Count Prefix and Suffix Pairs I

Count Prefix and Suffix Pairs I - Problem

You are given a 0-indexed string array words.

Let's define a boolean function isPrefixAndSuffix that takes two strings, str1 and str2:

isPrefixAndSuffix(str1, str2) returns true if str1 is both a prefix and a suffix of str2, and false otherwise.

For example, isPrefixAndSuffix("aba", "ababa") is true because "aba" is a prefix of "ababa" and also a suffix, but isPrefixAndSuffix("abc", "abcd") is false.

Return an integer denoting the number of index pairs (i, j) such that i < j, and isPrefixAndSuffix(words[i], words[j]) is true.

Input & Output

Example 1 — Basic Case

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

› Output: 4

💡 Note: Valid pairs: (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", (1,2) "aba" is prefix and suffix of "ababa". Total: 4 pairs.

Example 2 — No Valid Pairs

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

› Output: 2

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

Example 3 — Single Characters

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

› Output: 0

💡 Note: No valid pairs since "abab" is longer than "ab", and "ab" is not both prefix and suffix of "abab".

Constraints

1 ≤ words.length ≤ 50
1 ≤ words[i].length ≤ 10
words[i] consists only of lowercase English letters

Visualization

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Asked in

G Google 15 M Microsoft 12

The key insight is that a string can only be both prefix and suffix of another string if it's shorter than the target string. The brute force approach checks all pairs (i,j) where i < j and verifies both prefix and suffix conditions. An optimized version adds early length checking to skip impossible cases. Time: O(n² × m), Space: O(1).

Common Approaches

✓ Brute Force

⏱️ Time: O(n² × m) Space: O(1)

For each pair of indices (i,j) where i < j, check if words[i] is both a prefix and suffix of words[j]. Use built-in string methods to verify prefix and suffix conditions.

Length-Based Optimization

⏱️ Time: O(n² × m) Space: O(1)

Before checking prefix and suffix conditions, first verify that str1 length is less than or equal to str2 length. This eliminates impossible cases early and reduces string operations.

Brute Force — Algorithm Steps

Iterate through all pairs (i,j) where i < j
For each pair, check if words[i] is prefix of words[j]
Check if words[i] is suffix of words[j]
Count pairs where both conditions are true

Visualization

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Step-by-Step Walkthrough

Compare Pairs

Check all combinations where i < j

Prefix & Suffix Check

Verify if str1 is both prefix and suffix of str2

Count Valid Pairs

Increment counter for matching 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 (strcmp(str2 + len2 - len1, str1) != 0) return 0;
    
    return 1;
}

int solution(char words[][101], 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() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    char words[50][101];
    int n = 0;
    
    // Simple JSON parsing for string array
    char* ptr = strchr(line, '[');
    ptr++;
    char* token = strtok(ptr, ",]");
    while (token != NULL && n < 50) {
        // Remove quotes and whitespace
        while (*token == ' ' || *token == '"') token++;
        int len = strlen(token);
        while (len > 0 && (token[len-1] == ' ' || token[len-1] == '"' || token[len-1] == '\n')) {
            token[len-1] = '\0';
            len--;
        }
        strcpy(words[n], token);
        n++;
        token = strtok(NULL, ",]");
    }
    
    int result = solution(words, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n² × m)

Nested loops create O(n²) pairs, each string comparison takes O(m) where m is average string length

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

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Code -

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