Sum of Scores of Built Strings - Problem

You are building a string s of length n one character at a time, prepending each new character to the front of the string. The strings are labeled from 1 to n, where the string with length i is labeled s_i.

For example, for s = "abaca", s₁ == "a", s₂ == "ca", s₃ == "aca", etc.

The score of s_i is the length of the longest common prefix between s_i and s_n (Note that s == s_n).

Given the final string s, return the sum of the score of every s_i.

Input & Output

Example 1 — Basic Case

$ Input: s = "abaca"

› Output: 7

💡 Note: s1="a" scores 1, s2="ca" scores 0, s3="aca" scores 1, s4="baca" scores 0, s5="abaca" scores 5. Total: 1+0+1+0+5=7

Example 2 — Single Character

$ Input: s = "a"

› Output: 1

💡 Note: Only s1="a" which has 1 character matching with itself. Score: 1

Example 3 — All Different Characters

$ Input: s = "abc"

› Output: 3

💡 Note: s1="c" scores 0, s2="bc" scores 0, s3="abc" scores 3. Total: 0+0+3=3

Constraints

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

Visualization

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Asked in

G Google 12 M Microsoft 8 a Amazon 6

The key insight is that each intermediate string si corresponds to a suffix of the final string. The optimal approach uses the Z-algorithm to compute longest common prefix lengths for all suffixes in O(n) time. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force Comparison

⏱️ Time: O(n³) Space: O(n²)

Build each intermediate string by prepending characters, then compare each one with the final string to find the longest common prefix length.

Direct Suffix Comparison

⏱️ Time: O(n²) Space: O(1)

Instead of building intermediate strings, recognize that si corresponds to s[n-i:]. Compare each suffix with the original string to find common prefix lengths.

Z-Algorithm

⏱️ Time: O(n) Space: O(n)

The Z-algorithm efficiently computes the length of the longest substring starting from each position that matches a prefix of the string. Since each si is a suffix of s, we can use Z-algorithm directly.

Brute Force Comparison — Algorithm Steps

Build all intermediate strings s1, s2, ..., sn
For each si, compare with sn character by character
Sum up all prefix lengths

Visualization

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Step-by-Step Walkthrough

Build Strings

Create s1="a", s2="ca", s3="aca", etc.

Compare

Find common prefix length with final string

Sum

Add all prefix lengths together

Code -

solution.c — C

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

long long solution(char* s) {
    int n = strlen(s);
    long long totalScore = 0;
    
    // Build each intermediate string si
    for (int i = 1; i <= n; i++) {
        // si starts at position n-i (last i characters)
        char* si = s + (n - i);
        
        // Find longest common prefix between si and s
        int prefixLen = 0;
        int siLen = strlen(si);
        int minLen = (siLen < n) ? siLen : n;
        for (int j = 0; j < minLen; j++) {
            if (si[j] == s[j]) {
                prefixLen++;
            } else {
                break;
            }
        }
        
        totalScore += prefixLen;
    }
    
    return totalScore;
}

int main() {
    char s[100001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    long long result = solution(s);
    printf("%lld\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Build n strings (O(n²) total) and compare each with final string (O(n²) comparisons)

⚠ Quadratic Growth

Space Complexity

O(n²)

Store all n intermediate strings, total length O(n²)

✓ Linear Space

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Medium Frequency

~35 min Avg. Time

890 Likes

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Sum of Scores of Built Strings - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Comparison — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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