Program to find minimum possible maximum value after k operations in python

Sometimes we need to minimize the maximum value in a list by performing a limited number of decrease operations. This problem requires finding the minimum possible maximum value after k operations where each operation subtracts 1 from any element.

Problem Understanding

Given a list of numbers and k operations, we can subtract 1 from any element k times. The goal is to minimize the maximum value in the resulting list.

Example

If we have nums = [3, 4, 6, 5] and k = 6 operations ?

• Decrease 6 three times: [3, 4, 3, 5]
• Decrease 5 two times: [3, 4, 3, 3]
• Decrease 4 one time: [3, 3, 3, 3]

The maximum value becomes 3, which is optimal.

Algorithm Approach

The strategy is to always reduce the largest elements first to minimize the overall maximum ?

• Sort numbers in descending order
• Process elements from largest to smallest
• For each unique value, calculate how many elements need reduction
• Use available operations efficiently

Implementation

class Solution:
    def solve(self, nums, k):
        nums.sort(reverse=True)
        i = 0
        curr = nums[0]
        
        while k > 0:
            # Count elements with current maximum value
            while i < len(nums) and nums[i] == curr:
                i += 1
            
            # If we have enough operations to reduce all maximum elements
            if k >= i:
                k -= i
                curr -= 1
            else:
                # Not enough operations to reduce all maximum elements
                return curr
        
        return curr

# Test the solution
ob = Solution()
nums = [3, 4, 6, 5]
k = 6
result = ob.solve(nums, k)
print(f"Minimum possible maximum value: {result}")

Minimum possible maximum value: 3

Step-by-Step Execution

Let's trace through the algorithm with nums = [3, 4, 6, 5] and k = 6 ?

def solve_with_trace(nums, k):
    nums.sort(reverse=True)
    print(f"Sorted array: {nums}")
    
    i = 0
    curr = nums[0]
    operations_left = k
    
    while operations_left > 0:
        # Count elements with current value
        start_i = i
        while i < len(nums) and nums[i] == curr:
            i += 1
        
        elements_to_reduce = i
        print(f"Current max: {curr}, Elements to reduce: {elements_to_reduce}")
        print(f"Operations left: {operations_left}")
        
        if operations_left >= elements_to_reduce:
            operations_left -= elements_to_reduce
            curr -= 1
            print(f"Reduced to: {curr}, Operations remaining: {operations_left}")
        else:
            print(f"Cannot reduce further, returning: {curr}")
            return curr
    
    return curr

# Test with tracing
nums = [3, 4, 6, 5]
k = 6
result = solve_with_trace(nums.copy(), k)
print(f"\nFinal result: {result}")

Sorted array: [6, 5, 4, 3]
Current max: 6, Elements to reduce: 1
Operations left: 6
Reduced to: 5, Operations remaining: 5
Current max: 5, Elements to reduce: 2
Operations left: 5
Reduced to: 4, Operations remaining: 3
Current max: 4, Elements to reduce: 3
Operations left: 3
Reduced to: 3, Operations remaining: 0

Final result: 3

How It Works

The algorithm works by processing elements in groups of equal values. At each step, it counts how many elements have the current maximum value and determines if we have enough operations to reduce all of them by 1.

This greedy approach is optimal because reducing the largest elements first always leads to the minimum possible maximum value.

Conclusion

This greedy algorithm efficiently finds the minimum possible maximum value by always targeting the largest elements first. The time complexity is O(n log n) due to sorting, and it optimally uses the available k operations.