Write a user-defined function in Python named Puzzle(W, N) which takes the argument W as an English word and N as an integer and returns the string where every Nth alphabet of the word W is replaced with an underscore ("_").
Example: If W contains the word "TELEVISION" and N is 3, then the function should return the string "TE_EV_SI_N". Likewise, for the word "TELEVISION" if N is 4, the function should return "TEL_VIS_ON".
def Puzzle(W, N): # Function definition
result = "" # Initialize an empty string to store the result
for i in range(len(W)): # Loop through each character in the word
if (i + 1) % N == 0: # Check if the (i+1)th character is the Nth
result += "_" # Replace the Nth character with an underscore
else:
result += W[i] # Otherwise, keep the character as is
return result # Return the resulting string
# Example usage
W = "TELEVISION"
N = 3
print(Puzzle(W, N)) # Output: "TE_EV_SI_N"
N = 4
print(Puzzle(W, N)) # Output: "TEL_VIS_ON"
Explanation:
The function Puzzle(W, N) iterates through each character of the input string W.
Using the condition (i + 1) % N == 0, it checks whether the position of the character (1-based index) is a multiple of N.
If true, the character is replaced with an underscore ("_"), otherwise, the original character is retained.
The resulting string is built incrementally and returned as output.
The SELECT statement when combined with \(\_\_\_\_\_\_\) clause, returns records without repetition.
print(16 * 5 / 4 * 2 / 5 - 8)
myStr[:4] extracts the first 4 characters, which are "MISS".myStr[-5:] extracts the last 5 characters, which are "SIPPI"."#" in between, resulting in "MISS#SIPPI".In SQL, the aggregate function which will display the cardinality of the table is \(\_\_\_\_\_\).
The SELECT statement when combined with \(\_\_\_\_\_\_\) clause, returns records without repetition.
In SQL, the aggregate function which will display the cardinality of the table is \(\_\_\_\_\_\).
myStr = "MISSISSIPPI"
print(myStr[:4] + "#" + myStr[-5:])