Program to find minimum difference of max and mins after updating elements at most three times in Python

Suppose we have a list of numbers called nums, and we can perform an operation to update any element to any value. We can perform at most 3 such operations, and we need to find the resulting minimum difference between the max and min values in nums.

So, if the input is like nums = [2, 3, 4, 5, 6, 7], then the output will be 2. We can change the list to [4, 3, 4, 5, 4, 4] and then 5 - 3 = 2.

Algorithm

To solve this problem, we follow these steps ?

• If size of nums ? 4, then return 0 (we can make all elements equal)
• Sort the list nums
• Try all possible combinations of removing up to 3 elements from either end
• Return the minimum difference between max and min for the remaining elements

Example

Let's implement the solution step by step ?

class Solution:
    def solve(self, nums):
        if len(nums) <= 4:
            return 0
        
        nums.sort()
        n = len(nums)
        
        # Try all combinations of removing at most 3 elements from either end
        min_diff = float('inf')
        for i in range(4):
            # Remove i elements from left, (3-i) elements from right
            left = i
            right = n - 1 - (3 - i)
            min_diff = min(min_diff, nums[right] - nums[left])
        
        return min_diff

# Test the solution
ob = Solution()
nums = [2, 3, 4, 5, 6, 7]
result = ob.solve(nums)
print(f"Input: {nums}")
print(f"Output: {result}")

Input: [2, 3, 4, 5, 6, 7]
Output: 2

How It Works

The key insight is that with 3 operations, we can effectively remove up to 3 elements from consideration. After sorting the array, we try all possible ways to remove elements:

• Remove 0 from left, 3 from right: Keep elements from index 0 to n-4
• Remove 1 from left, 2 from right: Keep elements from index 1 to n-3
• Remove 2 from left, 1 from right: Keep elements from index 2 to n-2
• Remove 3 from left, 0 from right: Keep elements from index 3 to n-1

Another Example

Let's test with a different array to see how the algorithm works ?

nums = [1, 5, 0, 10, 14]
ob = Solution()
result = ob.solve(nums)
print(f"Input: {nums}")
print(f"After sorting: {sorted(nums)}")
print(f"Minimum difference: {result}")

Input: [1, 5, 0, 10, 14]
After sorting: [0, 1, 5, 10, 14]
Minimum difference: 1

Comparison of Approaches

Conclusion

The solution sorts the array and tries all possible ways to remove up to 3 elements from either end. This ensures we find the minimum possible difference between max and min values after at most 3 operations.