How do you solve Count Partitions With Max-Min Difference at Most K on LeetCode?

Count Partitions With Max-Min Difference at Most K (LeetCode #3578) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Dynamic programming allows us to build solutions incrementally.

What is the time complexity of Count Partitions With Max-Min Difference at Most K?

The optimal solution for Count Partitions With Max-Min Difference at Most K runs in O(n) time and uses O(n) space. The time complexity is O(n) as we traverse the array once with two pointers, making it efficient.

What is the brute force solution for Count Partitions With Max-Min Difference at Most K?

The brute force approach for Count Partitions With Max-Min Difference at Most K is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible partitions of the array and check each segment for the max-min condition. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Count Partitions With Max-Min Difference at Most K?

Explain your thought process clearly. Discuss edge cases and constraints. Practice coding under time constraints.

What are common mistakes when solving Count Partitions With Max-Min Difference at Most K?

Not considering all segments when checking max-min difference. Forgetting to apply modulo operation on the result.

#3578

Count Partitions With Max-Min Difference at Most K

Medium

ArrayDynamic ProgrammingQueueSliding WindowPrefix SumMonotonic QueueHash MapArray

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

We can use dynamic programming to efficiently count valid partitions by maintaining a sliding window of valid segments.

⚙️

Algorithm

3 steps

1Step 1: Initialize dp array where dp[i] is the count of partitions ending at index i.
2Step 2: Use two pointers to maintain a window where the max-min difference is at most k.
3Step 3: Update dp[i] based on valid partitions from previous indices.

solution.py12 lines

1def countPartitions(nums, k):
2    n = len(nums)
3    dp = [0] * n
4    dp[0] = 1
5    total = 1
6    left = 0
7    for right in range(n):
8        while nums[right] - nums[left] > k:
9            left += 1
10        total = (total + dp[left - 1]) % (10**9 + 7)
11        dp[right] = total
12    return total

ℹ

Complexity note: The time complexity is O(n) as we traverse the array once with two pointers, making it efficient.

1Dynamic programming allows us to build solutions incrementally.
2Using two pointers helps maintain valid segments efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.