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Number of Pairs of Strings With Concatenation Equal to Target

Medium

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ArraysStringsTwo Pointers

Given an array of digit strings nums and a digit string target, return the number of pairs of indices (i, j) (where i != j) such that the concatenation of nums[i] + nums[j] equals target.

Example 1:

Input: nums = ["777","7","77","77"], target = "7777"
Output: 4
Explanation: Valid pairs are:
- (0, 1): "777" + "7"
- (1, 0): "7" + "777"
- (2, 3): "77" + "77"
- (3, 2): "77" + "77"

Example 2:

Input: nums = ["123","4","12","34"], target = "1234"
Output: 2
Explanation: Valid pairs are:
- (0, 1): "123" + "4"
- (2, 3): "12" + "34"

Example 3:

Input: nums = ["1","1","1"], target = "11"
Output: 6
Explanation: Valid pairs are:
- (0, 1): "1" + "1"
- (1, 0): "1" + "1"
- (0, 2): "1" + "1"
- (2, 0): "1" + "1"
- (1, 2): "1" + "1"
- (2, 1): "1" + "1"

Constraints:

2 <= nums.length <= 100
1 <= nums[i].length <= 100
2 <= target.length <= 100
nums[i] and target consist of digits.
nums[i] and target do not have leading zeros.

Try coding it in LeetCode

Number of Pairs of Strings With Concatenation Equal to Target

Medium

Asked by:

7 views

Topics:

ArraysStringsTwo Pointers

Example 1:

Input: nums = ["777","7","77","77"], target = "7777"
Output: 4
Explanation: Valid pairs are:
- (0, 1): "777" + "7"
- (1, 0): "7" + "777"
- (2, 3): "77" + "77"
- (3, 2): "77" + "77"

Example 2:

Input: nums = ["123","4","12","34"], target = "1234"
Output: 2
Explanation: Valid pairs are:
- (0, 1): "123" + "4"
- (2, 3): "12" + "34"

Example 3:

Input: nums = ["1","1","1"], target = "11"
Output: 6
Explanation: Valid pairs are:
- (0, 1): "1" + "1"
- (1, 0): "1" + "1"
- (0, 2): "1" + "1"
- (2, 0): "1" + "1"
- (1, 2): "1" + "1"
- (2, 1): "1" + "1"

Constraints:

2 <= nums.length <= 100
1 <= nums[i].length <= 100
2 <= target.length <= 100
nums[i] and target consist of digits.
nums[i] and target do not have leading zeros.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the length of the `nums` array and the length of each string within it?
Can the input strings contain leading or trailing spaces, or any special characters?
Is the target string guaranteed to be non-empty, and what are the possible lengths of the target string?
If multiple pairs concatenate to the target, should I return all such pairs or just the count of pairs?
Are the strings in the input array guaranteed to be unique, or can there be duplicates that concatenate to form the target?

Brute Force Solution

Approach

The brute force approach to this problem means checking every possible pairing of words to see if they form the target word. It's like trying every combination until you find the ones that work. Essentially we will go through all words and see if any two, when stuck together, equal the target.

Here's how the algorithm would work step-by-step:

Pick the very first word in the list.
Now, go through every other word in the list and stick it to the end of the first word.
See if the combined word equals the target word. If it does, count it as a match.
Repeat this process of sticking the first word to the beginning of every other word.
Once you've checked all combinations with the first word, move to the second word and repeat the same process, checking it with every other word in the list.
Continue this until you have checked every word in the list against every other word.
The final count of matches represents the total number of pairs that, when stuck together, equal the target word.

Code Implementation

def number_of_pairs_of_strings_with_concatenation_equal_to_target(words, target):
    number_of_matching_pairs = 0
    number_of_words = len(words)

    for first_word_index in range(number_of_words):
        for second_word_index in range(number_of_words):
            # Need to avoid comparing the same word to itself
            if first_word_index == second_word_index:
                continue

            # Check if the concatenation order is correct
            concatenated_string = words[first_word_index] + words[second_word_index]

            # Check to see if this concatenation equals our target
            if concatenated_string == target:
                number_of_matching_pairs += 1

    return number_of_matching_pairs

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n words in the input array. For each word, it concatenates it with every other word in the array, which requires another n iterations. The string concatenation and comparison operations themselves take constant time. Therefore, we have a nested loop structure resulting in approximately n * n operations. This simplifies to a time complexity of O(n²).

Space Complexity

O(1) – The provided brute force approach iterates through the input array, comparing concatenated strings. It doesn't use any auxiliary data structures like lists, hash maps, or sets to store intermediate results. Only a few constant space variables are required for loop counters and storing the concatenation, whose length is bounded by the length of the target string. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The most efficient way to solve this problem involves strategically counting the number of times parts of the target string appear in the given list of strings. Instead of checking every possible pair, we can use a trick to quickly find the right combinations. We can leverage a technique similar to counting how many times each string appears.

Here's how the algorithm would work step-by-step:

First, count how many times each string appears in the given list.
Then, go through each string in the list and check if it could be the start of our target string. If it is, note the remaining part of the target we need to find.
For each remaining part, check if it exists in our counts of strings. If it does, that means we have found a possible pair that makes the target string.
If the start and end string are the same, be careful not to double-count the number of times it contributes to forming the target. Adjust count accordingly.
Add all the valid pairs together to get the total number of pairs that can form the target string.

Code Implementation

def number_of_pairs(string_list, target_string):
    string_counts = {}
    for string in string_list:
        string_counts[string] = string_counts.get(string, 0) + 1

    number_of_valid_pairs = 0
    for first_string in string_list:
        # Check if the current string can be a prefix of the target
        if target_string.startswith(first_string):

            remaining_part = target_string[len(first_string):]

            # Find the needed remaining part in the counts dictionary
            if remaining_part in string_counts:

                count_of_remaining_part = string_counts[remaining_part]

                # Adjust count if strings are the same to avoid overcounting
                if first_string == remaining_part:

                    count_of_remaining_part -= 1
                    if count_of_remaining_part >= 0:
                        number_of_valid_pairs += count_of_remaining_part

                else:

                    number_of_valid_pairs += count_of_remaining_part

    return number_of_valid_pairs

Big(O) Analysis

Time Complexity

O(n*m) – Let n be the number of strings in the input array and m be the length of the target string. First, we iterate through the array of n strings to count their occurrences which takes O(n) time. Then we iterate through the n strings again. Inside this loop, we check if a string is a prefix of the target string (length m operation). If it is, we compute the remaining suffix of the target (length m operation) and look up this suffix in our count map (O(1) amortized). In the worst case, the prefix check will take O(m) time, the suffix extraction takes O(m) time, and this is done for each of the n strings. Therefore, the overall time complexity will be O(n*m).

Space Complexity

O(N) – The algorithm's space complexity is primarily determined by the frequency count of strings from the input list, which is stored in a hash map. This hash map's size can grow linearly with the number of unique strings in the input, at most N, where N represents the number of strings in the input list. The counts of strings will thus take up O(N) space. Therefore, the overall auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty 'nums' array	Return 0 immediately since no pairs can be formed.
Null or empty 'target' string	Return 0 immediately as concatenation with an empty string will rarely result in a match, simplifying logic.
'nums' array contains empty strings	Consider empty strings in 'nums' and their contribution to 'target' when concatenated.
Target string is very long	The solution must be efficient in comparing potentially very large strings using appropriate data structures like hashmaps to avoid excessive iteration.
All strings in 'nums' are identical and form the target when concatenated with themselves or each other.	Handles this case by correctly counting all valid pairs, ensuring distinct pairs are not double-counted.
No pairs in 'nums' concatenate to form the target string.	The algorithm should correctly return 0 in such cases.
Integer overflow potential when counting pairs	Use a data type (e.g., long or equivalent in the language) large enough to store the count of pairs without overflow.
Extremely large input array size	Use Hashmap approach, which has O(n) average time complexity, instead of nested loops which could be O(n^2).

Null or empty 'nums' array

How to Handle:

Return 0 immediately since no pairs can be formed.

Null or empty 'target' string

How to Handle:

Return 0 immediately as concatenation with an empty string will rarely result in a match, simplifying logic.

'nums' array contains empty strings

How to Handle:

Consider empty strings in 'nums' and their contribution to 'target' when concatenated.

Target string is very long

How to Handle:

The solution must be efficient in comparing potentially very large strings using appropriate data structures like hashmaps to avoid excessive iteration.

All strings in 'nums' are identical and form the target when concatenated with themselves or each other.

How to Handle:

Handles this case by correctly counting all valid pairs, ensuring distinct pairs are not double-counted.

No pairs in 'nums' concatenate to form the target string.

How to Handle:

The algorithm should correctly return 0 in such cases.

Integer overflow potential when counting pairs

How to Handle:

Use a data type (e.g., long or equivalent in the language) large enough to store the count of pairs without overflow.

Extremely large input array size

How to Handle:

Use Hashmap approach, which has O(n) average time complexity, instead of nested loops which could be O(n^2).

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the length of the `nums` array and the length of each string within it?
Can the input strings contain leading or trailing spaces, or any special characters?
Is the target string guaranteed to be non-empty, and what are the possible lengths of the target string?
If multiple pairs concatenate to the target, should I return all such pairs or just the count of pairs?
Are the strings in the input array guaranteed to be unique, or can there be duplicates that concatenate to form the target?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Pick the very first word in the list.
Now, go through every other word in the list and stick it to the end of the first word.
See if the combined word equals the target word. If it does, count it as a match.
Repeat this process of sticking the first word to the beginning of every other word.
Once you've checked all combinations with the first word, move to the second word and repeat the same process, checking it with every other word in the list.
Continue this until you have checked every word in the list against every other word.
The final count of matches represents the total number of pairs that, when stuck together, equal the target word.

Code Implementation

def number_of_pairs_of_strings_with_concatenation_equal_to_target(words, target):
    number_of_matching_pairs = 0
    number_of_words = len(words)

    for first_word_index in range(number_of_words):
        for second_word_index in range(number_of_words):
            # Need to avoid comparing the same word to itself
            if first_word_index == second_word_index:
                continue

            # Check if the concatenation order is correct
            concatenated_string = words[first_word_index] + words[second_word_index]

            # Check to see if this concatenation equals our target
            if concatenated_string == target:
                number_of_matching_pairs += 1

    return number_of_matching_pairs

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, count how many times each string appears in the given list.
Then, go through each string in the list and check if it could be the start of our target string. If it is, note the remaining part of the target we need to find.
For each remaining part, check if it exists in our counts of strings. If it does, that means we have found a possible pair that makes the target string.
If the start and end string are the same, be careful not to double-count the number of times it contributes to forming the target. Adjust count accordingly.
Add all the valid pairs together to get the total number of pairs that can form the target string.

Code Implementation

def number_of_pairs(string_list, target_string):
    string_counts = {}
    for string in string_list:
        string_counts[string] = string_counts.get(string, 0) + 1

    number_of_valid_pairs = 0
    for first_string in string_list:
        # Check if the current string can be a prefix of the target
        if target_string.startswith(first_string):

            remaining_part = target_string[len(first_string):]

            # Find the needed remaining part in the counts dictionary
            if remaining_part in string_counts:

                count_of_remaining_part = string_counts[remaining_part]

                # Adjust count if strings are the same to avoid overcounting
                if first_string == remaining_part:

                    count_of_remaining_part -= 1
                    if count_of_remaining_part >= 0:
                        number_of_valid_pairs += count_of_remaining_part

                else:

                    number_of_valid_pairs += count_of_remaining_part

    return number_of_valid_pairs

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty 'nums' array	Return 0 immediately since no pairs can be formed.
Null or empty 'target' string	Return 0 immediately as concatenation with an empty string will rarely result in a match, simplifying logic.
'nums' array contains empty strings	Consider empty strings in 'nums' and their contribution to 'target' when concatenated.
Target string is very long	The solution must be efficient in comparing potentially very large strings using appropriate data structures like hashmaps to avoid excessive iteration.
All strings in 'nums' are identical and form the target when concatenated with themselves or each other.	Handles this case by correctly counting all valid pairs, ensuring distinct pairs are not double-counted.
No pairs in 'nums' concatenate to form the target string.	The algorithm should correctly return 0 in such cases.
Integer overflow potential when counting pairs	Use a data type (e.g., long or equivalent in the language) large enough to store the count of pairs without overflow.
Extremely large input array size	Use Hashmap approach, which has O(n) average time complexity, instead of nested loops which could be O(n^2).

Null or empty 'nums' array

How to Handle:

Return 0 immediately since no pairs can be formed.

Null or empty 'target' string

How to Handle:

Return 0 immediately as concatenation with an empty string will rarely result in a match, simplifying logic.

'nums' array contains empty strings

How to Handle:

Consider empty strings in 'nums' and their contribution to 'target' when concatenated.

Target string is very long

How to Handle:

The solution must be efficient in comparing potentially very large strings using appropriate data structures like hashmaps to avoid excessive iteration.

All strings in 'nums' are identical and form the target when concatenated with themselves or each other.

How to Handle:

Handles this case by correctly counting all valid pairs, ensuring distinct pairs are not double-counted.

No pairs in 'nums' concatenate to form the target string.

How to Handle:

The algorithm should correctly return 0 in such cases.

Integer overflow potential when counting pairs

How to Handle:

Use a data type (e.g., long or equivalent in the language) large enough to store the count of pairs without overflow.

Extremely large input array size

How to Handle:

Use Hashmap approach, which has O(n) average time complexity, instead of nested loops which could be O(n^2).