Program to split lists into strictly increasing sublists of size greater than k in Python

Sometimes we need to split a list into sublists where each sublist is strictly increasing and has a minimum size. This problem asks us to determine if such a split is possible for a given list and minimum size k.

The key insight is that duplicate elements cannot be in the same strictly increasing sublist. If we have too many duplicates, we might not be able to form enough sublists of the required minimum size.

Algorithm

To solve this problem, we follow these steps ?

• Count the frequency of each element in the list
• Find the maximum frequency among all elements
• Check if max_frequency * k ≤ total_elements

The logic is that if an element appears max_frequency times, we need at least max_frequency different sublists. Each sublist needs at least k elements, so we need at least max_frequency * k total elements.

Example

from collections import Counter

class Solution:
    def solve(self, nums, k):
        c = Counter(nums)
        max_count = max(c.values())
        return max_count * k <= len(nums)

# Test the solution
ob = Solution()
nums = [6, 7, 5, 10, 13]
k = 2
result = ob.solve(nums, k)
print(f"Can split {nums} with k={k}: {result}")

Can split [6, 7, 5, 10, 13] with k=2: True

Step-by-Step Explanation

Let's trace through the example ?

from collections import Counter

nums = [6, 7, 5, 10, 13]
k = 2

# Count frequencies
c = Counter(nums)
print("Element frequencies:", dict(c))

# Find maximum frequency
max_count = max(c.values())
print("Maximum frequency:", max_count)

# Check condition
total_elements = len(nums)
required_elements = max_count * k
print(f"Required elements: {max_count} * {k} = {required_elements}")
print(f"Available elements: {total_elements}")
print(f"Can split: {required_elements <= total_elements}")

Element frequencies: {6: 1, 7: 1, 5: 1, 10: 1, 13: 1}
Maximum frequency: 1
Required elements: 1 * 2 = 2
Available elements: 5
Can split: True

Another Example with Duplicates

from collections import Counter

class Solution:
    def solve(self, nums, k):
        c = Counter(nums)
        max_count = max(c.values())
        return max_count * k <= len(nums)

# Test with duplicates
ob = Solution()
nums = [1, 1, 1, 2, 3]
k = 3
result = ob.solve(nums, k)
print(f"Can split {nums} with k={k}: {result}")

# Show the reasoning
c = Counter(nums)
max_count = max(c.values())
print(f"Element 1 appears {max_count} times")
print(f"We need {max_count} sublists, each with {k} elements")
print(f"Total required: {max_count * k}, Available: {len(nums)}")

Can split [1, 1, 1, 2, 3] with k=3: False
Element 1 appears 3 times
We need 3 sublists, each with 3 elements
Total required: 9, Available: 5

Possible Split Visualization

For the first example [6, 7, 5, 10, 13] with k=2, a valid split could be ?

• Sublist 1: [5, 6] (length 2, strictly increasing)
• Sublist 2: [7, 10, 13] (length 3, strictly increasing)

Conclusion

This algorithm efficiently determines if a list can be split into strictly increasing sublists of minimum size k by counting element frequencies. The key insight is that the maximum frequency determines the minimum number of sublists needed.