Count Prefix and Suffix Pairs II - Practice Coding Problems

Count Prefix and Suffix Pairs II - Problem

Count Prefix and Suffix Pairs II

You're given an array of strings called words. Your task is to find all pairs of strings where one string is both a prefix and a suffix of another string that comes later in the array.

More specifically, you need to implement a function isPrefixAndSuffix(str1, str2) that returns true if:
• str1 is a prefix of str2 (appears at the beginning)
• str1 is a suffix of str2 (appears at the end)

Examples:
• isPrefixAndSuffix("aba", "ababa") → true because "aba" appears at both the start and end of "ababa"
• isPrefixAndSuffix("abc", "abcd") → false because "abc" is only a prefix, not a suffix

Return the total count of valid index pairs (i, j) where i < j and isPrefixAndSuffix(words[i], words[j]) is true.

Input & Output

example_1.py — Basic Case

$ Input: ["a","aba","ababa","aa"]

› Output: 4

💡 Note: Valid pairs are: (0,1) "a" is prefix/suffix of "aba", (0,2) "a" is prefix/suffix of "ababa", (0,3) "a" is prefix/suffix of "aa", and (1,2) "aba" is prefix/suffix of "ababa"

example_2.py — No Matches

$ Input: ["pa","papa","ma","mama"]

› Output: 2

💡 Note: Valid pairs are: (0,1) "pa" is prefix/suffix of "papa", and (2,3) "ma" is prefix/suffix of "mama"

example_3.py — Single Characters

$ Input: ["abab","ab"]

› Output: 0

💡 Note: "ab" is a prefix of "abab" but not a suffix, so no valid pairs exist

Visualization

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Understanding the Visualization

Catalog Programs

Create an index of all play names and their structure

Pattern Matching

For each program, check if any previous play names bookend it

Efficient Lookup

Use the Trie structure to quickly find matching patterns

Count Relationships

Tally all the valid prefix-suffix relationships found

Key Takeaway

🎯 Key Insight: A string can only be both prefix and suffix if it appears at the exact beginning and end of another string. The Trie structure lets us efficiently organize and search these patterns without redundant comparisons.

Time & Space Complexity

Time Complexity

⏱️

O(n²×m)

n² for all pairs, m for string comparison operations

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

Constraints

1 ≤ words.length ≤ 10⁵
1 ≤ words[i].length ≤ 10⁵
Sum of lengths of all words ≤ 5 × 10⁵
words[i] consists only of lowercase English letters

Asked in

G Google 35 f Meta 28 a Amazon 22 ⊞ Microsoft 18

The optimal approach uses a Trie data structure to efficiently find strings that are both prefix and suffix of other strings. While the brute force solution checks every pair in O(n²×m) time, the Trie-based solution reduces this to O(n×m²) by avoiding redundant string comparisons and leveraging the prefix tree structure for fast lookups.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n²×m)	O(1)	Check every pair of strings to see if one is both prefix and suffix of the other
Trie-Based Solution (Optimal)	O(n×m²)	O(n×m)	Use a Trie data structure to efficiently find prefix-suffix matches

Brute Force (Nested Loops) — Algorithm Steps

Iterate through all pairs (i,j) where i < j
For each pair, check if words[i].length <= words[j].length
Check if words[j] starts with words[i] (prefix check)
Check if words[j] ends with words[i] (suffix check)
If both conditions are true, increment the count

Visualization

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Step-by-Step Walkthrough

Initialize

Start with first string as potential prefix-suffix

Compare

Check against each subsequent string

Validate

Test both prefix and suffix conditions

Count

Increment counter for valid pairs

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int isPrefixAndSuffix(const char* str1, const char* str2) {
    int len1 = strlen(str1);
    int len2 = strlen(str2);
    
    if (len1 > len2) return 0;
    
    // Check prefix
    if (strncmp(str2, str1, len1) != 0) return 0;
    
    // Check suffix
    if (strncmp(str2 + len2 - len1, str1, len1) != 0) return 0;
    
    return 1;
}

int countPrefixSuffixPairs(char** words, int wordsSize) {
    int count = 0;
    
    for (int i = 0; i < wordsSize; i++) {
        for (int j = i + 1; j < wordsSize; j++) {
            if (isPrefixAndSuffix(words[i], words[j])) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char* words[] = {"a", "aba", "ababa", "aa"};
    printf("%d\n", countPrefixSuffixPairs(words, 4));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²×m)

n² for all pairs, m for string comparison operations

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

Constraints

1 ≤ words.length ≤ 10⁵
1 ≤ words[i].length ≤ 10⁵
Sum of lengths of all words ≤ 5 × 10⁵
words[i] consists only of lowercase English letters

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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