5 Best Ways to Find the Starting Index of the Child Who Receives the Last Balloon in Python

Rate this post

πŸ’‘ Problem Formulation: In a game where balloons are distributed to children sitting in a circle, a programmer must calculate the index of the child who receives the last balloon. We assume n children are sitting around a circle and an integer k, which denotes the step count (i.e., balloons are distributed one by one to each k-th child in sequence). The input is the total number of children n and the step count k. The desired output is the starting index of the child who gets the last balloon.

Method 1: Simulate the Distribution

An intuitive method to solve this is by simulating the distribution process. In Python, we can use a list to represent the children sitting in a circle and a loop to simulate the distribution of balloons. The function specification is to create a function that takes the number of children and the step count as inputs, and returns the index of the child receiving the last balloon.

Here’s an example:

def find_last_balloon_index(n, k):
    children = list(range(n))
    index = 0
    while len(children) > 1:
        index = (index + (k - 1)) % len(children)
        children.pop(index)
    return children[0]

print(find_last_balloon_index(5, 2))

Output:

2

In this code snippet, we construct a list of children and iterate through it, popping out children at every k-th position until only one child is left. The function returns the index of the last remaining child. This method directly reflects the problem statement, but may be inefficient for large inputs due its O(n*k) time complexity.

Method 2: Use the Josephus Problem Formula

This problem is a variant of the famous Josephus problem. There is a mathematical formula to solve this problem: the index of the last remaining person is given by the equation (n - 1 + k) % n, where n is the starting number of people, and k is the step count, applied recursively. This method provides an O(n) time complexity solution.

Here’s an example:

def find_last_balloon_index(n, k):
    if n == 1:
        return 0
    else:
        return (find_last_balloon_index(n - 1, k) + k) % n

print(find_last_balloon_index(5, 2))

Output:

2

This recursive function calculates the position of the last child remaining after each round until only one child is left, thereby determining the last balloon’s recipient. While this is an elegant and efficient solution, understanding and debugging recursion can be tricky for beginners.

Method 3: Iterative Josephus Solution

The recursive solution can be transformed into an iterative one, which avoids potential stack overflow issues with recursion for large numbers of children. The iterative version of the Josephus problem solution loops through from 1 to n to simulate the recursion stack iteratively.

Here’s an example:

def find_last_balloon_index(n, k):
    result = 0
    for i in range(2, n + 1):
        result = (result + k) % i
    return result

print(find_last_balloon_index(5, 2))

Output:

2

The code defines a loop running from 2 until n, and at each step, it applies the Josephus problem formula to find the safe position. This is beneficial because it has the same O(n) time complexity as the recursive solution but without the risk of stack overflow.

Method 4: Using Queue Data Structure

Another approach is to utilize a queue to simulate the round-robin distribution of balloons. The queue will be used to maintain the order of children, and children will be dequeued and enqueued to simulate passing the balloon until one remains.

Here’s an example:

from collections import deque

def find_last_balloon_index(n, k):
    q = deque(range(n))
    while len(q) > 1:
        q.rotate(-k+1)
        q.popleft()
    return q[0]

print(find_last_balloon_index(5, 2))

Output:

2

This code snippet uses the deque data structure from the collections module to manipulate the order of children efficiently. By rotating the queue and popping the child at each iteration, we simulate the distribution until the last child is determined.

Bonus One-Liner Method 5: Using Python’s Reduce Function

The last method exploits Python’s reduce function from the functools module, which accumulates a result across a list of values. This concise one-liner directly implements the iterative version of the Josephus problem solution.

Here’s an example:

from functools import reduce

find_last_balloon_index = lambda n, k: reduce(lambda x, y: (x + k) % y, range(1, n+1))
print(find_last_balloon_index(5, 2))

Output:

2

This one-liner defines a lambda function that uses reduce to apply the solution iteratively with the use of a lambda for the combiner function. This method is highly succinct but might be less readable, especially for those unfamiliar with reduce or lambda functions in Python.

Summary/Discussion

  • Method 1: Simulation. Easy to understand. Inefficient for large datasets.
  • Method 2: Recursive Josephus Formula. O(n) complexity. Difficult to debug for beginners.
  • Method 3: Iterative Josephus Solution. Avoids recursion issues. Same efficiency as method 2.
  • Method 4: Queue Simulation. Efficiently simulates the problem. Requires more memory than previous methods.
  • Method 5: Reduce Function One-Liner. Compact and clever. Potentially confusing for novices.