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Let’s consider the problem statement as follows:

Given an array of integers, your task is to find the maximum number of unique operations (i.e., pairs of indices) such that the sum of the elements at those indices is the same for all chosen pairs. In other words, determine the maximum count of such valid operations where the sums are equal.

Clarifying Questions

To fully understand the problem, here are a few clarifying questions:

1. Input Constraints: Are there any constraints on the size of the array or the values within the array?
2. Output: Should the solution return the maximum count of such operations, or should it return the actual pairs or sums?
3. Edge Cases: How to handle cases where the array is empty or has very few elements (less than 2)?
4. Duplicates: Does the array have duplicate elements? Does this affect the pairing?

For this exercise, let’s assume:

• The input array can have up to (10^5) elements.
• The array can have both positive and negative integers.
• The output should be the maximum number of unique operations.
• Duplicates are allowed and should be considered in the pairs.

Strategy

Here’s a step-by-step strategy to solve the problem:

1. Dictionary for Pair Sums: Use a dictionary to store the count of each possible sum of pairs.
2. Iterate Through Pairs: Iterate through all possible pairs of indices and calculate their sums.
3. Update Dictionary: For each sum, update its count in the dictionary.
4. Find Maximum: After processing all pairs, find the sum with the maximum count of pairs.

Code

from collections import defaultdict

def max_operations_with_same_sum(arr):
    if len(arr) < 2:
        return 0
    
    sum_count = defaultdict(int)
    
    for i in range(len(arr)):
        for j in range(i + 1, len(arr)):
            pair_sum = arr[i] + arr[j]
            sum_count[pair_sum] += 1
    
    max_operations = max(sum_count.values(), default=0)
    
    return max_operations

# Test cases
print(max_operations_with_same_sum([1, 2, 3, 4]))  # Expected output: 2 (possible pairs: (1,3), (2,2))
print(max_operations_with_same_sum([-1, 1, -1, 1]))  # Expected output: 6 (pair sums: 0)
print(max_operations_with_same_sum([1]))  # Expected output: 0 (not enough elements to form pairs)

Time Complexity

• Pair Generation: Generating all pairs for an array of length ( n ) would take ( O(n^2) ) time.
• Dictionary Updates: Each sum computation and dictionary update takes ( O(1) ).

Thus, the overall time complexity is ( O(n^2) ), where ( n ) is the number of elements in the array.

Conclusion

This approach leverages nested loops to compute all possible pairs and uses a dictionary to count the frequency of each sum. The solution is efficient for moderate-size arrays but might need further optimization for very large arrays.

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