Longest Common Subsequence in TypeScript

Longest Common Subsequence

in TypeScript

Welcome to the Longest Common Subsequence 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

"use strict";

function printUsage(): void {
  console.error(
    'Usage: please provide two lists in the format "1, 2, 3, 4, 5"',
  );
  process.exit(1);
}

function parseInput(input: string | undefined): number[] | null {
  if (!input || input.trim() === "") return null;

  const parts = input.split(",").map((s) => s.trim());
  const arr: number[] = [];

  for (const p of parts) {
    if (p === "") return null;

    const val = Number.parseInt(p, 10);
    if (Number.isNaN(val)) return null;

    arr.push(val);
  }

  return arr;
}

function lcs(arr1: number[], arr2: number[]): void {
  const n = arr1.length;
  const m = arr2.length;

  const dp: number[][] = Array.from({ length: n + 1 }, () =>
    Array(m + 1).fill(0),
  );

  for (let i = 1; i <= n; i++) {
    for (let j = 1; j <= m; j++) {
      if (arr1[i - 1] === arr2[j - 1]) {
        dp[i][j] = 1 + dp[i - 1][j - 1];
      } else {
        dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
      }
    }
  }

  const result: number[] = [];
  let i = n;
  let j = m;

  while (i > 0 && j > 0) {
    if (arr1[i - 1] === arr2[j - 1]) {
      result.push(arr1[i - 1]);
      i--;
      j--;
    } else if (dp[i - 1][j] >= dp[i][j - 1]) {
      i--;
    } else {
      j--;
    }
  }

  result.reverse();
  console.log(result.join(", "));
}

function main(): void {
  const input1 = process.argv[2];
  const input2 = process.argv[3];

  const arr1 = parseInput(input1);
  const arr2 = parseInput(input2);

  if (!arr1 || !arr2) {
    printUsage();
  }

  lcs(arr1!, arr2!);
}

main();

Longest Common Subsequence 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

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How to Run the Solution

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