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Count Sequences to K - Solution & Explanation

HardArrayMathDynamic ProgrammingMemoization11 min readAsked at: Google, LinkedIn
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Problem Statement

You are given an integer array nums, and an integer k.

Start with an initial value val = 1 and process nums from left to right. At each index i, you must choose exactly one of the following actions:

  • Multiply val by nums[i].
  • Divide val by nums[i].
  • Leave val unchanged.

After processing all elements, val is considered equal to k only if its final rational value exactly equals k.

Return the count of distinct sequences of choices that result in val == k.

Note: Division is rational (exact), not integer division. For example, 2 / 4 = 1 / 2.

 

Example 1:

Input: nums = [2,3,2], k = 6

Output: 2

Explanation:

The following 2 distinct sequences of choices result in val == k:

Sequence Operation on nums[0] Operation on nums[1] Operation on nums[2] Final val
1 Multiply: val = 1 * 2 = 2 Multiply: val = 2 * 3 = 6 Leave val unchanged 6
2 Leave val unchanged Multiply: val = 1 * 3 = 3 Multiply: val = 3 * 2 = 6 6

Example 2:

Input: nums = [4,6,3], k = 2

Output: 2

Explanation:

The following 2 distinct sequences of choices result in val == k:

Sequence Operation on nums[0] Operation on nums[1] Operation on nums[2] Final val
1 Multiply: val = 1 * 4 = 4 Divide: val = 4 / 6 = 2 / 3 Multiply: val = (2 / 3) * 3 = 2 2
2 Leave val unchanged Multiply: val = 1 * 6 = 6 Divide: val = 6 / 3 = 2 2

Example 3:

Input: nums = [1,5], k = 1

Output: 3

Explanation:

The following 3 distinct sequences of choices result in val == k:

Sequence Operation on nums[0] Operation on nums[1] Final val
1 Multiply: val = 1 * 1 = 1 Leave val unchanged 1
2 Divide: val = 1 / 1 = 1 Leave val unchanged 1
3 Leave val unchanged Leave val unchanged 1

 

Constraints:

  • 1 <= nums.length <= 19
  • 1 <= nums[i] <= 6
  • 1 <= k <= 1015

Approach Overview

Problem Overview: You need to count how many sequences can be formed so that the resulting value equals K under specific mathematical constraints. The challenge is that the number of possible sequences grows quickly, so a naive enumeration will time out. The key observation is that valid transitions are governed by mathematical relationships between factors of K.

Approach 1: Brute Force Sequence Enumeration (Exponential Time, Exponential Space)

The most direct idea is to recursively build every possible sequence and check whether it produces K. At each step you append a valid number and update the running value. If the value exceeds K or violates the constraints, you stop exploring that branch. This approach quickly becomes infeasible because the branching factor is large and the recursion explores many repeated states. Time complexity grows exponentially with sequence length, and recursion depth determines the space usage.

Approach 2: Memoization Search with Factor Transitions (O(k · d(k)) time, O(k) space)

A better strategy is to treat the problem as a dynamic programming state defined by the current value contributing to K. Instead of recomputing the number of sequences for the same intermediate value, store the result in a memo table. From a value x, iterate over valid next numbers that keep the sequence mathematically consistent with reaching K. In practice this often means iterating over divisors or multiples related to K, which dramatically reduces the search space.

The recursion becomes dp(x) = number of sequences that can reach K starting from state x. When a state is revisited, return the cached result instead of recomputing the subtree. Because each value up to K is solved once and transitions depend on its divisors, the runtime becomes roughly O(k · d(k)), where d(k) is the number of divisors. Space complexity is O(k) for the memo table and recursion stack.

This optimization relies heavily on insights from dynamic programming and memoization. Efficient divisor iteration uses ideas from number theory, which prevents exploring impossible states.

Recommended for interviews: Interviewers expect the memoized search or DP formulation. Showing the brute force first demonstrates understanding of the state space, but recognizing repeated subproblems and caching results is the key step that reduces the complexity to a manageable level.

Solution

We define a function dfs(i, p, q) that represents the number of different choice sequences when processing at index i with the current rational value being \frac{p}{q}. Initially, dfs(0, 1, 1) represents starting from the initial value of 1.

For each index i, we have three choices:

  1. Keep it unchanged, i.e., dfs(i + 1, p, q).
  2. Multiply by nums[i], i.e., dfs(i + 1, p cdot nums[i], q).
  3. Divide by nums[i], i.e., dfs(i + 1, p, q cdot nums[i]).

To avoid excessively large numbers, we simplify the numerator and denominator after each multiplication or division. Finally, when i equals n, if \frac{p}{q} exactly equals k, we return 1; otherwise, we return 0.

The time complexity is O(n^4 + log k), and the space complexity is O(n^4), where n is the length of the array nums.

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Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Brute Force BacktrackingExponentialExponentialUseful for understanding the state space or verifying small inputs
Dynamic Programming with MemoizationO(k · d(k))O(k)General optimal approach when transitions depend on divisors or mathematical constraints
Bottom-Up DP on FactorsO(k · d(k))O(k)Preferred when iterative DP is easier to implement than recursion

Video Solution

Leetcode 3850 | Count Sequences to K | DFS | Beginner Friendly | Leetcode Weekly Contest 490 • CodeWithMeGuys • 477 views views

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Frequently Asked Questions

Is Count Sequences to K easy or hard?
Count Sequences to K is considered a Hard problem because it requires combining mathematical insight with dynamic programming. Recognizing that transitions depend on factors of K and applying memoization to avoid recomputation is the main difficulty.
Count Sequences to K Python/Java solution
The typical implementation uses a recursive DFS with memoization. Python solutions often rely on dictionaries or functools.lru_cache, while Java implementations use arrays or HashMap for caching DP states. Both follow the same divisor-based transition logic.
How to solve Count Sequences to K in O(k · d(k))?
Define a recursive function dp(x) that returns the number of sequences that can eventually form K from value x. Enumerate valid next values using divisors or mathematically valid transitions and sum their results. Store results in a memo table so each state is computed only once.
What is the best approach for Count Sequences to K?
The most efficient approach uses dynamic programming with memoization. Treat each intermediate value related to K as a state and cache the number of sequences that can lead to K from that state. By iterating over valid factor transitions, the algorithm avoids recomputation and runs in about O(k · d(k)) time.
Is Count Sequences to K asked at Google/Amazon/Meta?
Problems combining dynamic programming with number theory frequently appear in interviews at companies like Google and Amazon. Variants involving factor transitions, divisor enumeration, or DP with memoization are common in high‑difficulty interview rounds.
What data structure is used in Count Sequences to K?
The core structure is a memoization table or hash map that stores computed DP states. Additional helper structures may include arrays for divisor lists or factor precomputation to speed up transitions.
What is the time complexity of Count Sequences to K?
The optimized memoization solution runs in roughly O(k · d(k)), where d(k) is the number of divisors of k. Each state up to k is computed once and transitions iterate through valid factors or multiples. Space complexity is O(k) for the memo table.

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