Sum of Scores of Built Strings - Practice Coding Problems

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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Understanding the Visualization

Input

Final string s = "abaca"

Build Process

Prepend characters: a → ca → aca → baca → abaca

Score Calculation

Compare each intermediate string with final string

Key Takeaway

🎯 Key Insight: Each intermediate string is a suffix of the final string - use string matching algorithms like Z-algorithm for optimal performance

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

Approach	Time	Space	Notes
✓ Z-Algorithm	O(n)	O(n)	Use Z-algorithm to compute longest common prefix for all suffixes in linear time
Brute Force Comparison	O(n³)	O(n²)	For each intermediate string, compare character by character with the final string
Direct Suffix Comparison	O(n²)	O(1)	Recognize that each si is a suffix of the final string and compare directly

Z-Algorithm — Algorithm Steps

Apply Z-algorithm to find Z-array
Z[i] gives longest common prefix between s[i:] and s
Sum all Z[i] values

Visualization

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Step-by-Step Walkthrough

Initialize

Set Z[0] = n, maintain left and right pointers

Compute Z-values

Use previous computations to avoid redundant work

Sum Array

Sum all Z-values for final answer

Code -

solution.c — C

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

long long solution(char* s) {
    int n = strlen(s);
    if (n == 0) return 0;
    
    // Compute Z-array using Z-algorithm
    int* z = (int*)calloc(n, sizeof(int));
    z[0] = n;  // First element is always n
    
    int l = 0, r = 0;
    for (int i = 1; i < n; i++) {
        if (i <= r) {
            z[i] = (r - i + 1 < z[i - l]) ? r - i + 1 : z[i - l];
        }
        
        while (i + z[i] < n && s[z[i]] == s[i + z[i]]) {
            z[i]++;
        }
        
        if (i + z[i] - 1 > r) {
            l = i;
            r = i + z[i] - 1;
        }
    }
    
    long long sum = 0;
    for (int i = 0; i < n; i++) {
        sum += z[i];
    }
    
    free(z);
    return sum;
}

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(n)

Z-algorithm processes each character at most twice using two pointers

✓ Linear Growth

Space Complexity

O(n)

Store Z-array of size n

⚡ Linearithmic Space

23.4K Views

Medium Frequency

~35 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Z-Algorithm — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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