# Calculating Moves to Win in the Repeated Integer Deletion Game with Python

Rate this post

π‘ Problem Formulation: This article is dedicated to solving a puzzle where the objective is to calculate the minimum number of moves required to win a game, by writing a program in Python. The game involves an array of integers where, within each move, one can delete any integer that appears more than once. Winning means no integer is repeated. For example, given the array [1, 3, 3, 1], one solution could be to delete both instances of 1 or 3 resulting in 2 moves. The output, therefore, would be `2`.

## Method 1: Greedy Algorithm with Dictionary Counting

By leveraging a greedy algorithm and dictionary counting, this approach iteratively counts the occurrences of each integer and then deletes the most frequent integers first to minimize the number of moves. The function defines a dictionary to keep track of integer occurrences and proceeds with the deletion process accordingly.

Here’s an example:

```def min_deletion_moves(arr):
counts = {}
for num in arr:
counts[num] = counts.get(num, 0) + 1
moves = 0
while max(counts.values()) > 1:
moves += 1
most_common = max(counts, key=counts.get)
counts[most_common] -= 2 if counts[most_common] > 1 else 1
return moves

# Example usage:
print(min_deletion_moves([1, 3, 3, 1]))```

The output of this code snippet is:

`2`

This snippet defines a function `min_deletion_moves()` which first populates the dictionary `counts` with the frequency of each number in the given array. The `while` loop continues to delete pairs of the most frequent number until no duplicates are left, counting the moves made.

## Method 2: Sort and Iterate

In this method, the approach is to first sort the array, then iterate over it to determine the required moves by detecting consecutive duplicates. The efficiency of this method shines when dealing with large datasets where sorting can be done quickly, and linear iteration can efficiently provide the solution.

Here’s an example:

```def min_deletion_moves(arr):
arr.sort()
last_num = None
moves = 0
for num in arr:
if num == last_num:
moves += 1
last_num = None
else:
last_num = num
return moves

# Example usage:
print(min_deletion_moves([1, 3, 3, 1]))```

The output of this code snippet is:

`2`

The `min_deletion_moves()` function sorts the array upfront. The loop iterates through the sorted list, only counting a move if a number is identical to the one before it. By design, the loop only counts every second occurrence of a number because that signals the completion of a move.

## Method 3: Using Counter from Collections Module

The Counter class from Python’s collections module is specifically designed for counting hashable objects. It makes counting the occurrences of each number in the array a more straightforward task. This method works well by providing clean and concise code and relies on the robustness of standard libraries.

Here’s an example:

```from collections import Counter

def min_deletion_moves(arr):
counter = Counter(arr)
moves = sum(val // 2 for val in counter.values())
return moves

# Example usage:
print(min_deletion_moves([1, 3, 3, 1]))```

The output of this code snippet is:

`2`

By utilizing `Counter` from the collections, the `min_deletion_moves()` function simplifies the counting process drastically. The function calculates the sum of half the frequency of each number since each deletion move decreases the frequency by two.

## Method 4: Optimized Brute Force with Sets

This strategy involves converting the array to a set to eliminate duplicates initially and using a brute force approach to determine the number of moves. The simplicity and effective use of set properties make this method a good candidate for small to medium-sized arrays.

Here’s an example:

```def min_deletion_moves(arr):
unique_numbers = set(arr)
moves = (len(arr) - len(unique_numbers)) // 2
return moves

# Example usage:
print(min_deletion_moves([1, 3, 3, 1]))```

The output of this code snippet is:

`2`

This `min_deletion_moves()` function leverages the efficiency of sets to filter duplicates and then calculates the number of moves as half of the difference between the length of the original array and the set of unique elements.

## Bonus One-Liner Method 5: Functional Approach with Lambda

A one-liner solution using Python’s functional capabilities with lambda expressions and the reduce function from functools can provide an elegant alternative that potentially wins for brevity and cleverness, though at the potential cost of readability for those unfamiliar with functional programming concepts.

Here’s an example:

```from functools import reduce

# Example usage:
print(reduce(lambda moves, x: moves + (x[1] // 2), Counter([1, 3, 3, 1]).items(), 0))```

The output of this code snippet is:

`2`

Utilizing a lambda function, this line succinctly combines `reduce` to accumulate the number of moves by halving the frequency of each integer. It achieves the same result as Method 3 but in a single, albeit dense, line of code.

## Summary/Discussion

• Method 1: Greedy Algorithm with Dictionary Counting. This approach is intuitive and easily adaptable to similar problems. However, it may not be the most efficient for very large arrays as it iterates through the dictionary multiple times.
• Method 2: Sort and Iterate. Sorting the array allows a simple iteration strategy. This approach is particularly efficient on large sorted datasets but requires the overhead of sorting initially.
• Method 3: Using Counter from Collections Module. Utilizes robust standard libraries for a clear and performance-optimized solution. It’s very efficient but does abstract away some understanding of what’s happening under the hood.
• Method 4: Optimized Brute Force with Sets. The use of sets simplifies the process, making it a good choice for quick implementation with minimal code. It may not scale well to very large datasets.
• Bonus Method 5: Functional Approach with Lambda. Showcases Python’s functional programming features. While elegant, it can be less accessible for those not familiar with lambda and `reduce`.