Unique Substrings in Wraparound String

Unique Substrings in Wraparound String - Problem

We define the string base to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so base will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd...."

Given a string s, return the number of unique non-empty substrings of s that are present in base.

A substring is a contiguous sequence of characters within a string. For example, "abc" is a substring of "abcde" but "aec" is not.

Input & Output

Example 1 — Basic Consecutive Sequence

$ Input: s = "abc"

› Output: 6

💡 Note: Substrings are: "a", "b", "c", "ab", "bc", "abc". All are consecutive in wraparound string, so answer is 6.

Example 2 — Wraparound Case

$ Input: s = "zabc"

› Output: 10

💡 Note: Includes wraparound: "z" → "a". Valid substrings: "z", "a", "b", "c", "za", "ab", "bc", "zab", "abc", "zabc". Total: 10.

Example 3 — Duplicate Characters

$ Input: s = "cac"

› Output: 2

💡 Note: Valid substrings are "c" and "a". Even though "c" appears twice, we only count unique substrings.

Constraints

1 ≤ s.length ≤ 10⁵
s consists of lowercase English letters.

Visualization

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Asked in

G Google 25 f Facebook 20 a Amazon 15

The key insight is to use dynamic programming to track the maximum length of valid wraparound sequence ending at each letter. Best approach uses O(n) time by maintaining current sequence length and updating max lengths. Time: O(n), Space: O(1).

Common Approaches

✓ Brute Force - Check All Substrings

⏱️ Time: O(n³) Space: O(n²)

Generate every possible substring of s and check if it forms a valid consecutive sequence in the wraparound alphabet. Use a set to avoid counting duplicates.

Dynamic Programming - Track Max Length per Character

⏱️ Time: O(n) Space: O(1)

Use dynamic programming to track the longest valid wraparound sequence ending at each letter. The key insight is that for each letter, we only need to know the maximum length sequence ending with that letter.

Brute Force - Check All Substrings — Algorithm Steps

Generate all possible substrings
For each substring, check if it's consecutive in wraparound alphabet
Use set to store unique valid substrings
Return size of set

Visualization

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Step-by-Step Walkthrough

Generate Substrings

Create all possible substrings from input string

Validate Each

Check if each substring is consecutive in wraparound alphabet

Count Unique

Use set to count only unique valid substrings

Code -

solution.c — C

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

int isValidSubstring(char* sub, int len) {
    if (len <= 1) return 1;
    for (int i = 1; i < len; i++) {
        if ((sub[i] - sub[i-1] + 26) % 26 != 1) {
            return 0;
        }
    }
    return 1;
}

int solution(char* s) {
    int n = strlen(s);
    char substrings[10000][105];
    int count = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j <= n; j++) {
            int len = j - i;
            char substring[105];
            strncpy(substring, s + i, len);
            substring[len] = '\0';
            
            if (isValidSubstring(substring, len)) {
                int exists = 0;
                for (int k = 0; k < count; k++) {
                    if (strcmp(substrings[k], substring) == 0) {
                        exists = 1;
                        break;
                    }
                }
                if (!exists) {
                    strcpy(substrings[count], substring);
                    count++;
                }
            }
        }
    }
    
    return count;
}

int main() {
    char s[105];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    int result = solution(s);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

O(n²) substrings, each taking O(n) time to validate

⚠ Quadratic Growth

Space Complexity

O(n²)

Set can store up to O(n²) unique substrings

⚠ Quadratic Space

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Input & Output

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Related Problems

Common Approaches

Brute Force - Check All Substrings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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