Watch 7 video solutions for Count Sequences to K, a hard level problem involving Array, Math, Dynamic Programming. This walkthrough by CodeWithMeGuys has 434 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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:
val by nums[i].val by nums[i].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 <= 191 <= nums[i] <= 61 <= k <= 1015Problem 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.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Backtracking | Exponential | Exponential | Useful for understanding the state space or verifying small inputs |
| Dynamic Programming with Memoization | O(k · d(k)) | O(k) | General optimal approach when transitions depend on divisors or mathematical constraints |
| Bottom-Up DP on Factors | O(k · d(k)) | O(k) | Preferred when iterative DP is easier to implement than recursion |