return count
```
In this case, the number of complementary pairs that can be formed is 3. The pairs are:
1. ("abc", "abcd") -> "abcabcd" can be rearranged to form a palindrome: "abcdcba"
2. ("abc", "bc") -> "abcbc" can be rearranged to form a palindrome: "bcacb"
3. ("abcd", "adc") -> "abcdadc" can be rearranged to form a palindrome: "acdbdca"
Therefore, the answer is 3.
The problem asks us to find the number of complementary pairs that can be formed from an array of strings. A complementary pair is defined as a pair of strings whose concatenation can be rearranged to form a palindrome.
To solve this problem, we can follow these steps:
1. Initialize a variable, let's call it `count`, to keep track of the number of complementary pairs found.
2. Iterate through each pair of strings in the array. Since pairs formed by indices `(i, j)` and `(j, i)` are considered the same, we can start the inner loop from `i + 1` instead of `0`.
3. Concatenate the two strings in the pair.
4. Check if the concatenated string can be rearranged to form a palindrome. We can do this by comparing the frequency of each character in the string. If all characters have an even frequency, or at most one character has an odd frequency, then the string can be rearranged to form a palindrome.
5. If the concatenated string can be rearranged to form a palindrome, increment the `count` variable by 1.
6. After the loops are finished, return the value of `count` as the number of complementary pairs found.
Let's apply this algorithm to the given example:
```python
stringData = ["abc", "abcd", "bc", "adc"]
count = 0
for i in range(len(stringData)):
for j in range(i + 1, len(stringData)):
concatenated = stringData[i] + stringData[j]
frequency = {}
for char in concatenated:
if char in frequency:
frequency[char] += 1
else:
frequency[char] = 1
odd_count = 0
for value in frequency.values():
if value % 2 != 0:
odd_count += 1
if odd_count <= 1:
count += 1