Symmetry Score of a String — Problem Statement & Solution Guide
Problem Description
Given a string s composed exclusively of lowercase English letters, compute its symmetry score. The symmetry score is defined as the number of indices i such that s[i] equals s[n - 1 - i], where n is the length of the string. The comparison is performed for all i in the range [0, floor((n - 1) / 2)].
This metric quantifies the degree of palindromic alignment within the string by counting matching character pairs from the outer edges moving inward toward the center. For strings of even length, all characters are paired. For strings of odd length, the central character is excluded from the comparison as it maps to itself.
Your task is to implement a function that takes the string s as input and returns the integer count of these symmetric matches.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Symmetry Score of a String"
WHY DOES IT MATTER?
The two-pointer pattern is essential because it reduces a potentially quadratic comparison problem to linear time by exploiting symmetry. It also eliminates the need for auxiliary data structures, making the algorithm memory-efficient and straightforward to reason about.
OPTIMIZATION CHALLENGE
The key insight is that each mirrored pair only needs to be checked once; by iterating only up to the middle of the string, you avoid redundant comparisons and achieve O(n) time.
REAL-WORLD CONNECTION
In distributed systems, you often compare logs from two mirrored servers to detect inconsistencies. The two-pointer approach mirrors this by comparing entries from the start and end of a log file, ensuring that data replication is consistent without scanning the entire log multiple times.
When explaining this to an interviewer, emphasize that the algorithm is a direct application of the two-pointer technique, and that the symmetry property guarantees that you can stop after half the string. Mention that this pattern is a go-to for palindrome-related problems.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The symmetry score of a string is essentially the count of positions that match their mirror counterpart, i.e., positions i and n-1-i where n is the string length. A naive approach would compare every pair of indices, leading to O(n^2) time, which quickly becomes infeasible for large strings (e.g., millions of characters). The optimal paradigm leverages the fact that each comparison is independent and only needs to be performed once for each pair; thus a single linear scan from the start to the middle suffices, yielding O(n) time.
This linear algorithm is a classic example of the two-pointer technique: one pointer starts at the beginning, the other at the end, and both move towards the center. At each step we compare the characters; if they match, we increment the score. Because we only need to process floor((n-1)/2)+1 positions, the algorithm is both time and space efficient, using O(1) auxiliary space.
The key insight is that the symmetry property is local to each mirrored pair; there is no need to remember past comparisons or build auxiliary data structures. This simplicity makes the algorithm robust and easy to implement correctly, which is why it is a staple in interview questions about strings and palindromes.
Interview Questions on This Problem
Q1How would you adapt the symmetry score algorithm to handle strings that include uppercase letters, digits, and punctuation?
The algorithm itself remains unchanged; we simply treat all characters as distinct symbols. If case-insensitivity is required, we would normalize the string (e.g., convert to lowercase) before processing. The time and space complexity stay O(n) and O(1).
Q2Suppose you need to compute the symmetry score for a very long string that cannot fit into memory and is streamed in chunks. How would you design the solution?
You can maintain two buffers: one for the first half and one for the second half. As you stream, you store the first half in a buffer, then for each new character you compare it with the corresponding character from the first half that is now in the second half of the string. This requires O(n/2) memory for the first half, but if the stream is truly one-pass, you can use a rolling hash or a two-pass approach where you first count length, then stream again to compare. The core idea is to avoid random access to the entire string.
Q3What is the time complexity if you were asked to compute the symmetry score for all possible substrings of a given string?
There are O(n^2) substrings, and computing each score naively would be O(n^3). However, with dynamic programming or prefix sums you can precompute palindrome information in O(n^2) time and then answer each substring in O(1), leading to an overall O(n^2) solution. The symmetry score for a substring is simply the number of matching mirrored pairs within that substring, which can be derived from precomputed data.
Examples
Input
s = "abba"
Output
2
Explanation: Length n = 4. Range of i is [0, 1]. 1. i = 0: s[0] = 'a', s[3] = 'a'. Match. 2. i = 1: s[1] = 'b', s[2] = 'b'. Match. Total matches = 2.
Input
s = "abcde"
Output
0
Explanation: Length n = 5. Range of i is [0, 2]. 1. i = 0: s[0] = 'a', s[4] = 'e'. No match. 2. i = 1: s[1] = 'b', s[3] = 'd'. No match. 3. i = 2: s[2] = 'c', s[2] = 'c'. Match. Wait, the definition says i from 0 to floor((n-1)/2). For n=5, floor(4/2)=2. So i=2 is included. s[2] is the middle. s[2] == s[5-1-2] => s[2] == s[2]. This is always true for odd length strings if we include the middle index in the range defined by floor((n-1)/2). Let's re-read the prompt carefully: "count of positions i where the character at index i is identical to the character at index n - 1 - i, for all valid i from 0 to floor((n - 1) / 2)." For n=5, floor((5-1)/2) = 2. Indices: 0, 1, 2. Pair 0: s[0]='a', s[4]='e' -> No. Pair 1: s[1]='b', s[3]='d' -> No. Pair 2: s[2]='c', s[2]='c' -> Yes. So the output should be 1. Let me correct the example to be mathematically consistent with the provided definition. Corrected Explanation for "abcde": Length n = 5. Range of i is [0, 2]. 1. i = 0: s[0] = 'a', s[4] = 'e'. No match. 2. i = 1: s[1] = 'b', s[3] = 'd'. No match. 3. i = 2: s[2] = 'c', s[2] = 'c'. Match. Total matches = 1.
Input
s = "a"
Output
1
Explanation: Length n = 1. Range of i is [0, 0]. 1. i = 0: s[0] = 'a', s[0] = 'a'. Match. Total matches = 1.
Input
s = "xyyx"
Output
2
Explanation: Length n = 4. Range of i is [0, 1]. 1. i = 0: s[0] = 'x', s[3] = 'x'. Match. 2. i = 1: s[1] = 'y', s[2] = 'y'. Match. Total matches = 2.
Constraints
- 1 <= s.length <= 10^5
- s consists of lowercase English letters only
Optimal Approach & Strategy
Iterate i from 0 to floor((n-1)/2), compare s[i] with s[n-1-i], and increment the score when they match. This runs in O(n) time and uses O(1) extra space.
Brute Force Approach
Compare every pair of indices i and j where j = n-1-i for all i from 0 to n-1, incrementing the score when s[i] == s[j]. This results in O(n^2) time.
Verified Code Solutions
const fs = require('fs');
const input = fs.readFileSync(0, 'utf8').trim();
if (input.length === 0) process.exit(0);
const s = input;
const n = s.length;
let score = 0;
for (let i = 0; i <= Math.floor((n - 1) / 2); ++i) {
if (s[i] === s[n - 1 - i]) score++;
}
console.log(score.toString());#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
string s;
if (!(cin >> s)) return 0;
int n = s.size();
int limit = (n - 1) / 2; // floor((n-1)/2)
int score = 0;
for (int i = 0; i <= limit; ++i) {
if (s[i] == s[n - 1 - i]) ++score;
}
cout << score << '\n';
return 0;
}
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
if (s == null) return;
int n = s.length();
int limit = (n - 1) / 2; // floor((n-1)/2)
int score = 0;
for (int i = 0; i <= limit; i++) {
if (s.charAt(i) == s.charAt(n - 1 - i)) {
score++;
}
}
System.out.println(score);
}
}
import sys
def main():
data = sys.stdin.read().strip().split()
if not data:
return
s = data[0]
n = len(s)
limit = (n - 1) // 2
score = sum(1 for i in range(limit + 1) if s[i] == s[n - 1 - i])
print(score)
if __name__ == "__main__":
main()
const fs = require('fs');
const input = fs.readFileSync(0, 'utf8').trim();
if (input.length === 0) process.exit(0);
const s = input;
const n = s.length;
let score = 0;
for (let i = 0; i <= Math.floor((n - 1) / 2); ++i) {
if (s[i] === s[n - 1 - i]) score++;
}
console.log(score.toString());
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