LeetCode #3583 — MEDIUM

Count Special Triplets

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

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums.

A special triplet is defined as a triplet of indices (i, j, k) such that:

0 <= i < j < k < n, where n = nums.length
nums[i] == nums[j] * 2
nums[k] == nums[j] * 2

Return the total number of special triplets in the array.

Since the answer may be large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [6,3,6]

Output: 1

Explanation:

The only special triplet is (i, j, k) = (0, 1, 2), where:

nums[0] = 6, nums[1] = 3, nums[2] = 6
nums[0] = nums[1] * 2 = 3 * 2 = 6
nums[2] = nums[1] * 2 = 3 * 2 = 6

Example 2:

Input: nums = [0,1,0,0]

Output: 1

Explanation:

The only special triplet is (i, j, k) = (0, 2, 3), where:

nums[0] = 0, nums[2] = 0, nums[3] = 0
nums[0] = nums[2] * 2 = 0 * 2 = 0
nums[3] = nums[2] * 2 = 0 * 2 = 0

Example 3:

Input: nums = [8,4,2,8,4]

Output: 2

Explanation:

There are exactly two special triplets:

(i, j, k) = (0, 1, 3)
- nums[0] = 8, nums[1] = 4, nums[3] = 8
- nums[0] = nums[1] * 2 = 4 * 2 = 8
- nums[3] = nums[1] * 2 = 4 * 2 = 8
(i, j, k) = (1, 2, 4)
- nums[1] = 4, nums[2] = 2, nums[4] = 4
- nums[1] = nums[2] * 2 = 2 * 2 = 4
- nums[4] = nums[2] * 2 = 2 * 2 = 4

Constraints:

3 <= n == nums.length <= 10⁵
0 <= nums[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. A special triplet is defined as a triplet of indices (i, j, k) such that: 0 <= i < j < k < n, where n = nums.length nums[i] == nums[j] * 2 nums[k] == nums[j] * 2 Return the total number of special triplets in the array. Since the answer may be large, return it modulo 109 + 7.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[6,3,6]

Example 2

[0,1,0,0]

Example 3

[8,4,2,8,4]

Step 02

Core Insight

What unlocks the optimal approach

Use frequency arrays or maps, e.g. <code>freqPrev</code> and <code>freqNext</code>—to track how many times each value appears before and after the current index.
For each index <code>j</code> in the triplet (<code>i</code>,<code>j</code>,<code>k</code>), compute its contribution to the answer using your freqPrev and freqNext counts.

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 #3583: Count Special Triplets
class Solution {
    public int specialTriplets(int[] nums) {
        Map<Integer, Integer> left = new HashMap<>();
        Map<Integer, Integer> right = new HashMap<>();
        for (int x : nums) {
            right.merge(x, 1, Integer::sum);
        }
        long ans = 0;
        final int mod = (int) 1e9 + 7;
        for (int x : nums) {
            right.merge(x, -1, Integer::sum);
            ans = (ans + 1L * left.getOrDefault(x * 2, 0) * right.getOrDefault(x * 2, 0) % mod)
                % mod;
            left.merge(x, 1, Integer::sum);
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #3583: Count Special Triplets
func specialTriplets(nums []int) int {
	left := make(map[int]int)
	right := make(map[int]int)
	for _, x := range nums {
		right[x]++
	}
	ans := int64(0)
	mod := int64(1e9 + 7)
	for _, x := range nums {
		right[x]--
		ans = (ans + int64(left[x*2])*int64(right[x*2])%mod) % mod
		left[x]++
	}
	return int(ans)
}

# Accepted solution for LeetCode #3583: Count Special Triplets
class Solution:
    def specialTriplets(self, nums: List[int]) -> int:
        left = Counter()
        right = Counter(nums)
        ans = 0
        mod = 10**9 + 7
        for x in nums:
            right[x] -= 1
            ans = (ans + left[x * 2] * right[x * 2] % mod) % mod
            left[x] += 1
        return ans

// Accepted solution for LeetCode #3583: Count Special Triplets
use std::collections::HashMap;

impl Solution {
    pub fn special_triplets(nums: Vec<i32>) -> i32 {
        let mut left: HashMap<i32, i64> = HashMap::new();
        let mut right: HashMap<i32, i64> = HashMap::new();

        for &x in &nums {
            *right.entry(x).or_insert(0) += 1;
        }

        let modulo: i64 = 1_000_000_007;
        let mut ans: i64 = 0;

        for &x in &nums {
            if let Some(v) = right.get_mut(&x) {
                *v -= 1;
            }

            let t = x * 2;

            let l = *left.get(&t).unwrap_or(&0);
            let r = *right.get(&t).unwrap_or(&0);

            ans = (ans + (l * r) % modulo) % modulo;

            *left.entry(x).or_insert(0) += 1;
        }

        ans as i32
    }
}

// Accepted solution for LeetCode #3583: Count Special Triplets
function specialTriplets(nums: number[]): number {
    const left = new Map<number, number>();
    const right = new Map<number, number>();
    for (const x of nums) {
        right.set(x, (right.get(x) || 0) + 1);
    }
    let ans = 0;
    const mod = 1e9 + 7;
    for (const x of nums) {
        right.set(x, (right.get(x) || 0) - 1);
        const lx = left.get(x * 2) || 0;
        const rx = right.get(x * 2) || 0;
        ans = (ans + ((lx * rx) % mod)) % mod;
        left.set(x, (left.get(x) || 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(n)

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.