algoadvance

Given a 0-indexed integer array nums of length n and an integer k, return the number of pairs (i, j) where 0 <= i < j < n, such that nums[i] == nums[j] and (i * j) is divisible by k.

Clarifying Questions:

1. Input Constraints:
   • What is the range of values for n?
     • Typically, n could be any size within [1, 1000] based on similar problems, but the exact constraints should be checked.
   • What is the range of integers in the nums array?
     • Usually, it could be any integers within [1, 10^6], but again, verify from problem constraints.
   • Could there be negative integers in the nums array?
     • This typically shouldn’t happen based on standard constraints, unless specified.
2. Output:
   • The output should be a single integer, which is the count of valid pairs (i, j).
3. Edge Cases:
   • Small arrays, where n = 1 or 2.
   • Values of k that are larger than the length of the array.
   • Arrays with distinct elements where no pair is equal.

Strategy:

1. Brute Force:
   • Iterate through every possible pair (i, j) with 0 <= i < j < n.
   • Check if nums[i] == nums[j].
   • Check if (i * j) % k == 0.
   • Increment the count if both conditions are met.
2. Optimization Insight:
   • A brute force approach might be acceptable given the constraints, especially if n is relatively small (under a few thousand).
   • However, as an extra step, we might consider pre-checking for pairs based on nums values reducing unnecessary multiplicative checks.
3. Implementation:
   • Iterate through all pairs and apply the stated conditions.
   • Maintain a count of valid pairs.

Code:

def countPairs(nums, k):
    n = len(nums)
    count = 0

    # Iterate through every pair (i, j) with 0 <= i < j < n
    for i in range(n):
        for j in range(i + 1, n):
            if nums[i] == nums[j] and (i * j) % k == 0:
                count += 1

    return count

# Example usage
nums = [3, 1, 2, 2, 2, 1, 3]
k = 2
print(countPairs(nums, k))  # Output should be based on problem description

Time Complexity:

• Brute Force Approach:
  • The time complexity of the above solution is (O(n^2)) due to the nested loops iterating through all pairs.
  • The space complexity is (O(1)), as we are only using a few additional variables.

Given the constraints typically provided in such problems, (O(n^2)) should be efficient enough for (n \leq 1000). However, if (n) is significantly larger, alternative strategies might be required.

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