Data Center Symmetry Correction — Problem Statement & Solution Guide

StringsMediumTwo Pointers

TimeO(n)

SpaceO(1)

Problem Description

A data center transmits packets as strings of lowercase English letters. Determine if the given string is currently a palindrome or if it can be transformed into a palindrome by performing exactly one adjacent character swap.

DSA Pattern Breakdown

"Data Center Symmetry Correction"

medium

WHY DOES IT MATTER?

This pattern is essential for understanding how local modifications can affect global string properties. It tests the ability to reason about constraints and optimize for minimal changes, which is a common theme in system design and algorithmic problems.

OPTIMIZATION CHALLENGE

The key insight is that a single adjacent swap can only fix mismatches that are close to each other. By focusing on the first mismatch and its immediate neighbors, we avoid checking all possible swaps, reducing the time complexity from O(n^2) to O(n).

REAL-WORLD CONNECTION

In distributed systems, data consistency checks often involve verifying if data is symmetric or can be made symmetric with minimal corrections. This problem mirrors the logic used in error-correcting codes and data validation protocols.

During an interview, clearly articulate why you are only checking specific swaps. Emphasize that a single adjacent swap cannot fix mismatches that are far apart, which justifies the limited scope of your checks.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem of determining if a string is a palindrome or can become one via a single adjacent swap relies on the structural properties of palindromes. A palindrome reads the same forwards and backwards, meaning for every index i, s[i] must equal s[n-1-i]. When we introduce the constraint of exactly one adjacent swap, we are essentially asking if the string is 'one edit distance' away from being a palindrome, but restricted to local modifications. The naive approach of checking all possible swaps is O(n^2), which is inefficient for large strings. Instead, we can leverage the fact that a single adjacent swap can only fix mismatches that are close to each other or involve the center of the string.

Interview Questions on This Problem

Q1At a fintech platform, how would you validate if a transaction ID string is symmetric or can be made symmetric with a minimal change to ensure data integrity during transmission?

I would first check if the string is already a palindrome. If not, I would identify the first mismatch from the left and right. If swapping the two mismatched characters makes the string a palindrome, the answer is true. If not, I would check if swapping either of the mismatched characters with its adjacent neighbor fixes the symmetry. This ensures O(n) time complexity.

Q2In a high-growth startup, how can you optimize the validation of user-generated content strings to ensure they meet symmetry requirements for a specific feature, given high throughput?

I would implement a two-pointer approach to check for palindromic properties. If a mismatch is found, I would evaluate the impact of a single adjacent swap on the symmetry. By limiting the swap checks to the immediate neighbors of the mismatched characters, I can maintain O(n) time complexity and O(1) space complexity, which is crucial for high-throughput systems.

Q3At a global product company, how would you handle edge cases where the string length is odd or even when determining if a single adjacent swap can create a palindrome?

For odd-length strings, the middle character can be swapped with its left or right neighbor. For even-length strings, the two middle characters can be swapped. I would ensure that my two-pointer approach correctly handles these cases by checking the center indices specifically when the pointers meet or cross.

Examples

Example 1

Input

abcba

Output

true

Explanation: Step-by-step: with input 'abcba', we check if it's a palindrome, which it is, so we return true

Example 2

Input

abca

Output

true

Explanation: Step-by-step: with input 'abca', we check if it's a palindrome, which it's not, but we can swap 'c' and 'a' to get 'abac' and then another swap is not needed, however 'abca' can be transformed into 'acba' which is a palindrome by swapping 'b' and 'c', so we return true

Constraints

2 <= s.length <= 10^5
s consists of lowercase English letters.

Optimal Approach & Strategy

Use a two-pointer approach to find the first mismatch. Check if swapping the mismatched characters with their immediate neighbors results in a palindrome. This approach has a time complexity of O(n) and O(1) space complexity.

Brute Force Approach

Generate all possible strings by performing every possible adjacent swap and check if any of them is a palindrome. This approach has a time complexity of O(n^2) and is inefficient for large strings.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(s) {
       if (s === s.split('').reverse().join('')) return true;
       for (let i = 0; i < s.length - 1; i++) {
           let arr = s.split('');
           let temp = arr[i];
           arr[i] = arr[i + 1];
           arr[i + 1] = temp;
           if (arr.join('') === arr.join('').split('').reverse().join('')) return true;
           arr[i + 1] = arr[i];
           arr[i] = temp;
       }
       return false;
   }

class Solution {
   public:
       bool solution(string s) {
           if (s == string(s.rbegin(), s.rend())) return true;
           for (int i = 0; i < s.length() - 1; i++) {
               swap(s[i], s[i + 1]);
               if (s == string(s.rbegin(), s.rend())) return true;
               swap(s[i], s[i + 1]);
           }
           return false;
       }
   };

class Solution {
       public boolean solution(String s) {
           if (s.equals(new StringBuilder(s).reverse().toString())) return true;
           char[] arr = s.toCharArray();
           for (int i = 0; i < arr.length - 1; i++) {
               char temp = arr[i];
               arr[i] = arr[i + 1];
               arr[i + 1] = temp;
               if (new String(arr).equals(new StringBuilder(new String(arr)).reverse().toString())) return true;
               arr[i + 1] = arr[i];
               arr[i] = temp;
           }
           return false;
       }
   }

def solution(s):
       if s == s[::-1]: return True
       s = list(s)
       for i in range(len(s) - 1):
           s[i], s[i + 1] = s[i + 1], s[i]
           if s == s[::-1]: return True
           s[i], s[i + 1] = s[i + 1], s[i]
       return False

function solution(s) {
       if (s === s.split('').reverse().join('')) return true;
       for (let i = 0; i < s.length - 1; i++) {
           let arr = s.split('');
           let temp = arr[i];
           arr[i] = arr[i + 1];
           arr[i + 1] = temp;
           if (arr.join('') === arr.join('').split('').reverse().join('')) return true;
           arr[i + 1] = arr[i];
           arr[i] = temp;
       }
       return false;
   }

Asked in Top Tech Interviews

Cred

Solve in Interative Editor

Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.

Data Center Symmetry Correction Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Strings

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Data Center Symmetry Correction — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor