Problem Description

You are given a 0-indexed array of positive integers nums.

A triplet of three distinct indices (i, j, k) is called a single divisor triplet if the sum nums[i] + nums[j] + nums[k] is divisible by exactly one of the three values: nums[i], nums[j], or nums[k].

Your task is to return the total number of single divisor triplets in the array nums.

For example, if you have indices (i, j, k) where:

• nums[i] = 4
• nums[j] = 6
• nums[k] = 7
• Sum = 4 + 6 + 7 = 17

You would check if 17 is divisible by 4 (No), by 6 (No), and by 7 (No). Since none of them divide 17, this wouldn't be a single divisor triplet. You need exactly one of the three numbers to divide the sum.

The indices must be distinct, meaning i ≠ j, j ≠ k, and i ≠ k.

Intuition

The key observation is that the values in the array are limited to the range [1, 100]. This small range allows us to think about the problem differently - instead of iterating through all possible index triplets (which would be O(n³)), we can focus on the unique values and their frequencies.

Since we only care about the values at the three indices (not the indices themselves), we can count how many times each value appears in the array. Then, we can enumerate all possible combinations of three values from the range [1, 100].

For each combination of three values (a, b, c), we check if their sum a + b + c is divisible by exactly one of them. If this condition is met, we need to count how many ways we can pick indices from the array to form this value combination.

The counting becomes interesting when some values are repeated:

• If all three values are different, we can freely choose any index with value a, any index with value b, and any index with value c. This gives us count(a) × count(b) × count(c) ways.
• If two values are the same (say a = b), we need to pick two different indices with value a and one index with value c. The number of ways to pick two different indices from count(a) occurrences is count(a) × (count(a) - 1), giving us count(a) × (count(a) - 1) × count(c) total ways.

This approach reduces the time complexity significantly because we're only iterating through at most 100 × 100 × 100 = 1,000,000 combinations of values, regardless of the size of the input array.

Learn more about Math patterns.

Solution Approach

We implement the solution using counting and enumeration:

1. Count the frequency of each value: First, we use a Counter to count how many times each unique value appears in the array nums. This gives us a dictionary where keys are the values and values are their frequencies.

2. Enumerate all possible value combinations: We iterate through all possible combinations of three values (a, b, c) from the counter. Since we're using the counter's keys, we only consider values that actually exist in the array.

3. Check the single divisor condition: For each combination, we calculate the sum s = a + b + c and count how many of the three values divide the sum evenly. We use the expression sum(s % v == 0 for v in (a, b, c)) to count divisors. If exactly one value divides the sum, we proceed to count the triplets.

4. Count valid triplets based on value equality:

   • When a == b: We have two indices with the same value a and one with value c. The number of ways to choose 2 different indices from x occurrences of a is x × (x - 1) (ordered pairs), multiplied by z ways to choose c, giving us x × (x - 1) × z triplets.

   • When a == c: Similar logic applies, giving us x × (x - 1) × y triplets.

   • When b == c: We get x × y × (y - 1) triplets.

   • When all values are different: We can freely choose one index for each value, giving us x × y × z triplets.

5. Accumulate the result: We add the count of valid triplets for each combination to our answer ans.

The algorithm handles the distinct indices requirement automatically through the counting logic - when two values are the same, we use count × (count - 1) to ensure we pick different indices for those values.

This approach has a time complexity of O(m³) where m is the number of unique values in the array (at most 100), which is much more efficient than checking all O(n³) index triplets directly.

Example Walkthrough

Let's walk through a small example with nums = [4, 6, 7, 3, 2].

Step 1: Count frequencies

• Value 4 appears 1 time
• Value 6 appears 1 time
• Value 7 appears 1 time
• Value 3 appears 1 time
• Value 2 appears 1 time

Step 2: Enumerate value combinations

Let's check a few combinations:

Combination (4, 6, 7):

• Sum = 4 + 6 + 7 = 17
• Check divisibility: 17 % 4 = 1 (not divisible), 17 % 6 = 5 (not divisible), 17 % 7 = 3 (not divisible)
• Number of divisors = 0 (not exactly 1, so skip)

Combination (4, 6, 2):

• Sum = 4 + 6 + 2 = 12
• Check divisibility: 12 % 4 = 0 (divisible), 12 % 6 = 0 (divisible), 12 % 2 = 0 (divisible)
• Number of divisors = 3 (not exactly 1, so skip)

Combination (4, 7, 3):

• Sum = 4 + 7 + 3 = 14
• Check divisibility: 14 % 4 = 2 (not divisible), 14 % 7 = 0 (divisible), 14 % 3 = 2 (not divisible)
• Number of divisors = 1 (exactly 1! This is valid)
• All three values are different: count = 1 × 1 × 1 = 1 triplet

Combination (6, 7, 2):

• Sum = 6 + 7 + 2 = 15
• Check divisibility: 15 % 6 = 3 (not divisible), 15 % 7 = 1 (not divisible), 15 % 2 = 1 (not divisible)
• Number of divisors = 0 (not exactly 1, so skip)

Combination (3, 3, 6): (if we had duplicate 3s)

• If nums had two 3s, say nums = [3, 3, 6, 7, 2]
• Sum = 3 + 3 + 6 = 12
• Check divisibility: 12 % 3 = 0 (divisible), 12 % 6 = 0 (divisible)
• Number of divisors = 2 (not exactly 1, so skip)

In this example, we found 1 valid single divisor triplet formed by indices containing values (4, 7, 3). The actual indices would be (0, 2, 3) since nums[0]=4, nums[2]=7, nums[3]=3.

The key insight is that instead of checking all C(5,3) = 10 possible index triplets, we only checked the unique value combinations, making the algorithm much more efficient for arrays with repeated values or when the value range is small.

Solution Implementation

Time and Space Complexity

The time complexity is O(M^3), where M is the number of distinct elements in the array nums (which corresponds to the size of the Counter dictionary). This is because we have three nested loops, each iterating through all unique values in the Counter dictionary. Even though the reference mentions "the range of elements," the actual implementation iterates through the distinct elements present in nums, not the entire possible range.

The space complexity is O(M), where M is the number of distinct elements in nums. This space is primarily used to store the Counter dictionary cnt, which maintains a count for each unique element in the input array.

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

Common Pitfalls

1. Overcounting When Values Are Equal

The most critical pitfall in this problem is incorrectly handling cases where some values in the triplet are equal. When two or three values are the same, naive multiplication of frequencies leads to overcounting because we're counting the same index multiple times.

Incorrect approach:

# This would count the same index twice when value_a == value_b
if value_a == value_b:
    result += count_a * count_a * count_c  # WRONG!

Why it's wrong: If value_a == value_b and both refer to the same value with frequency count_a, we need to choose 2 distinct indices from count_a positions. Simply multiplying count_a * count_a would allow choosing the same index twice.

Correct approach:

if value_a == value_b:
    # Choose 2 distinct indices from count_a positions
    result += count_a * (count_a - 1) * count_c

2. Missing the Case Where All Three Values Are the Same

Another subtle pitfall occurs when value_a == value_b == value_c. The current code handles this correctly through the first condition (value_a == value_b), but developers might accidentally write separate logic that double-counts this scenario.

Potential mistake:

# Adding extra conditions that overlap
if value_a == value_b == value_c:
    result += count_a * (count_a - 1) * (count_a - 2)
# Then also checking value_a == value_b separately would double count!

Solution: The existing cascading if-elif structure correctly handles this case without duplication. When all three values are equal, only the first condition (value_a == value_b) executes, computing count_a * (count_a - 1) * count_a.

3. Integer Division vs Modulo Confusion

Some developers might confuse the divisibility check and use integer division instead of modulo:

Incorrect:

# This checks if division result is non-zero, not divisibility
divisor_count = sum(
    triplet_sum // value != 0  # WRONG!
    for value in (value_a, value_b, value_c)
)

Correct:

# Modulo operator checks if remainder is zero (evenly divisible)
divisor_count = sum(
    triplet_sum % value == 0
    for value in (value_a, value_b, value_c)
)

4. Not Considering Order Matters for Index Triplets

The problem asks for ordered triplets of indices (i, j, k). Some might think combinations are needed and divide by 6 (3!), but the counting logic already handles ordered triplets correctly. Adding unnecessary division would undercount valid triplets.

Key insight: When choosing indices for values (a, b, c), the order of indices matters. The formula count_a * count_b * count_c for distinct values already gives ordered selections, which is what we want.