LeetCode #3850 — HARD

Count Sequences to K

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

[2,3,2]
6

Example 2

[4,6,3]
2

Example 3

[1,5]
1

Step 02

Core Insight

What unlocks the optimal approach

Represent numbers by their prime exponents as <code>(2^x, 3^y, 5^z)</code>.
Use Dynamic Programming on the prime exponents: let <code>dp(idx, x, y, z)</code> be the number of ways to reach exponents <code>(x,y,z)</code> after processing index <code>idx</code>.
When <code>idx == nums.length</code>, compare <code>(x, y, z)</code> to the prime exponent decomposition of <code>k</code> and count matches.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3850: Count Sequences to K
class Solution {

    record State(int i, long p, long q) {
    }

    private Map<State, Integer> f;
    private int[] nums;
    private int n;
    private long k;

    public int countSequences(int[] nums, long k) {
        this.nums = nums;
        this.n = nums.length;
        this.k = k;
        this.f = new HashMap<>();
        return dfs(0, 1L, 1L);
    }

    private int dfs(int i, long p, long q) {
        if (i == n) {
            return (p == k && q == 1L) ? 1 : 0;
        }

        State key = new State(i, p, q);
        if (f.containsKey(key)) {
            return f.get(key);
        }

        int res = dfs(i + 1, p, q);

        long x = nums[i];

        long g1 = gcd(p * x, q);
        res += dfs(i + 1, (p * x) / g1, q / g1);

        long g2 = gcd(p, q * x);
        res += dfs(i + 1, p / g2, (q * x) / g2);

        f.put(key, res);
        return res;
    }

    private long gcd(long a, long b) {
        while (b != 0) {
            long t = a % b;
            a = b;
            b = t;
        }
        return a;
    }
}

// Accepted solution for LeetCode #3850: Count Sequences to K
func countSequences(nums []int, k int64) int {
	n := len(nums)
	type state struct {
		i int
		p int64
		q int64
	}
	f := make(map[state]int)

	var gcd func(int64, int64) int64
	gcd = func(a, b int64) int64 {
		for b != 0 {
			a, b = b, a%b
		}
		return a
	}

	var dfs func(int, int64, int64) int
	dfs = func(i int, p int64, q int64) int {
		if i == n {
			if p == k && q == 1 {
				return 1
			}
			return 0
		}

		key := state{i, p, q}
		if v, ok := f[key]; ok {
			return v
		}

		res := dfs(i+1, p, q)

		x := int64(nums[i])

		g1 := gcd(p*x, q)
		res += dfs(i+1, (p*x)/g1, q/g1)

		g2 := gcd(p, q*x)
		res += dfs(i+1, p/g2, (q*x)/g2)

		f[key] = res
		return res
	}

	return dfs(0, 1, 1)
}

# Accepted solution for LeetCode #3850: Count Sequences to K
class Solution:
    def countSequences(self, nums: List[int], k: int) -> int:
        @cache
        def dfs(i: int, p: int, q: int) -> int:
            if i == n:
                return 1 if p == k and q == 1 else 0
            x = nums[i]
            res = dfs(i + 1, p, q)
            g = gcd(p * x, q)
            res += dfs(i + 1, p * x // g, q // g)
            g = gcd(p, q * x)
            res += dfs(i + 1, p // g, q * x // g)
            return res

        n = len(nums)
        ans = dfs(0, 1, 1)
        dfs.cache_clear()
        return ans

// Accepted solution for LeetCode #3850: Count Sequences to K
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3850: Count Sequences to K
// class Solution {
// 
//     record State(int i, long p, long q) {
//     }
// 
//     private Map<State, Integer> f;
//     private int[] nums;
//     private int n;
//     private long k;
// 
//     public int countSequences(int[] nums, long k) {
//         this.nums = nums;
//         this.n = nums.length;
//         this.k = k;
//         this.f = new HashMap<>();
//         return dfs(0, 1L, 1L);
//     }
// 
//     private int dfs(int i, long p, long q) {
//         if (i == n) {
//             return (p == k && q == 1L) ? 1 : 0;
//         }
// 
//         State key = new State(i, p, q);
//         if (f.containsKey(key)) {
//             return f.get(key);
//         }
// 
//         int res = dfs(i + 1, p, q);
// 
//         long x = nums[i];
// 
//         long g1 = gcd(p * x, q);
//         res += dfs(i + 1, (p * x) / g1, q / g1);
// 
//         long g2 = gcd(p, q * x);
//         res += dfs(i + 1, p / g2, (q * x) / g2);
// 
//         f.put(key, res);
//         return res;
//     }
// 
//     private long gcd(long a, long b) {
//         while (b != 0) {
//             long t = a % b;
//             a = b;
//             b = t;
//         }
//         return a;
//     }
// }

// Accepted solution for LeetCode #3850: Count Sequences to K
function countSequences(nums: number[], k: number): number {
    const n = nums.length;
    const f = new Map<string, number>();

    function gcd(a: number, b: number): number {
        while (b !== 0) {
            const t = a % b;
            a = b;
            b = t;
        }
        return a;
    }

    function dfs(i: number, p: number, q: number): number {
        if (i === n) {
            return p === k && q === 1 ? 1 : 0;
        }

        const key = `${i},${p},${q}`;
        if (f.has(key)) return f.get(key)!;

        let res = dfs(i + 1, p, q);

        const x = nums[i];

        const g1 = gcd(p * x, q);
        res += dfs(i + 1, (p * x) / g1, q / g1);

        const g2 = gcd(p, q * x);
        res += dfs(i + 1, p / g2, (q * x) / g2);

        f.set(key, res);
        return res;
    }

    return dfs(0, 1, 1);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^4 + log k)

Space

O(n^4)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.