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Problem Description

Given an array of integers arr and three integers a, b, and c, find the number of good triplets.

A triplet (arr[i], arr[j], arr[k]) is good if it satisfies the following conditions:

Clarifying Questions

  1. Is the input size constrained?
    • Yes, we should assume the constraints based on typical coding challenges.
  2. Can there be any negative numbers in the array?
    • Yes, in general integer arrays can have negative numbers.
  3. What is the range of a, b, and c?
    • They are typical integer values as part of the problem.

Strategy

We need to iterate over all possible triplets (i, j, k) in the array and check if they satisfy the given conditions. A brute force approach involves nested loops to examine all triplets.

Steps:

  1. Use three nested loops to iterate over all triplets.
  2. For each triplet, check if it satisfies the conditions.
  3. Count the triplets that satisfy all conditions.

Time Complexity

The brute force method works in O(n^3) time complexity because we are using three nested loops to iterate over the array elements. Given the constraints are typically low (for example, arr.length <= 100), this complexity is acceptable for this problem.

Code

Here’s the implementation of the described strategy:

def count_good_triplets(arr, a, b, c):
    count = 0
    n = len(arr)
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if abs(arr[i] - arr[j]) <= a and abs(arr[j] - arr[k]) <= b and abs(arr[i] - arr[k]) <= c:
                    count += 1
    return count

# Example usage
arr = [3,0,1,1,9,7]
a = 7
b = 2
c = 3
print(count_good_triplets(arr, a, b, c))  # Output: 4

Conclusion

This solution iterates through each triplet and checks the conditions directly, guaranteeing that all potential triplets are examined. While the time complexity is O(n^3), it is adequate for the typical constraints of such problems. If larger constraints exist, optimization techniques or different algorithms would be required.

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