This problem requires quickly identifying all index pairs (i, j) where i does not equal j and nums[i] + nums[j] equals the target string. By using hash maps to store string frequencies and scanning the array efficiently, you can compute the total number of valid concatenations without redundant checks. The approach balances speed and simplicity for arrays of modest size.
Problem Statement
Given an array of digit strings nums and a digit string target, determine how many ordered pairs of indices (i, j) with i not equal to j exist such that concatenating nums[i] and nums[j] forms target.
Return the total number of such pairs. For example, given nums = ["777","7","77","77"] and target = "7774", you must count each valid combination where the first string plus the second string equals target exactly.
Examples
Example 1
Input: nums = ["777","7","77","77"], target = "7777"
Output: 4
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
Valid pairs are:
- (0, 1): "123" + "4"
- (2, 3): "12" + "34"
Example 3
Input: nums = ["1","1","1"], target = "11"
Output: 6
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.
Solution Approach
Use a Hash Map for Counting
Create a hash map storing the frequency of each string in nums. Then for each string, check if the target starts with it and whether the corresponding suffix exists in the map. Multiply the occurrences appropriately and exclude self-pairing if needed.
Scan Array with Prefix-Suffix Check
Iterate over nums, treating each element as a potential prefix. For each prefix, calculate the required suffix by removing the prefix from target. Lookup this suffix in the hash map to count valid pairs efficiently without nested loops for all pairs.
Handle Duplicate Strings Carefully
Since nums can contain duplicates, ensure that when counting pairs where the prefix equals the suffix, you subtract the current index to avoid counting a string with itself. Accumulate counts across all elements to get the final answer.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity is O(n * k) where n is the number of strings and k is the average string length, dominated by substring and hash lookups. Space complexity is O(n) for the hash map storing string counts.
What Interviewers Usually Probe
- Looking for use of hash map to reduce naive O(n^2) pair checks.
- Interested in handling duplicate strings correctly without overcounting.
- Expect candidates to separate prefix and suffix logic clearly.
Common Pitfalls or Variants
Common pitfalls
- Forgetting to exclude self-pairs when prefix equals suffix.
- Miscounting duplicates by not using frequency map properly.
- Using nested loops over all pairs which can be slow for larger arrays.
Follow-up variants
- Count pairs for a target concatenation with at most one character difference.
- Find pairs of strings whose concatenation is a palindrome.
- Return the actual list of index pairs instead of just the count.
How GhostInterview Helps
- Automatically detects the prefix-suffix pattern in string arrays to suggest hash-based counting.
- Highlights potential overcounting mistakes when duplicates exist in nums.
- Provides step-by-step guidance for implementing efficient array scanning plus hash lookup.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the main approach for Number of Pairs of Strings With Concatenation Equal to Target?
Use a hash map to store string frequencies and scan the array checking each string as a prefix, then count matching suffixes.
Can we use nested loops for this problem?
Yes, but nested loops yield O(n^2) complexity and can be inefficient; hash-based scanning is faster and safer.
How do duplicates in nums affect counting?
Duplicates require careful handling to avoid counting the same index with itself and to multiply counts correctly using the frequency map.
Does the order of concatenation matter?
Yes, (i, j) and (j, i) are considered different pairs if nums[i] + nums[j] and nums[j] + nums[i] both equal the target.
What is the pattern this problem emphasizes?
It emphasizes the array scanning plus hash lookup pattern for counting valid string concatenations efficiently.
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