Problem Description

You are given an integer array nums.

The problem asks you to find all special triplets in the array. A special triplet consists of three indices (i, j, k) that satisfy the following conditions:

1. The indices must be in strictly increasing order: 0 <= i < j < k < n, where n is the length of the array
2. The element at index i must equal twice the element at index j: nums[i] == nums[j] * 2
3. The element at index k must also equal twice the element at index j: nums[k] == nums[j] * 2

In other words, you need to find triplets where the first and third elements are both exactly double the middle element, and they appear in that specific order in the array.

For example, if you have an array like [4, 2, 4], this forms a special triplet with indices (0, 1, 2) because:

• nums[0] = 4 = 2 * 2 = nums[1] * 2
• nums[2] = 4 = 2 * 2 = nums[1] * 2

Your task is to count the total number of such special triplets in the given array. Since the answer can be very large, you should return the result modulo 10^9 + 7.

Intuition

The key insight is to think about this problem from the perspective of the middle element rather than trying to find all valid triplets directly.

If we fix the middle element nums[j], then we need to find:

• How many elements equal to nums[j] * 2 appear before index j (these can be our nums[i])
• How many elements equal to nums[j] * 2 appear after index j (these can be our nums[k])

The total number of special triplets with nums[j] as the middle element would be the product of these two counts, since we can pair any valid i with any valid k.

To efficiently track these counts, we can use two hash tables:

• A left counter to track elements we've already seen (to the left of current position)
• A right counter to track elements we haven't processed yet (to the right of current position)

As we traverse the array from left to right, for each element that could serve as the middle element:

1. We remove it from the right counter (since it's no longer to the right of our current position)
2. We check how many times nums[j] * 2 appears in both left and right counters
3. We multiply these counts to get the number of valid triplets with this middle element
4. We add the current element to the left counter for future iterations

This approach cleverly transforms a seemingly complex three-pointer problem into a single-pass solution with hash tables, reducing the time complexity while maintaining correctness.

Solution Approach

The solution implements the enumerate-middle-element strategy using hash tables to efficiently count valid triplets.

Data Structures Used:

• left: A Counter (hash table) that tracks the frequency of elements to the left of the current position
• right: A Counter that initially contains all elements and tracks frequencies to the right of the current position

Algorithm Steps:

1. Initialize the counters:

   • Create an empty left counter
   • Create a right counter initialized with all elements from nums
   • Set ans = 0 to accumulate the result
   • Define mod = 10^9 + 7 for the modulo operation

2. Iterate through each element as the middle element: For each element x in nums:

   a. Update right counter: Remove the current element from right by decrementing its count: right[x] -= 1

   • This ensures x is not counted as being to the right of itself

   b. Calculate triplets with x as middle:

   • The number we're looking for on both sides is x * 2
   • Count of valid i indices: left[x * 2] (elements equal to x * 2 before current position)
   • Count of valid k indices: right[x * 2] (elements equal to x * 2 after current position)
   • Add their product to the answer: ans = (ans + left[x * 2] * right[x * 2]) % mod

   c. Update left counter: Add the current element to left by incrementing its count: left[x] += 1

   • This makes x available for future iterations as a left element

3. Return the final answer: After processing all elements, return ans

Time Complexity: O(n) where n is the length of the array, as we make a single pass through the array with constant-time hash table operations.

Space Complexity: O(n) for storing the two hash tables.

The beauty of this approach is that it avoids the naive O(n^3) solution of checking all possible triplets by fixing the middle element and using multiplication principle to count valid combinations.

Example Walkthrough

Let's walk through the solution with the array nums = [1, 2, 4, 1, 2].

We're looking for triplets where nums[i] = nums[j] * 2 and nums[k] = nums[j] * 2.

Initial Setup:

• left = {} (empty counter)
• right = {1: 2, 2: 2, 4: 1} (frequency of all elements)
• ans = 0

Step 1: Process index 0, x = 1

• Remove from right: right = {1: 1, 2: 2, 4: 1}
• Looking for 1 * 2 = 2 on both sides:
  • left[2] = 0 (no 2's to the left)
  • right[2] = 2 (two 2's to the right)
  • Add to answer: ans += 0 * 2 = 0
• Add to left: left = {1: 1}

Step 2: Process index 1, x = 2

• Remove from right: right = {1: 1, 2: 1, 4: 1}
• Looking for 2 * 2 = 4 on both sides:
  • left[4] = 0 (no 4's to the left)
  • right[4] = 1 (one 4 to the right)
  • Add to answer: ans += 0 * 1 = 0
• Add to left: left = {1: 1, 2: 1}

Step 3: Process index 2, x = 4

• Remove from right: right = {1: 1, 2: 1, 4: 0}
• Looking for 4 * 2 = 8 on both sides:
  • left[8] = 0 (no 8's to the left)
  • right[8] = 0 (no 8's to the right)
  • Add to answer: ans += 0 * 0 = 0
• Add to left: left = {1: 1, 2: 1, 4: 1}

Step 4: Process index 3, x = 1

• Remove from right: right = {1: 0, 2: 1, 4: 0}
• Looking for 1 * 2 = 2 on both sides:
  • left[2] = 1 (one 2 to the left at index 1)
  • right[2] = 1 (one 2 to the right at index 4)
  • Add to answer: ans += 1 * 1 = 1
• Add to left: left = {1: 2, 2: 1, 4: 1}

Step 5: Process index 4, x = 2

• Remove from right: right = {1: 0, 2: 0, 4: 0}
• Looking for 2 * 2 = 4 on both sides:
  • left[4] = 1 (one 4 to the left at index 2)
  • right[4] = 0 (no 4's to the right)
  • Add to answer: ans += 1 * 0 = 0
• Add to left: left = {1: 2, 2: 2, 4: 1}

Final Result: ans = 1

The special triplet found is (1, 3, 4) with values (2, 1, 2) where both nums[1] = 2 and nums[4] = 2 equal 2 * nums[3] = 2 * 1.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. The algorithm iterates through the array once with a single for loop. Within each iteration, all operations are constant time: decrementing a counter value, performing arithmetic operations and modulo calculations, accessing hash map values, and incrementing a counter value. Since Counter operations (access, increment, decrement) are O(1) on average, the overall time complexity is linear.

The space complexity is O(n), where n is the length of the array nums. The algorithm uses two Counter objects: left and right. In the worst case where all elements in nums are unique, left will eventually contain all n distinct elements (built up one by one), and right initially contains all n distinct elements. Therefore, the total space used by both Counters is O(n). The additional variables ans and mod use constant space O(1), which doesn't affect the overall space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow When Not Applying Modulo Properly

The Pitfall: A common mistake is to delay the modulo operation until the very end or apply it incorrectly during intermediate calculations. When multiplying left_counter[double_value] * right_counter[double_value], this product can exceed the integer limits before applying modulo, especially with large input arrays where frequencies can be high.

Incorrect approach:

# WRONG - might overflow before modulo
result += left_counter[double_value] * right_counter[double_value]
result = result % MOD  # Applied too late

Correct approach:

# Apply modulo to the product immediately
triplet_count = (left_counter[double_value] * right_counter[double_value]) % MOD
result = (result + triplet_count) % MOD

2. Misunderstanding the Problem Constraints

The Pitfall: Developers often misinterpret the relationship between the three elements. The problem states that nums[i] == nums[j] * 2 and nums[k] == nums[j] * 2, which means both the first and third elements should be double the middle element. A common mistake is thinking we need nums[i] * 2 == nums[j] instead.

Wrong interpretation leads to:

# WRONG - looking for wrong pattern
target_value = current_num // 2  # This would find (x/2, x, x/2) pattern
triplet_count = left_counter[target_value] * right_counter[target_value]

Correct interpretation:

# Looking for (2x, x, 2x) pattern where x is the middle element
double_value = current_num * 2
triplet_count = left_counter[double_value] * right_counter[double_value]

3. Forgetting to Handle Zero or Negative Values

The Pitfall: If the array contains zeros or negative numbers, special care is needed. When current_num = 0, then double_value = 0, which means we're looking for triplets of form (0, 0, 0). The counter logic handles this correctly, but developers might add unnecessary special cases that break the solution.

Solution: The current implementation handles all cases uniformly without special-casing, which is the correct approach. The Counter automatically returns 0 for non-existent keys, handling edge cases gracefully.

4. Counter Update Order Mistakes

The Pitfall: The order of updating the left and right counters is crucial. A common mistake is updating them in the wrong sequence:

Wrong order:

# WRONG - adds current element to left before processing
left_counter[current_num] += 1  # Should be done AFTER counting
right_counter[current_num] -= 1
triplet_count = left_counter[double_value] * right_counter[double_value]

Correct order:

# First remove from right
right_counter[current_num] -= 1
# Then count valid triplets
triplet_count = left_counter[double_value] * right_counter[double_value]
# Finally add to left
left_counter[current_num] += 1

This ensures the current element is neither in the left nor right counter when counting triplets with it as the middle element.