Asked in: UKG
Image of the Question
All Testcases Passed ✔
Solution
def getTripletCount(a, d):
# Write your code here
a = [i%d for i in a]
h = {}
// ... rest of solution available after purchaseExplanation
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To solve the problem of counting distinct triplets (i, j, k) such that i < j < k and the sum of a[i], a[j], and a[k] is divisible by a given integer d, it’s crucial to first deeply understand the nature of the problem and the constraints, then develop a clear strategy that balances correctness with efficiency.
Start by considering the naive approach: iterating over all possible triplets (i, j, k) in the array, checking if their sum is divisible by d, and counting how many satisfy this condition. Although this brute-force approach is straightforward and guaranteed to be correct, its time complexity is O(n^3), where n is the size of the array. This is often impractical for large n due to performance issues.
To go beyond brute force, focus on the properties of modular arithmetic since the divisibility condition relates directly to the remainder when sums are divided by d. The key insight is to examine the array elements modulo d rather than their absolute values. This reduces the problem to looking at the sums of remainders modulo d.
Here’s the step-by-step thought process:
1. **Understand the modular condition:**
The condition (a[i] + a[j] + a[k]) % d == 0 means the sum of the remainders of a[i], a[j], and a[k] when divided by d must be congruent to zero modulo d. So, if you denote r[i] = a[i] % d, the problem simplifies to finding triplets of indices where (r[i] + r[j] + r[k]) % d == 0.
2. **Group elements by their remainders:**
Consider creating a frequency distribution of how many elements correspond to each possible remainder from 0 to d-1. Since d can be large (up to 10^9), a direct frequency array might not be feasible for memory, but this grouping is conceptually useful. If d were smaller, you could store counts of each remainder and then count valid triplets based on combinations of remainders that sum to 0 modulo d.
3. **Reducing complexity through remainder combinations:**
The goal is to find all combinations of three remainders (r1, r2, r3) such that (r1 + r2 + r3) % d == 0. Each triplet of remainders corresponds to choosing elements from the respective groups. Count how many ways you can pick triplets from these remainder groups.
4. **Handling large d:**
Since d can be very large (up to 10^9), storing frequency for every remainder isn’t practical. Thus, think about alternative approaches that don't require full enumeration of all remainder classes. Consider iterating over the array, and for each pair of elements, calculate their combined remainder and check if there exists an element with a remainder that complements them to a total divisible by d. This requires an efficient way to track remainder counts dynamically.
5. **Two-pointer or hash-based approach:**
Imagine you fix the third element k and want to find pairs (i, j) with i < j < k whose remainders sum to a value that makes the total divisible by d. Maintain a data structure (such as a hash map) to store counts of remainders for elements before k, and iterate over pairs efficiently.
6. **Combinatorial counting and avoiding double counting:**
When counting triplets, be mindful of ordering constraints (i < j < k). Your method must ensure that the indices adhere to this order. Also, be careful to count each triplet exactly once. Often, breaking the problem into stages helps:
- First, fix the third element k.
- Then count the number of pairs (i, j) with i < j < k that complete the triplet.
7. **Edge cases and validation:**
Consider arrays with repeated elements and very large values, ensuring modular arithmetic doesn’t overflow and indexing constraints are respected. Also, verify the case where d = 1, meaning any sum is divisible, to confirm your logic handles trivial divisibility.
8. **Performance considerations:**
Aim to reduce the complexity ideally to O(n^2) by fixing one element and checking pairs in a controlled manner, since O(n^3) is usually too slow for large inputs. If possible, explore optimization strategies like prefix computations or frequency maps of partial sums modulo d.
9. **Summary of the approach:**
- Transform the array elements to their remainders modulo d.
- For each element chosen as the third element k, count how many pairs (i, j) with indices less than k have remainders summing to the value needed for the total sum to be divisible by d.
- Use a data structure to store counts of remainders of elements before k to enable efficient queries.
- Sum these counts for all valid k.
By following these directions, you focus on the modular arithmetic properties, reduce unnecessary computations, and leverage data structures for efficient counting, leading to a scalable solution to the problem of counting divisible triplets.
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