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You are given a list of strings. You need to find the maximum number of pairs (i, j) such that the concatenation of the two strings words[i] + words[j] is equal to the concatenation of the two strings in reverse order words[j] + words[i].

In other words, find the maximum number of pairs (i, j) such that words[i] + words[j] == words[j] + words[i].

Here is the function signature you need to implement:

def maximumNumberOfStringPairs(words: List[str]) -> int:
    pass

Clarifying Questions

1. Are there any constraints on the length of each string in the list?
   • Typically, the problem statement will provide constraints, but if not, assume the strings are of reasonable length (e.g., up to 100 characters).
2. Can the same string be used to form multiple pairs?
   • Based on the typical problem constraints, a string should only be used once to form a pair with another string.
3. What is the expected range for the list size?
   • This usually would be given in the problem constraints, but for reasonable algorithm complexity, assume the list can have up to 1000 elements.
4. Are empty strings possible in the list?
   • This depends on constraints but usually consider handling such edge cases unless specified otherwise.

Strategy

1. Explanation:
   • To find pairs (i, j) such that words[i] + words[j] == words[j] + words[i], the core idea is to identify pairs of words that when concatenated form the same result in both possible orders.
2. Approach:
   • If words[i] + words[j] == words[j] + words[i], it implies that words[i] == words[j] because the lengths of the strings ensure the concatenation is symmetrical only if they are equal.
   • Use a hashmap (dictionary) to count the occurrences of each word.
   • The maximum number of pairs is determined by counting how many times we can form pairs using the available words.
3. Steps:
   • Initialize a counter to keep track of word occurrences.
   • Iterate through the list of words and count each occurrence.
   • Iterate through the counter and compute how many pairs can be formed from the counts.
4. Constraints:
   • Ensure that the function handles edge cases like empty input list, single element list, and no valid pairs.

Code

from typing import List
from collections import Counter

def maximumNumberOfStringPairs(words: List[str]) -> int:
    if not words:  # edge case: empty list
        return 0
    
    count = Counter(words)
    pairs = 0
    
    for word in count:
        pairs += count[word] // 2
    
    return pairs

Time Complexity

• Time Complexity: O(n), where n is the number of words in the list. This complexity arises from the need to iterate through the list to generate the counts and then iterate through the counts to determine the pairs.
• Space Complexity: O(n), which accounts for the space used by the Counter to store the occurrences of each word.

By following this code and approach, you should be able to solve the problem of finding the maximum number of string pairs effectively.

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