j],a[i]

# It recursively sorts a bitonic sequence in ascending order,
# if dir = 1, and in descending order otherwise (means dir=0).
# The sequence to be sorted starts at index position low,
# the parameter cnt is the number of elements to be sorted.
def bitonicMerge(a, low, cnt, dire):
	if cnt > 1:
		k = cnt/2
		for i in range(low , low+k):
			compAndSwap(a, i, i+k, dire)
		bitonicMerge(a, low, k, dire)
		bitonicMerge(a, low+k, k, dire)

# This funcion first produces a bitonic sequence by recursively
# sorting its two halves in opposite sorting orders, and then
# calls bitonicMerge to make them in the same order
def bitonicSort(a, low, cnt,dire):
	if cnt > 1:
		k = cnt/2
		bitonicSort(a, low, k, 1)
		bitonicSort(a, low+k, k, 0)
		bitonicMerge(a, low, cnt, dire)

# Caller of bitonicSort for sorting the entire array of length N
# in ASCENDING order
def sort(a,N, up):
	bitonicSort(a,0, N, up)

# Driver code to test above
a = [3, 7, 4, 8, 6, 2, 1, 5]
n = len(a)
up = 1

sort(a, n, up)
print ("\n\nSorted array is")
for i in range(n):
	print("%d" %a[i]),

# Python binary search function
def binary_search(arr, target):
    left = 0
    right = len(arr) - 1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

# Usage
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
target = 7
result = binary_search(arr, target)
if result != -1:
    print(f"Element is present at index {result}")
else:
    print("Element is not present in array")

# Python code to demonstrate
# method to remove i'th character
# Naive Method

# Initializing String
test_str = "CodeCatch"

# Printing original string
print ("The original string is : " + test_str)

# Removing char at pos 3
# using loop
new_str = ""

for i in range(len(test_str)):
	if i != 2:
		new_str = new_str + test_str[i]

# Printing string after removal
print ("The string after removal of i'th character : " + new_str)

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