LeetCode #2223 — HARD

Sum of Scores of Built Strings

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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.

Example 1:

Input: s = "babab"
Output: 9
Explanation:
For s₁ == "b", the longest common prefix is "b" which has a score of 1.
For s₂ == "ab", there is no common prefix so the score is 0.
For s₃ == "bab", the longest common prefix is "bab" which has a score of 3.
For s₄ == "abab", there is no common prefix so the score is 0.
For s₅ == "babab", the longest common prefix is "babab" which has a score of 5.
The sum of the scores is 1 + 0 + 3 + 0 + 5 = 9, so we return 9.

Example 2:

Input: s = "azbazbzaz"
Output: 14
Explanation: 
For s₂ == "az", the longest common prefix is "az" which has a score of 2.
For s₆ == "azbzaz", the longest common prefix is "azb" which has a score of 3.
For s₉ == "azbazbzaz", the longest common prefix is "azbazbzaz" which has a score of 9.
For all other s_i, the score is 0.
The sum of the scores is 2 + 3 + 9 = 14, so we return 14.

Constraints:

1 <= s.length <= 10⁵
s consists of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: 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 si. For example, for s = "abaca", s1 == "a", s2 == "ca", s3 == "aca", etc. The score of si is the length of the longest common prefix between si and sn (Note that s == sn). Given the final string s, return the sum of the score of every si.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Binary Search · String Matching

Example 1

"babab"

Example 2

"azbazbzaz"

Core Insight

What unlocks the optimal approach

Each s_i is a suffix of the string s, so consider algorithms that can determine the longest prefix that is also a suffix.
Could you use the Z array from the Z algorithm to find the score of each s_i?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2223: Sum of Scores of Built Strings
class Solution {
  public long sumScores(String s) {
    final int n = s.length();
    // https://cp-algorithms.com/string/z-function.html#implementation
    int[] z = new int[n];
    // [l, r] := the indices of the rightmost segment match
    int l = 0;
    int r = 0;

    for (int i = 1; i < n; ++i) {
      if (i < r)
        z[i] = Math.min(r - i, z[i - l]);
      while (i + z[i] < n && s.charAt(z[i]) == s.charAt(i + z[i]))
        ++z[i];
      if (i + z[i] > r) {
        l = i;
        r = i + z[i];
      }
    }

    return Arrays.stream(z).asLongStream().sum() + n;
  }
}

// Accepted solution for LeetCode #2223: Sum of Scores of Built Strings
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #2223: Sum of Scores of Built Strings
// class Solution {
//   public long sumScores(String s) {
//     final int n = s.length();
//     // https://cp-algorithms.com/string/z-function.html#implementation
//     int[] z = new int[n];
//     // [l, r] := the indices of the rightmost segment match
//     int l = 0;
//     int r = 0;
// 
//     for (int i = 1; i < n; ++i) {
//       if (i < r)
//         z[i] = Math.min(r - i, z[i - l]);
//       while (i + z[i] < n && s.charAt(z[i]) == s.charAt(i + z[i]))
//         ++z[i];
//       if (i + z[i] > r) {
//         l = i;
//         r = i + z[i];
//       }
//     }
// 
//     return Arrays.stream(z).asLongStream().sum() + n;
//   }
// }

# Accepted solution for LeetCode #2223: Sum of Scores of Built Strings
class Solution:
  def sumScores(self, s: str) -> int:
    n = len(s)
    # https://cp-algorithms.com/string/z-function.html#implementation
    z = [0] * n
    # [l, r] := the indices of the rightmost segment match
    l = 0
    r = 0

    for i in range(1, n):
      if i < r:
        z[i] = min(r - i, z[i - l])
      while i + z[i] < n and s[z[i]] == s[i + z[i]]:
        z[i] += 1
      if i + z[i] > r:
        l = i
        r = i + z[i]

    return sum(z) + n

// Accepted solution for LeetCode #2223: Sum of Scores of Built Strings
/**
 * [2223] Sum of Scores of Built Strings
 *
 * 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 si.
 *
 * 	For example, for s = "abaca", s1 == "a", s2 == "ca", s3 == "aca", etc.
 *
 * The score of si is the length of the longest common prefix between si and sn (Note that s == sn).
 * Given the final string s, return the sum of the score of every si.
 *  
 * Example 1:
 *
 * Input: s = "babab"
 * Output: 9
 * Explanation:
 * For s1 == "b", the longest common prefix is "b" which has a score of 1.
 * For s2 == "ab", there is no common prefix so the score is 0.
 * For s3 == "bab", the longest common prefix is "bab" which has a score of 3.
 * For s4 == "abab", there is no common prefix so the score is 0.
 * For s5 == "babab", the longest common prefix is "babab" which has a score of 5.
 * The sum of the scores is 1 + 0 + 3 + 0 + 5 = 9, so we return 9.
 * Example 2:
 *
 * Input: s = "azbazbzaz"
 * Output: 14
 * Explanation:
 * For s2 == "az", the longest common prefix is "az" which has a score of 2.
 * For s6 == "azbzaz", the longest common prefix is "azb" which has a score of 3.
 * For s9 == "azbazbzaz", the longest common prefix is "azbazbzaz" which has a score of 9.
 * For all other si, the score is 0.
 * The sum of the scores is 2 + 3 + 9 = 14, so we return 14.
 *
 *  
 * Constraints:
 *
 * 	1 <= s.length <= 10^5
 * 	s consists of lowercase English letters.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/sum-of-scores-of-built-strings/
// discuss: https://leetcode.com/problems/sum-of-scores-of-built-strings/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn sum_scores(s: String) -> i64 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2223_example_1() {
        let s = "babab".to_string();

        let result = 9;

        assert_eq!(Solution::sum_scores(s), result);
    }

    #[test]
    #[ignore]
    fn test_2223_example_2() {
        let s = "azbazbzaz".to_string();

        let result = 14;

        assert_eq!(Solution::sum_scores(s), result);
    }
}

// Accepted solution for LeetCode #2223: Sum of Scores of Built Strings
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2223: Sum of Scores of Built Strings
// class Solution {
//   public long sumScores(String s) {
//     final int n = s.length();
//     // https://cp-algorithms.com/string/z-function.html#implementation
//     int[] z = new int[n];
//     // [l, r] := the indices of the rightmost segment match
//     int l = 0;
//     int r = 0;
// 
//     for (int i = 1; i < n; ++i) {
//       if (i < r)
//         z[i] = Math.min(r - i, z[i - l]);
//       while (i + z[i] < n && s.charAt(z[i]) == s.charAt(i + z[i]))
//         ++z[i];
//       if (i + z[i] > r) {
//         l = i;
//         r = i + z[i];
//       }
//     }
// 
//     return Arrays.stream(z).asLongStream().sum() + n;
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.