Next tasks

Author: dev-ik

5. Longest Palindromic Substring/index.test.ts

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+import { longestPalindrome } from './index';
+
+describe('longestPalindrome', () => {
+  const testCases = [
+    {
+      name: 'Case 1',
+      input: 'babad',
+      expected: 'bab',
+    },
+    {
+      name: 'Case 2',
+      input: 'cbbd',
+      expected: 'bb',
+    },
+    {
+      name: 'Case 3',
+      input: 'aacabdkacaa',
+      expected: 'aca',
+    },
+  ];
+
+  for (const testCase of testCases) {
+    test(testCase.name, () => {
+      expect(longestPalindrome(testCase.input)).toBe(testCase.expected);
+    });
+  }
+});

5. Longest Palindromic Substring/index.ts

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+// Condition of the task:
+// Given a string s, return the longest palindromic substring  in s.
+
+const processString = (s: string): string => {
+  let res = '';
+
+  for (let i = 0; i < s.length; i++) {
+    res += `#${s[i]}`;
+  }
+
+  return res + '#';
+};
+
+const min = (a: number, b: number): number => (a < b ? a : b);
+export const longestPalindrome = (s: string): string => {
+  const processedString = processString(s);
+  const n = processedString.length;
+  const P = new Array(n).fill(0);
+  let C = 0,
+    R = 0;
+
+  for (let i = 0; i < n - 1; i++) {
+    const mirror = 2 * C - i;
+
+    if (i < R) {
+      P[i] = min(R - i, P[mirror]);
+    } else {
+      P[i] = 0;
+    }
+
+    while (
+      i + 1 + P[i] < n &&
+      i - 1 - P[i] >= 0 &&
+      processedString[i + 1 + P[i]] === processedString[i - 1 - P[i]]
+    ) {
+      P[i]++;
+    }
+
+    if (i + P[i] > R) {
+      C = i;
+      R = i + P[i];
+    }
+  }
+  let maxLen = 0;
+  let centerIndex = 0;
+  for (let i = 0; i < n; i++) {
+    if (P[i] > maxLen) {
+      maxLen = P[i];
+      centerIndex = i;
+    }
+  }
+
+  const start = (centerIndex - maxLen) / 2;
+  const end = start + maxLen;
+  return s.slice(start, end);
+};