Removing Minimum Number Of Magic Beans - Leetcode Solution

The problem "Removing Minimum Number Of Magic Beans" on LeetCode can be solved using a greedy approach.

Problem Statement:

You are given an array of integers arr of size n. You can perform the following operation any number of times:

Choose a non-empty subarray of arr and replace each integer in the subarray with its value modulo 2.

For example, if you have an array [1,2,3,4], you can choose the subarray [2,3] and replace it with [0,1], resulting in the array [1,0,1,4].

A subarray is a contiguous sequence of elements in the array.

Your task is to remove all the values in the array such that the array becomes empty. Return the minimum number of operations needed to achieve this.

Solution:

The problem requires us to perform subarray operations on the array, but in order to remove all values from the array, the sum of all even numbers and the sum of all odd numbers have to be equal. This is because when a subarray is replaced by its values modulo 2, the change in parity (even/odd) of the numbers is what matters, not their actual values.

Therefore, we can iterate over the array and keep a count of the number of even and odd numbers. If the count of both is equal, we can return 0 as the array is already empty. If not, we can perform operations on the subarray(s) with the greater count of numbers of one parity until the counts are equal.

We can achieve this with a simple greedy approach. At each step, we choose the subarray with the highest count of numbers of one parity, and replace it with its values modulo 2. This guarantees that the count of even and odd numbers will decrease, but the parity sum will remain the same. We repeat this until the count of even and odd numbers becomes equal and the array becomes empty.

Pseudo Code:

1. Initialize evenCount = 0, oddCount = 0
2. Iterate over arr and increment evenCount if arr[i] is even, oddCount otherwise
3. If evenCount == oddCount, return 0
4. Compare evenCount and oddCount and set maxCount to the count of the greater parity
5. Initialize operationsCount = 0
6. While evenCount != oddCount: a. If maxCount == evenCount, find the subarray with the highest sum of even numbers and replace it with its values modulo 2 b. Else, find the subarray with the highest sum of odd numbers and replace it with its values modulo 2 c. Increment operationsCount d. Recalculate evenCount and oddCount e. Set maxCount to the count of the greater parity
7. Return operationsCount

Time Complexity:

The algorithm iterates over the array once, then repeatedly performs operations on the subarray(s) until the array becomes empty. Each operation involves iterating over the subarray, so the time complexity of the algorithm is O(n^2), where n is the length of arr.

Space Complexity:

The algorithm uses only constant extra space, so the space complexity is O(1).

Complete Solution:

class Solution:
    def minOperations(self, arr: List[int]) -> int:
        evenCount, oddCount = 0, 0
        
        # Count the number of even and odd numbers in the array
        for num in arr:
            if num % 2 == 0:
                evenCount += 1
            else:
                oddCount += 1
        
        # If the counts are equal, array is already empty
        if evenCount == oddCount:
            return 0
        
        operationsCount = 0
        
        # Perform operations until array becomes empty
        while len(arr) > 0:
            # Choose the parity with the highest count
            if evenCount > oddCount:
                maxCount = evenCount
            else:
                maxCount = oddCount
            
            if maxCount == evenCount:
                # Find the subarray with highest sum of even numbers and replace it with its values mod 2
                maxEvenSum = 0
                maxSubarray = []
                for i in range(len(arr)):
                    if arr[i] % 2 == 0:
                        if len(maxSubarray) == 0 or arr[i] > maxSubarray[-1]:
                            maxSubarray.append(arr[i])
                            maxEvenSum += arr[i]
                for i in range(len(arr)):
                    if arr[i] % 2 == 0:
                        if len(maxSubarray) == 0 or arr[i] > maxSubarray[-1]:
                            continue
                        else:
                            arr[i] = arr[i] % 2
                    else:
                        arr[i] = arr[i] % 2
                
                evenCount = len(list(filter(lambda x: x%2==0,arr)))
                oddCount = len(arr) - evenCount
                
            else:
                # Find the subarray with highest sum of odd numbers and replace it with its values mod 2
                maxOddSum = 0
                maxSubarray = []
                for i in range(len(arr)):
                    if arr[i] % 2 == 1:
                        if len(maxSubarray) == 0 or arr[i] > maxSubarray[-1]:
                            maxSubarray.append(arr[i])
                            maxOddSum += arr[i]
                for i in range(len(arr)):
                    if arr[i] % 2 == 1:
                        if len(maxSubarray) == 0 or arr[i] > maxSubarray[-1]:
                            continue
                        else:
                            arr[i] = arr[i] % 2
                    else:
                        arr[i] = arr[i] % 2
                
                evenCount = len(list(filter(lambda x: x%2==0,arr)))
                oddCount = len(arr) - evenCount
            
            operationsCount += 1
        
        return operationsCount

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