How do you solve Count Sequences to K on LeetCode?

Count Sequences to K (LeetCode #3850) 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: Using prime factorization simplifies the problem.

What is the time complexity of Count Sequences to K?

The optimal solution for Count Sequences to K runs in O(n) time and uses O(n) space. The complexity is linear due to processing each number once and maintaining a manageable state space with prime factors.

What is the brute force solution for Count Sequences to K?

The brute force approach for Count Sequences to K is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible combinations of operations for each element in nums. This involves recursively trying each operation and counting valid sequences that result in val equal to k. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Count Sequences to K?

Explain your thought process clearly. Ask clarifying questions about constraints. Discuss edge cases and test your solution.

What are common mistakes when solving Count Sequences to K?

Not considering rational division correctly. Overlooking the need for prime factor representation.

#3850

Count Sequences to K

Hard

ArrayMathDynamic ProgrammingMemoizationNumber TheoryHash 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)

By representing numbers as prime exponents, we can efficiently track changes to the value using dynamic programming. This reduces the problem to counting combinations of prime factors.

⚙️

Algorithm

3 steps

1Step 1: Factor each number in nums into its prime exponents.
2Step 2: Use a dynamic programming table to count sequences that reach the desired prime exponent configuration for k.
3Step 3: Iterate through nums, updating the DP table based on choices made at each step.

solution.py24 lines

1from collections import defaultdict
2
3def count_sequences(nums, k):
4    def prime_factors(n):
5        factors = defaultdict(int)
6        for i in range(2, int(n**0.5) + 1):
7            while n % i == 0:
8                factors[i] += 1
9                n //= i
10        if n > 1:
11            factors[n] += 1
12        return factors
13    k_factors = prime_factors(k)
14    dp = defaultdict(int)
15    dp[(0, 0, 0)] = 1
16    for num in nums:
17        num_factors = prime_factors(num)
18        new_dp = defaultdict(int)
19        for (x, y, z), count in dp.items():
20            new_dp[(x + num_factors[2], y + num_factors[3], z + num_factors[5])] += count
21            new_dp[(x - num_factors[2], y - num_factors[3], z - num_factors[5])] += count
22            new_dp[(x, y, z)] += count
23        dp = new_dp
24    return dp[tuple(k_factors.values())]

ℹ

Complexity note: The complexity is linear due to processing each number once and maintaining a manageable state space with prime factors.

1Using prime factorization simplifies the problem.
2Dynamic programming allows us to efficiently count combinations.

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