Count Prefix and Suffix Pairs I - Practice Coding Problems

Count Prefix and Suffix Pairs I - Problem

Given a 0-indexed string array words, you need to find how many pairs of strings have a special relationship.

A string str1 is considered both a prefix and suffix of another string str2 if:

str1 appears at the beginning of str2 (prefix)
str1 appears at the end of str2 (suffix)

Example: "aba" is both a prefix and suffix of "ababa" because:

Prefix: "ababa" starts with "aba"
Suffix: "ababa" ends with "aba"

Your task is to count all valid pairs (i, j) where i < j and words[i] is both a prefix and suffix of words[j].

Input & Output

example_1.py — Basic case with multiple matches

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

› Output: 4

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

example_2.py — No valid pairs

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

› Output: 2

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

example_3.py — Single character strings

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

› Output: 0

💡 Note: No valid pairs because "abab" cannot be both prefix and suffix of "ab" (longer string cannot be prefix/suffix of shorter string)

Visualization

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Understanding the Visualization

Identify Candidates

Look for shorter strings that could potentially frame longer ones

Check Prefix

Verify if the shorter string appears at the beginning of the longer string

Check Suffix

Verify if the same shorter string appears at the end of the longer string

Count Matches

Increment counter for each valid prefix-suffix pair found

Key Takeaway

🎯 Key Insight: For a string to be both prefix and suffix, the target string must be at least as long and have matching characters at both ends

Time & Space Complexity

Time Complexity

⏱️

O(n² × m)

We check all n² pairs, and each comparison takes O(m) time where m is the average string length

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables for counting, no additional data structures needed

✓ Linear Space

Constraints

1 ≤ words.length ≤ 50
1 ≤ words[i].length ≤ 10
words[i] consists only of lowercase English letters
All strings are non-empty

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

The optimal approach uses a brute force method with nested loops to check all pairs (i,j) where i < j. For each pair, we verify if words[i] is both a prefix and suffix of words[j] using built-in string methods. The time complexity is O(n² × m) where n is the number of words and m is the average string length.

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

Brute Force (Nested Loops) — Algorithm Steps

Iterate through all pairs (i, j) where i < j
For each pair, check if words[i] is a prefix of words[j]
If it's a prefix, check if words[i] is also a suffix of words[j]
If both conditions are true, increment the counter
Return the total count

Visualization

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Step-by-Step Walkthrough

Initialize

Start with the first string and compare with all subsequent strings

Check Pair

For each pair (i,j), verify if words[i] is both prefix and suffix of words[j]

Count Valid

Increment counter when a valid prefix-suffix relationship is found

Code -

solution.c — C

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

int isPrefixAndSuffix(char* str1, 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 n) {
    int count = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (isPrefixAndSuffix(words[i], words[j])) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    int n;
    scanf("%d", &n);
    
    char** words = (char**)malloc(n * sizeof(char*));
    for (int i = 0; i < n; i++) {
        words[i] = (char*)malloc(100 * sizeof(char));
        scanf("%s", words[i]);
    }
    
    printf("%d\n", countPrefixSuffixPairs(words, n));
    
    for (int i = 0; i < n; i++) {
        free(words[i]);
    }
    free(words);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n² × m)

We check all n² pairs, and each comparison takes O(m) time where m is the average string length

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables for counting, no additional data structures needed

✓ Linear Space

Constraints

1 ≤ words.length ≤ 50
1 ≤ words[i].length ≤ 10
words[i] consists only of lowercase English letters
All strings are non-empty

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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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