What is the time complexity of Number of Pairs of Strings With Concatenation Equal to Target?

The optimal solution for Number of Pairs of Strings With Concatenation Equal to Target runs in O(n) time and uses O(n) space. This complexity is achieved because we only iterate through the list a couple of times, and the hash map operations (insert and lookup) are average O(1).

What is the brute force solution for Number of Pairs of Strings With Concatenation Equal to Target?

The brute force approach for Number of Pairs of Strings With Concatenation Equal to Target is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible pairs of strings in the array. For each pair, it concatenates the two strings and checks if the result matches the target string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Pairs of Strings With Concatenation Equal to Target?

Number of Pairs of Strings With Concatenation Equal to Target uses the following concepts: Array, Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Pairs of Strings With Concatenation Equal to Target?

Always clarify the problem statement and constraints. Think about edge cases, such as empty strings or maximum lengths. Discuss your thought process aloud to engage with the interviewer.

What are common mistakes when solving Number of Pairs of Strings With Concatenation Equal to Target?

Forgetting to check i != j when counting pairs. Not considering the case where the complement is the same as the current string.

#2023

Number of Pairs of Strings With Concatenation Equal to Target

Medium

ArrayHash TableStringCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking all pairs, we can use a hash map to store the frequency of each string. For each string, we can check if the required complement exists in the map to form the target.

⚙️

Algorithm

5 steps

1Step 1: Create a hash map to store the frequency of each string in nums.
2Step 2: Initialize a counter to zero for valid pairs.
3Step 3: Iterate through each string in nums, for each string, calculate its complement needed to form the target.
4Step 4: Check if the complement exists in the map. If it does, add its frequency to the counter.
5Step 5: Adjust the count to avoid counting pairs where i == j.

solution.py11 lines

1def countPairs(nums, target):
2    from collections import Counter
3    count = 0
4    freq = Counter(nums)
5    for num in nums:
6        complement = target[len(num):]
7        if complement in freq:
8            count += freq[complement]
9            if complement == num:
10                count -= 1
11    return count

ℹ

Complexity note: This complexity is achieved because we only iterate through the list a couple of times, and the hash map operations (insert and lookup) are average O(1).

1Using a hash map reduces the need for nested loops.
2Understanding string concatenation helps in identifying complements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.