#3578

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Count Partitions With Max-Min Difference at Most K

Count the number of valid ways to partition an array into contiguous segments where max-min difference is at most k.

array dynamic programming queue sliding window prefix sum

Problem Statement

Given an integer array nums and an integer k, you need to partition nums into one or more non-empty contiguous segments such that the difference between the maximum and minimum elements in each segment does not exceed k.

Return the total number of valid ways to partition the array under this condition, modulo 109 + 7.

Examples

Example 1

Input: nums = [9,4,1,3,7], k = 4

Output: 6

There are 6 valid partitions where the difference between the maximum and minimum elements in each segment is at most k = 4 :

Example 2

Input: nums = [3,3,4], k = 0

Output: 2

There are 2 valid partitions that satisfy the given conditions:

Constraints

2 <= nums.length <= 5 * 104
1 <= nums[i] <= 109
0 <= k <= 109

Solution Approach

Dynamic Programming with State Transitions

Use dynamic programming to maintain an array where each element represents the number of valid partitions ending at that index. Transition the state by considering previous indices where the difference between the maximum and minimum elements in the segment is at most k.

Sliding Window or Monotonic Queue for Efficiency

To efficiently determine the valid segments, use a sliding window or monotonic queue. These techniques allow the solution to avoid brute-force checking of all subarrays, ensuring the algorithm runs efficiently even for larger input sizes.

Modulo Operation

Since the number of valid partitions can be very large, take the result modulo 109 + 7 to prevent overflow and comply with problem constraints.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the final dynamic programming implementation, but generally, with efficient sliding window or monotonic queue techniques, it can be reduced to O(n) for both time and space, where n is the length of the array.

What Interviewers Usually Probe

Check if the candidate is familiar with dynamic programming and state transition problems.
Evaluate their ability to implement efficient sliding window or monotonic queue algorithms.
Assess how well they handle large inputs and optimize for time and space complexity.

Common Pitfalls or Variants

Common pitfalls

Not using an efficient approach like sliding window or monotonic queue, leading to a brute-force solution that doesn't scale well.
Failing to account for the modulo operation, which could cause overflow issues in the result.
Not properly managing the state transitions in dynamic programming, resulting in incorrect partition counts.

Follow-up variants

Limit the partition count to a fixed number of segments.
Allow additional conditions on the segment values, like sum constraints.
Provide a modified version with negative values in the array.

FAQ

What is the main algorithmic approach to solving the Count Partitions With Max-Min Difference at Most K problem?

The main approach is dynamic programming with state transitions, using techniques like sliding window or monotonic queues to efficiently manage subarray validity.

How do sliding window and monotonic queue techniques help in this problem?

These techniques help efficiently track the maximum and minimum values of segments, allowing the solution to check valid partitions in O(n) time instead of O(n^2).

What is the time complexity of the optimal solution?

The time complexity is O(n) for both time and space, using dynamic programming combined with sliding window or monotonic queue techniques.

How do I handle the modulo operation in the Count Partitions With Max-Min Difference at Most K problem?

Since the result can be large, always return the number of valid partitions modulo 109 + 7 to avoid overflow and adhere to the problem's constraints.

What are common pitfalls in solving the Count Partitions With Max-Min Difference at Most K problem?

Common pitfalls include using inefficient brute-force methods, failing to account for the modulo operation, and mishandling dynamic programming state transitions.

terminal

Solution

Solution 1: Dynamic Programming + Two Pointers + Ordered Set

We define $f[i]$ as the number of ways to partition the first $i$ elements. If an array satisfies that the difference between its maximum and minimum values does not exceed $k$, then any of its subarrays also satisfies this condition. Therefore, we can use two pointers to maintain a sliding window representing the current subarray.

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class Solution:
    def countPartitions(self, nums: List[int], k: int) -> int:
        mod = 10**9 + 7
        sl = SortedList()
        n = len(nums)
        f = [1] + [0] * n
        g = [1] + [0] * n
        l = 1
        for r, x in enumerate(nums, 1):
            sl.add(x)
            while sl[-1] - sl[0] > k:
                sl.remove(nums[l - 1])
                l += 1
            f[r] = (g[r - 1] - (g[l - 2] if l >= 2 else 0) + mod) % mod
            g[r] = (g[r - 1] + f[r]) % mod
        return f[n]

Continue Practicing

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