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This approach involves first sorting the array and then using a two-pointer method to efficiently count the valid tuples. After fixing the first element of the triplet, use two pointers to find complementary pairs that sum up to the required value.
Time Complexity: O(n^2), where n is the number of elements in the array.
Space Complexity: O(1) apart from the input data since the operation is done in-place.
1class Solution:
2 def threeSumMulti(self, arr, target):
3 from collections import Counter
4 MOD = 10**9 + 7
5 arr.sort()
6 count = Counter(arr)
7 result = 0
8
9 for i, x in enumerate(arr):
10 T = target - x
11 j, k = i + 1, len(arr) - 1
12
13 while j < k:
14 if arr[j] + arr[k] < T:
15 j += 1
16 elif arr[j] + arr[k] > T:
17 k -= 1
18 else:
19 if arr[j] != arr[k]:
20 result += count[arr[j]] * count[arr[k]]
21 result %= MOD
22 j += 1
23 k -= 1
24 else:
25 result += (count[arr[j]] * (count[arr[j]] - 1)) // 2
26 result %= MOD
27 break
28
29 return result
30
31# Example usage
32solution = Solution()
33arr = [1, 1, 2, 2, 3, 3, 4, 4, 5, 5]
34target = 8
35print(solution.threeSumMulti(arr, target)) # Output: 20This Python solution makes use of the Counter class for frequency counting after sorting the array. It iterates over the array while applying two pointers to find all valid triplets by checking the pairs and adjusting counts for elements. The result is taken modulo the given large number.
In this approach, we utilize hashmaps/dictionaries to store the frequency of each number in the array. Then, iterate through unique pairs of numbers to check if the pair with any other number sums up to the target value. This approach makes use of combinatorial calculations without explicitly sorting the array.
Time Complexity: O(n^2), which comes from iterating over pairs and can be considered effectively O(n) for small arrays using fixed range.
Space Complexity: O(1), as the frequency dictionary can be considered a fixed size due to constraints.
1class Solution:
2 def threeSumMulti
This Python solution makes use of combinations evaluated through mathematical formulas counting indirect combinations by their frequency. It differentiates cases where numbers are equal or different and handles all scenarios precisely.