Program to maximize the minimum value after increasing K sublists in Python

This problem requires finding the maximum possible minimum value in an array after performing K operations. Each operation increments a contiguous subarray of specified size by 1. We use binary search combined with a greedy approach to solve this efficiently.

Problem Understanding

Given an array nums, we can perform K operations where each operation:

• Selects a contiguous subarray of length size
• Increments every element in that subarray by 1

Our goal is to maximize the minimum value in the final array.

Example Walkthrough

For nums = [2, 5, 2, 2, 7], size = 3, k = 2:

• Initial: [2, 5, 2, 2, 7]
• Operation 1: Increment [2, 5, 2] ? [3, 6, 3, 2, 7]
• Operation 2: Increment [6, 3, 2] ? [3, 7, 4, 3, 7]
• Minimum value: 3

Algorithm Approach

We use binary search on the answer combined with a greedy validation:

Binary Search Strategy

We binary search on possible minimum values from 0 to 10^10, checking if each target minimum is achievable with K operations.

Greedy Validation

For a target minimum, we greedily apply operations from left to right, using a difference array to track cumulative increments efficiently.

Implementation

class Solution:
    def solve(self, nums, size, k):
        n = len(nums)
        
        def possible(target):
            events = [0] * n  # Difference array for tracking increments
            moves = 0  # Number of operations used
            cumulative = 0  # Current cumulative increment
            
            for i in range(n):
                cumulative += events[i]
                current_value = nums[i] + cumulative
                
                # If current value is less than target, we need to increment
                if current_value < target:
                    delta = target - current_value
                    moves += delta
                    cumulative += delta
                    
                    # Mark the end of this increment window
                    if i + size < n:
                        events[i + size] -= delta
            
            return moves <= k
        
        # Binary search on the answer
        left, right = 0, 10 ** 10
        
        while left < right:
            mid = (left + right + 1) // 2
            if possible(mid):
                left = mid
            else:
                right = mid - 1
        
        return left

# Test the solution
solution = Solution()
nums = [2, 5, 2, 2, 7]
size = 3
k = 2
result = solution.solve(nums, size, k)
print(f"Maximum minimum value: {result}")

Maximum minimum value: 3

How It Works

The algorithm uses two key techniques:

• Difference Array: Efficiently tracks cumulative increments without modifying the original array repeatedly
• Greedy Strategy: For each position, if the current value is below the target, we immediately apply the minimum required increments
• Binary Search: Searches for the maximum achievable minimum value in the range [0, 10^10]

Time Complexity

The time complexity is O(n log(max_value)) where n is the array length and max_value is 10^10. The space complexity is O(n) for the difference array.

Conclusion

This solution efficiently maximizes the minimum array value using binary search and greedy increment placement. The difference array technique ensures optimal time complexity for tracking cumulative operations across subarrays.