How do you solve Count the Number of Good Partitions on LeetCode?

Count the Number of Good Partitions (LeetCode #2963) 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: Understanding how to track the last occurrence of elements can significantly reduce complexity.

What is the brute force solution for Count the Number of Good Partitions?

The brute force approach for Count the Number of Good Partitions is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible way to partition the array and counts those that meet the 'good' criteria. This method is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count the Number of Good Partitions?

Count the Number of Good Partitions uses the following concepts: Array, Hash Table, Math, Combinatorics. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count the Number of Good Partitions?

Always clarify the problem statement and constraints before diving into solutions. Think about edge cases, such as arrays with repeated elements or all unique elements. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Count the Number of Good Partitions?

Failing to account for the last occurrence of numbers correctly, leading to incorrect counts. Not using modular arithmetic, which can cause integer overflow in large cases.

#2963

Count the Number of Good Partitions

Hard

ArrayHash TableMathCombinatoricsHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a two-pointer technique to efficiently partition the array while ensuring that no two subarrays contain the same number. This reduces the need to check all combinations.

⚙️

Algorithm

3 steps

1Step 1: Use a HashMap to track the last occurrence of each number.
2Step 2: Initialize a variable to count partitions and a pointer to track the end of the last valid partition.
3Step 3: Iterate through the array, updating the last occurrence of each number and counting valid partitions based on the last seen index.

solution.py15 lines

1def countGoodPartitions(nums):
2    MOD = 10**9 + 7
3    last_seen = {}
4    count = 1
5    total_partitions = 1
6
7    for i, num in enumerate(nums):
8        if num in last_seen:
9            count = (count * (i - last_seen[num])) % MOD
10        else:
11            count = (count * (i + 1)) % MOD
12        last_seen[num] = i
13        total_partitions = (total_partitions + count) % MOD
14
15    return total_partitions

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the array, and the space complexity is O(n) due to the HashMap storing the last seen indices.

1Understanding how to track the last occurrence of elements can significantly reduce complexity.
2Using combinatorial counting techniques can help in efficiently calculating the number of valid partitions.

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