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8. Complementary Pairs A pair of strings form a complementary pair if there is some permutation of their concatenation that is a palindrome. For example, the strings "abac" and "cab" form a complementary pair since their concatenation is "abaccab" which can be rearranged to form a palindrome, i.e., "bcaaacb". Given an array ofnstrings, find the number of complementary pairs that can be formed. Note: Pairs of strings formed by indices(i,j)and(j,i)are considered the same. Example Consider stringData=["abc," "abcd", "bc", "adc"]. The following complementary pairs can be formed: - ("abc", "abcd"), concatenated string = "abcabcd" arranged as a palindrome - "abcdcba". - ("abc". "bc"), concatenated string = "abcbc" -> "bcacb". ("abcd", "adc"), concatenated string = "abcdadc" -> "acdbdca". Return 3 , the number of complementary pairs. Function Description Complete the function countComplementaryPairs in the editor below. countComplementaryPairs has the following. parameter: string stringData[n]: the strings to pair Returns fong. int the number of complementary pairs that can be formed Constraints -1≤n≤10 5 -1≤length(stringData[i])≤3∗10 5 -1≤5umof the length of strings in stringData≤3∗10 5 - All strings consist of lowercase English characters only. Sample Output o 3 Explanation The following complementary pairs can be formed: "eiliblitar" - Tball", bally concatenated string = "balibar" -> "balliab" - Fall", "caliri concatenated string * "allealf ws "alichia".

User Npskirk
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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

User Meymann
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