Longest Palindromic Substring in TypeScript

Longest Palindromic

Substring in TypeScript

Welcome to the Longest Palindromic Substring in TypeScript page! Here, you'll find the source code for this program as well as a description of how the program works.

Current Solution

function printUsage(): void {
  console.error(
    "Usage: please provide a string that contains at least one palindrome",
  );
  process.exit(1);
}

function preprocess(s: string): string {
  // e.g. "abba" -> "^#a#b#b#a#$"
  return "^#" + s.split("").join("#") + "#$";
}

function longestPalindromicSubstring(s: string): string | null {
  if (!s) return null;

  const T = preprocess(s);
  const n = T.length;
  const P = new Array<number>(n).fill(0);

  let C = 0; // center
  let R = 0; // right boundary

  for (let i = 1; i < n - 1; i++) {
    const mirror = 2 * C - i;

    if (i < R) {
      P[i] = Math.min(R - i, P[mirror]);
    }

    // expand around center i
    while (T[i + 1 + P[i]] === T[i - 1 - P[i]]) {
      P[i]++;
    }

    // update center
    if (i + P[i] > R) {
      C = i;
      R = i + P[i];
    }
  }

  // find max palindrome
  let maxLen = 0;
  let centerIndex = 0;

  for (let i = 1; i < n - 1; i++) {
    if (P[i] > maxLen) {
      maxLen = P[i];
      centerIndex = i;
    }
  }

  if (maxLen === 0) return null;

  const start = (centerIndex - maxLen) >> 1;
  return s.substring(start, start + maxLen);
}

function main(): void {
  const input = process.argv[2];

  if (!input) {
    printUsage();
  }

  const result = longestPalindromicSubstring(input);
  if (!result || result?.length < 2) {
    printUsage();
  }

  console.log(result);
}

main();

Longest Palindromic Substring in TypeScript was written by:

• Ștefan-Iulian Alecu

If you see anything you'd like to change or update, please consider contributing.

How to Implement the Solution

No 'How to Implement the Solution' section available. Please consider contributing.

How to Run the Solution

No 'How to Run the Solution' section available. Please consider contributing.