LeetCode #3404 — MEDIUM

Count Special Subsequences

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array nums consisting of positive integers.

A special subsequence is defined as a subsequence of length 4, represented by indices (p, q, r, s), where p < q < r < s. This subsequence must satisfy the following conditions:

nums[p] * nums[r] == nums[q] * nums[s]
There must be at least one element between each pair of indices. In other words, q - p > 1, r - q > 1 and s - r > 1.

Return the number of different special subsequences in nums.

Example 1:

Input: nums = [1,2,3,4,3,6,1]

Output: 1

Explanation:

There is one special subsequence in nums.

(p, q, r, s) = (0, 2, 4, 6):
- This corresponds to elements (1, 3, 3, 1).
- nums[p] * nums[r] = nums[0] * nums[4] = 1 * 3 = 3
- nums[q] * nums[s] = nums[2] * nums[6] = 3 * 1 = 3

Example 2:

Input: nums = [3,4,3,4,3,4,3,4]

Output: 3

Explanation:

There are three special subsequences in nums.

(p, q, r, s) = (0, 2, 4, 6):
- This corresponds to elements (3, 3, 3, 3).
- nums[p] * nums[r] = nums[0] * nums[4] = 3 * 3 = 9
- nums[q] * nums[s] = nums[2] * nums[6] = 3 * 3 = 9
(p, q, r, s) = (1, 3, 5, 7):
- This corresponds to elements (4, 4, 4, 4).
- nums[p] * nums[r] = nums[1] * nums[5] = 4 * 4 = 16
- nums[q] * nums[s] = nums[3] * nums[7] = 4 * 4 = 16
(p, q, r, s) = (0, 2, 5, 7):
- This corresponds to elements (3, 3, 4, 4).
- nums[p] * nums[r] = nums[0] * nums[5] = 3 * 4 = 12
- nums[q] * nums[s] = nums[2] * nums[7] = 3 * 4 = 12

Constraints:

7 <= nums.length <= 1000
1 <= nums[i] <= 1000

Step 01

Brute Force Baseline

Problem summary: You are given an array nums consisting of positive integers. A special subsequence is defined as a subsequence of length 4, represented by indices (p, q, r, s), where p < q < r < s. This subsequence must satisfy the following conditions: nums[p] * nums[r] == nums[q] * nums[s] There must be at least one element between each pair of indices. In other words, q - p > 1, r - q > 1 and s - r > 1. Return the number of different special subsequences in nums.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math

Example 1

[1,2,3,4,3,6,1]

Example 2

[3,4,3,4,3,4,3,4]

Core Insight

What unlocks the optimal approach

Count pairs where <code>nums[p] / nums[q]</code> equals <code>nums[s] / nums[r]</code>, using GCD to handle ratios efficiently.
Try iterating over <code>(p, q)</code> pairs and efficiently count valid <code>(r, s)</code> pairs with the same ratio.

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

Upper-end input sizes

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 #3404: Count Special Subsequences
class Solution {
    public long numberOfSubsequences(int[] nums) {
        int n = nums.length;
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int r = 4; r < n - 2; ++r) {
            int c = nums[r];
            for (int s = r + 2; s < n; ++s) {
                int d = nums[s];
                int g = gcd(c, d);
                cnt.merge(((d / g) << 12) | (c / g), 1, Integer::sum);
            }
        }
        long ans = 0;
        for (int q = 2; q < n - 4; ++q) {
            int b = nums[q];
            for (int p = 0; p < q - 1; ++p) {
                int a = nums[p];
                int g = gcd(a, b);
                ans += cnt.getOrDefault(((a / g) << 12) | (b / g), 0);
            }
            int c = nums[q + 2];
            for (int s = q + 4; s < n; ++s) {
                int d = nums[s];
                int g = gcd(c, d);
                cnt.merge(((d / g) << 12) | (c / g), -1, Integer::sum);
            }
        }
        return ans;
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

// Accepted solution for LeetCode #3404: Count Special Subsequences
func numberOfSubsequences(nums []int) (ans int64) {
	n := len(nums)
	cnt := make(map[int]int)
	gcd := func(a, b int) int {
		for b != 0 {
			a, b = b, a%b
		}
		return a
	}
	for r := 4; r < n-2; r++ {
		c := nums[r]
		for s := r + 2; s < n; s++ {
			d := nums[s]
			g := gcd(c, d)
			key := ((d / g) << 12) | (c / g)
			cnt[key]++
		}
	}
	for q := 2; q < n-4; q++ {
		b := nums[q]
		for p := 0; p < q-1; p++ {
			a := nums[p]
			g := gcd(a, b)
			key := ((a / g) << 12) | (b / g)
			ans += int64(cnt[key])
		}
		c := nums[q+2]
		for s := q + 4; s < n; s++ {
			d := nums[s]
			g := gcd(c, d)
			key := ((d / g) << 12) | (c / g)
			cnt[key]--
		}
	}
	return
}

# Accepted solution for LeetCode #3404: Count Special Subsequences
class Solution:
    def numberOfSubsequences(self, nums: List[int]) -> int:
        n = len(nums)
        cnt = defaultdict(int)
        for r in range(4, n - 2):
            c = nums[r]
            for s in range(r + 2, n):
                d = nums[s]
                g = gcd(c, d)
                cnt[(d // g, c // g)] += 1
        ans = 0
        for q in range(2, n - 4):
            b = nums[q]
            for p in range(q - 1):
                a = nums[p]
                g = gcd(a, b)
                ans += cnt[(a // g, b // g)]
            c = nums[q + 2]
            for s in range(q + 4, n):
                d = nums[s]
                g = gcd(c, d)
                cnt[(d // g, c // g)] -= 1
        return ans

// Accepted solution for LeetCode #3404: Count Special Subsequences
// 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 #3404: Count Special Subsequences
// class Solution {
//     public long numberOfSubsequences(int[] nums) {
//         int n = nums.length;
//         Map<Integer, Integer> cnt = new HashMap<>();
//         for (int r = 4; r < n - 2; ++r) {
//             int c = nums[r];
//             for (int s = r + 2; s < n; ++s) {
//                 int d = nums[s];
//                 int g = gcd(c, d);
//                 cnt.merge(((d / g) << 12) | (c / g), 1, Integer::sum);
//             }
//         }
//         long ans = 0;
//         for (int q = 2; q < n - 4; ++q) {
//             int b = nums[q];
//             for (int p = 0; p < q - 1; ++p) {
//                 int a = nums[p];
//                 int g = gcd(a, b);
//                 ans += cnt.getOrDefault(((a / g) << 12) | (b / g), 0);
//             }
//             int c = nums[q + 2];
//             for (int s = q + 4; s < n; ++s) {
//                 int d = nums[s];
//                 int g = gcd(c, d);
//                 cnt.merge(((d / g) << 12) | (c / g), -1, Integer::sum);
//             }
//         }
//         return ans;
//     }
// 
//     private int gcd(int a, int b) {
//         return b == 0 ? a : gcd(b, a % b);
//     }
// }

// Accepted solution for LeetCode #3404: Count Special Subsequences
function numberOfSubsequences(nums: number[]): number {
    const n = nums.length;
    const cnt = new Map<number, number>();

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

    for (let r = 4; r < n - 2; r++) {
        const c = nums[r];
        for (let s = r + 2; s < n; s++) {
            const d = nums[s];
            const g = gcd(c, d);
            const key = ((d / g) << 12) | (c / g);
            cnt.set(key, (cnt.get(key) || 0) + 1);
        }
    }

    let ans = 0;
    for (let q = 2; q < n - 4; q++) {
        const b = nums[q];
        for (let p = 0; p < q - 1; p++) {
            const a = nums[p];
            const g = gcd(a, b);
            const key = ((a / g) << 12) | (b / g);
            ans += cnt.get(key) || 0;
        }
        const c = nums[q + 2];
        for (let s = q + 4; s < n; s++) {
            const d = nums[s];
            const g = gcd(c, d);
            const key = ((d / g) << 12) | (c / g);
            cnt.set(key, (cnt.get(key) || 0) - 1);
        }
    }

    return ans;
}

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)

Space

O(1)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.