{"id":10146,"date":"2025-09-27T16:59:10","date_gmt":"2025-09-27T11:29:10","guid":{"rendered":"https:\/\/namastedev.com\/blog\/?p=10146"},"modified":"2025-09-27T17:02:12","modified_gmt":"2025-09-27T11:32:12","slug":"longest-palindromic-substring","status":"publish","type":"post","link":"https:\/\/namastedev.com\/blog\/longest-palindromic-substring\/","title":{"rendered":"Longest Palindromic Substring"},"content":{"rendered":"\n<!-- PrismJS for Syntax Highlighting -->\n<link href=\"https:\/\/cdn.jsdelivr.net\/npm\/prismjs@1.29.0\/themes\/prism-tomorrow.min.css\" rel=\"stylesheet\">\n<script src=\"https:\/\/cdn.jsdelivr.net\/npm\/prismjs@1.29.0\/prism.min.js\"><\/script>\n<script src=\"https:\/\/cdn.jsdelivr.net\/npm\/prismjs@1.29.0\/plugins\/autoloader\/prism-autoloader.min.js\"><\/script>\n\n<style>\n.wp_blog_theme {\n  --primary: #E58C32;\n  --secondary: #030302;\n  --light-bg: #fef9f4;\n  --text-dark: #2d2d2d;\n  --tab-radius: 12px;\n  --shadow: 0 4px 12px rgba(0, 0, 0, 0.08);\n  --code-bg: #001f3f;\n  --code-text: #d4f1ff;\n}\n\n.wp_blog_container {\n  font-family: 'Segoe UI', sans-serif;\n  background: var(--light-bg);\n  margin: 0;\n  padding: 0;\n  color: var(--text-dark);\n}\n\n\/* Heading *\/\n.wp_blog_main-heading {\n  text-align: center;\n  font-size: 2.4rem;\n  color: var(--primary);\n  margin-top: 2.5rem;\n  font-weight: bold;\n}\n\n\/* Explanation Card *\/\n.wp_blog_explanation,\n.wp_blog_code-tabs-container {\n  max-width: 940px;\n  margin: 2rem auto;\n  padding: 2rem;\n  background: white;\n  border-radius: var(--tab-radius);\n  box-shadow: var(--shadow);\n}\n\n\/* Text and Visuals *\/\n.wp_blog_explanation h2 {\n  font-size: 1.4rem;\n  color: var(--primary);\n  margin-bottom: 0.5rem;\n}\n\n.wp_blog_explanation p,\n.wp_blog_explanation li {\n  font-size: 1.05rem;\n  line-height: 1.7;\n  margin: 0.5rem 0;\n}\n\n.wp_blog_explanation code {\n  background: #fef9f4; 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\/* makes it the reference for absolute children *\/\n}\n\n.wp_blog_toggle-btn {\n  position: absolute;\n  top: 1rem;\n  right: 1rem;\n  z-index: 9999;\n  padding: 0.5rem 0.8rem;\n  border-radius: 10%;\n  background: var(--primary);\n  color: white;\n  font-weight: bold;\n  cursor: pointer;\n  border: none;\n  box-shadow: var(--shadow);\n  transition: background 0.3s, transform 0.2s;\n}\n\n.wp_blog_toggle-btn:hover {\n  background: #cc772e;\n}\n\n.wp_blog_theme.dark-mode .wp_blog_code-tabs-container {\n  background: #1e1e1e;\n}\n<\/style>\n\n<div class=\"wp_blog_container wp_blog_theme\">\n  <button id=\"blogNotesThemeToggle\" class=\"wp_blog_toggle-btn\">\ud83c\udf19<\/button>\n  <h1 class=\"wp_blog_main-heading\"><\/h1>\n\n  <div class=\"wp_blog_explanation\">\n    <h2>Problem Statement:<\/h2>\n    <p>\n        Given a string <code>s<\/code>, return the longest <i>palindromic substring<\/i> in <code>s<\/code>.\n    <\/p>\n\n\n    <h3>Example 1:<\/h2>\n    <p><strong>Input:<\/strong> s = &#8220;babad&#8221;<\/p>\n    <p><strong>Output:<\/strong> &#8220;bab&#8221;<\/p>\n    <p><strong>Explanation:<\/strong> &#8220;aba&#8221; is also a valid answer.    <\/p>\n\n    <h3>Example 2:<\/h2>\n    <p><strong>Input:<\/strong> s = &#8220;cbbd&#8221;<\/p>\n    <p><strong>Output:<\/strong> &#8220;bb&#8221;<\/p>\n\n    <h3>Constraints<\/h3>\n    <ul>\n       <li><code>1 <= s.length <= 1000<\/code><\/li>\n        <li><code>s<\/code> consist of only digits and English letters.<\/li>\n    <\/ul>\n    \n<h2>Approach:<\/h2>\n<ul>\n    <li>Use <strong>Dynamic Programming (DP)<\/strong> to check if a <code>substring s[i...j]<\/code> is a palindrome.<\/li>\n    <li>Maintain a <code>dp[i][j]<\/code> table where:\n    <ul>\n        <li><code>dp[i][j] = true <\/code>if substring <code>s[i...j]<\/code> is a palindrome.<\/li>\n    <\/ul>\n    <\/li>\n\n    <li><strong>Base cases:<\/strong>\n    <ul>\n        <li>Single character substrings <code>(dp[i][i] = true)<\/code> are <strong>palindromes<\/strong>.<\/li>\n        <li>Two-character substrings <code>(s[i] == s[i+1])<\/code> are palindromes.<\/li>\n    <\/ul>\n    <\/li>\n\n    <li>For substrings of <code>length \u2265 3:<\/code>\n        <ul>\n            <li>A substring <code>s[i...j]<\/code> is a palindrome if <code>s[i] == s[j]<\/code> and <code>dp[i+1][j-1] <\/code>is <strong>true<\/strong>.<\/li>\n        <\/ul>\n    <\/li>\n\n    <li>Track the longest palindrome\u2019s start and end indices while filling DP<\/li>\n\n    <li><code>Return the longest palindrome<\/code> substring using substring <strong>(ans[0], ans[1] + 1)<\/strong>.<\/li>\n<\/ul>\n\n    <h2>Time & Space Complexity:<\/h2>\n    <p><strong>Time Complexity:<\/strong> <code>O(n<sup>2<\/sup>)<\/code><\/p>\n    <p><strong>Space Complexity:<\/strong> <code>O(n<sup>2<\/sup>)<\/code><\/p>\n\n<h2>Dry Run<\/h2>\n<div style=\"background: var(--light-bg); border-left: 4px solid var(--primary); padding: 1rem; border-radius: var(--tab-radius); margin: 1rem 0; color: var(--text-dark);\"> \n  <p><strong>Input:<\/strong> <code>s = \"babad\"<\/code><\/p> \n  \n  <pre style=\"white-space: pre-wrap; background: var(--code-bg); padding: 1rem; border-radius: 8px; overflow-x: auto; color: var(--code-text);\">\n\nStep 0: Start Function longestPalindrome(\"babad\")\nn = 5\ndp = 5x5 matrix filled with null\nans = [0, 0]\n\nBase cases (length 1 and length 2 checks):\n\ni = 0:\n- dp[0][0] = true            \/\/ \"b\"\n- Check 2-length: s[0] === s[1]? \"b\" === \"a\" \u2192 false\nState: ans = [0,0]\n\ni = 1:\n- dp[1][1] = true            \/\/ \"a\"\n- Check 2-length: s[1] === s[2]? \"a\" === \"b\" \u2192 false\nState: ans = [0,0]\n\ni = 2:\n- dp[2][2] = true            \/\/ \"b\"\n- Check 2-length: s[2] === s[3]? \"b\" === \"a\" \u2192 false\nState: ans = [0,0]\n\ni = 3:\n- dp[3][3] = true            \/\/ \"a\"\n- Check 2-length: s[3] === s[4]? \"a\" === \"d\" \u2192 false\nState: ans = [0,0]\n\ni = 4:\n- dp[4][4] = true            \/\/ \"d\"\n- No i+1 to check\nState: ans = [0,0]\n\nCurrent dp (T = true, null = unset):\n[\n [T, null, null, null, null],\n [ , T,    null, null, null],\n [ ,  , T,    null, null],\n [ ,  ,  , T,    null],\n [ ,  ,  ,  , T     ]\n]\n\nMain loop (len = 3 ... n):\n\nlen = 3:\n- i = 0, j = 2:\n  s[0] === s[2]? \"b\" === \"b\" \u2192 true\n  dp[1][1] is true \u2192 so dp[0][2] = true\n  ans = [0, 2]            \/\/ substring \"bab\"\n\n- i = 1, j = 3:\n  s[1] === s[3]? \"a\" === \"a\" \u2192 true\n  dp[2][2] is true \u2192 so dp[1][3] = true\n  ans = [1, 3]            \/\/ substring \"aba\" (updates ans)\n\n- i = 2, j = 4:\n  s[2] === s[4]? \"b\" === \"d\" \u2192 false\n  no update\n\ndp now (noting newly true entries):\n[\n [T, null, T,    null, null],\n [ , T,    null, T,    null],\n [ ,  , T,    null, null],\n [ ,  ,  , T,    null],\n [ ,  ,  ,  , T     ]\n]\n\nlen = 4:\n- i = 0, j = 3:\n  s[0] === s[3]? \"b\" === \"a\" \u2192 false\n\n- i = 1, j = 4:\n  s[1] === s[4]? \"a\" === \"d\" \u2192 false\n\n(no changes)\n\nlen = 5:\n- i = 0, j = 4:\n  s[0] === s[4]? \"b\" === \"d\" \u2192 false\n\n(no changes)\n\nStep 3: End\nFinal ans = [1, 3]\nReturn s.substring(1, 3 + 1) = \"aba\"\n  <\/pre> \n  \n  <p><strong>Output:<\/strong> <code>\"aba\"<\/code><\/p> \n<\/div>\n\n\n        <!-- <h2>Visualisation:<\/h2>\n        <img decoding=\"async\" src=\"https:\/\/namastedev.com\/blog\/wp-content\/uploads\/2025\/09\/Screenshot-2025-09-26-at-2.18.00\u202fPM.png\" alt=\"Stocks\" \/>\n    -->\n<\/div>\n\n\n  <div class=\"wp_blog_code-tabs-container\">\n    <div class=\"wp_blog_code-tabs-header\">\n      <button class=\"wp_blog_code-tab-button active\" data-lang=\"js\">JavaScript<\/button>\n      <button class=\"wp_blog_code-tab-button\" data-lang=\"py\">Python<\/button>\n      <button class=\"wp_blog_code-tab-button\" data-lang=\"java\">Java<\/button>\n      <button class=\"wp_blog_code-tab-button\" data-lang=\"cpp\">C++<\/button>\n      <button class=\"wp_blog_code-tab-button\" data-lang=\"c\">C<\/button>\n      <button class=\"wp_blog_code-tab-button\" data-lang=\"cs\">C#<\/button>\n    <\/div>\n\n    <div class=\"wp_blog_code-tab-content active\" data-lang=\"js\">\n      <pre><code class=\"language-javascript\">\nvar longestPalindrome = function(s) {\n    let n = s.length;\n    if (n === 0) return \"\";\n    \/\/ dp[i][j] will be true if s[i..j] is a palindrome\n    let dp = Array.from({ length: n }, () => Array(n).fill(false));\n    let ans = [0, 0];\n    \/\/ Base cases: length 1 and length 2\n    for (let i = 0; i < n; i++) {\n        dp[i][i] = true; \/\/ single char\n        if (i < n - 1 &#038;&#038; s[i] === s[i + 1]) {\n            dp[i][i + 1] = true;\n            ans = [i, i + 1];\n        }\n    }\n    \/\/ Check lengths from 3 to n\n    for (let len = 3; len <= n; len++) {\n        for (let i = 0; i <= n - len; i++) {\n            let j = i + len - 1;\n            if (s[i] === s[j] &#038;&#038; dp[i + 1][j - 1]) {\n                dp[i][j] = true;\n                ans = [i, j];\n            }\n        }\n    }\n    return s.substring(ans[0], ans[1] + 1);\n};\n<\/code><\/pre>\n    <\/div>\n    <div class=\"wp_blog_code-tab-content\" data-lang=\"py\">\n      <pre><code class=\"language-python\">\ndef longest_palindrome(s: str) -> str:\n    n = len(s)\n    if n == 0:\n        return \"\"\n\n    dp = [[False] * n for _ in range(n)]\n    start, end = 0, 0\n\n    for i in range(n):\n        dp[i][i] = True\n        if i < n - 1 and s[i] == s[i + 1]:\n            dp[i][i + 1] = True\n            start, end = i, i + 1\n\n    for length in range(3, n + 1):\n        for i in range(0, n - length + 1):\n            j = i + length - 1\n            if s[i] == s[j] and dp[i + 1][j - 1]:\n                dp[i][j] = True\n                start, end = i, j\n\n    return s[start:end + 1]\n      <\/code><\/pre>\n    <\/div>\n    <div class=\"wp_blog_code-tab-content\" data-lang=\"java\">\n      <pre><code class=\"language-java\">\npublic class Solution {\n    public static String longestPalindrome(String s) {\n        if (s == null || s.length() == 0) return \"\";\n        int n = s.length();\n        boolean[][] dp = new boolean[n][n];\n        int start = 0, end = 0;\n\n        for (int i = 0; i < n; i++) {\n            dp[i][i] = true;\n            if (i < n - 1 &#038;&#038; s.charAt(i) == s.charAt(i + 1)) {\n                dp[i][i + 1] = true;\n                start = i; end = i + 1;\n            }\n        }\n        for (int len = 3; len <= n; len++) {\n            for (int i = 0; i <= n - len; i++) {\n                int j = i + len - 1;\n                if (s.charAt(i) == s.charAt(j) &#038;&#038; dp[i + 1][j - 1]) {\n                    dp[i][j] = true;\n                    start = i; end = j;\n                }\n            }\n        }\n        return s.substring(start, end + 1);\n    }\n}\n    <\/code><\/pre>\n    <\/div>\n\n    <div class=\"wp_blog_code-tab-content\" data-lang=\"cpp\">\n      <pre><code class=\"language-cpp\">\n#include &lt;vector&gt;\n#include &lt;string&gt;\n\nstd::string longestPalindrome(const std::string &s) {\n    int n = (int)s.size();\n    if (n == 0) return \"\";\n    std::vector<std::vector<char>> dp(n, std::vector<char>(n, 0));\n    int start = 0, end = 0;\n    for (int i = 0; i < n; ++i) {\n        dp[i][i] = 1;\n        if (i < n - 1 &#038;&#038; s[i] == s[i + 1]) {\n            dp[i][i + 1] = 1;\n            start = i; end = i + 1;\n        }\n    }\n    for (int len = 3; len <= n; ++len) {\n        for (int i = 0; i <= n - len; ++i) {\n            int j = i + len - 1;\n            if (s[i] == s[j] &#038;&#038; dp[i + 1][j - 1]) {\n                dp[i][j] = 1;\n                start = i; end = j;\n            }\n        }\n    }\n    return s.substr(start, end - start + 1);\n}\n<\/code><\/pre>\n    <\/div>\n    <div class=\"wp_blog_code-tab-content\" data-lang=\"c\">\n      <pre><code class=\"language-c\">\n#include &lt;stdlib.h&gt;\n#include &lt;stdio.h&gt;\n#include &lt;string.h&gt;\n\nchar* longestPalindrome(const char* s) {\n    if (!s) return strdup(\"\");\n    int n = (int)strlen(s);\n    if (n == 0) return strdup(\"\");\n\n    \/\/ Use a flat array for dp: dp[i*n + j]\n    unsigned char* dp = (unsigned char*)calloc((size_t)n * n, sizeof(unsigned char));\n    if (!dp) return strdup(\"\");\n\n    int best_i = 0, best_j = 0;\n\n    \/\/ length 1 and 2\n    for (int i = 0; i < n; i++) {\n        dp[i * n + i] = 1;\n        if (i < n - 1 &#038;&#038; s[i] == s[i + 1]) {\n            dp[i * n + (i + 1)] = 1;\n            best_i = i; best_j = i + 1;\n        }\n    }\n    for (int len = 3; len <= n; len++) {\n        for (int i = 0; i <= n - len; i++) {\n            int j = i + len - 1;\n            if (s[i] == s[j] &#038;&#038; dp[(i + 1) * n + (j - 1)]) {\n                dp[i * n + j] = 1;\n                best_i = i; best_j = j;\n            }\n        }\n    }\n    int outLen = best_j - best_i + 1;\n    char* res = (char*)malloc((size_t)outLen + 1);\n    if (res) {\n        memcpy(res, s + best_i, outLen);\n        res[outLen] = '\\0';\n    }\n    free(dp);\n    return res;\n}\n <\/code><\/pre>\n    <\/div>\n\n    <div class=\"wp_blog_code-tab-content\" data-lang=\"cs\">\n      <pre><code class=\"language-csharp\">\nusing System;\npublic class Solution {\n    public static string LongestPalindrome(string s) {\n        int n = s?.Length ?? 0;\n        if (n == 0) return \"\";\n        bool[,] dp = new bool[n, n];\n        int start = 0, end = 0;\n\n        for (int i = 0; i < n; i++) {\n            dp[i, i] = true;\n            if (i < n - 1 &#038;&#038; s[i] == s[i + 1]) {\n                dp[i, i + 1] = true;\n                start = i; end = i + 1;\n            }\n        }\n        for (int len = 3; len <= n; len++) {\n            for (int i = 0; i <= n - len; i++) {\n                int j = i + len - 1;\n                if (s[i] == s[j] &#038;&#038; dp[i + 1, j - 1]) {\n                    dp[i, j] = true;\n                    start = i; end = j;\n                }\n            }\n        }\n        return s.Substring(start, end - start + 1);\n    }\n}\n      <\/code><\/pre>\n    <\/div>\n  <\/div>\n<\/div>\n\n<script>\ndocument.addEventListener(\"DOMContentLoaded\", () => {\n  const buttons = document.querySelectorAll(\".wp_blog_code-tab-button\");\n  const contents = document.querySelectorAll(\".wp_blog_code-tab-content\");\n\n  buttons.forEach((button) => {\n    button.addEventListener(\"click\", () => {\n      const lang = button.getAttribute(\"data-lang\");\n\n      buttons.forEach((btn) => btn.classList.remove(\"active\"));\n      contents.forEach((content) => content.classList.remove(\"active\"));\n\n      button.classList.add(\"active\");\n      document\n        .querySelector(`.wp_blog_code-tab-content[data-lang=\"${lang}\"]`)\n        .classList.add(\"active\");\n    });\n  });\n\n  const themeToggle = document.getElementById(\"blogNotesThemeToggle\");\n  const themeContainer = document.querySelector(\".wp_blog_theme\");\n\n  themeToggle.addEventListener(\"click\", () => {\n    themeContainer.classList.toggle(\"dark-mode\");\n    themeToggle.textContent =\n      themeContainer.classList.contains(\"dark-mode\") ? \"\u2600\ufe0f\" : \"\ud83c\udf19\";\n  });\n});\n<\/script>\n\n","protected":false},"excerpt":{"rendered":"<p>\ud83c\udf19 Problem Statement: Given a string s, return the longest palindromic substring in s. Example 1: Input: s = &#8220;babad&#8221; Output: &#8220;bab&#8221; Explanation: &#8220;aba&#8221; is also a valid answer. Example 2: Input: s = &#8220;cbbd&#8221; Output: &#8220;bb&#8221; Constraints 1<\/p>\n","protected":false},"author":108,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"categories":[210,322,260,176,175,211,811,810,174,172,173],"tags":[],"class_list":["post-10146","post","type-post","status-publish","format-standard","category-algorithms","category-algorithms-and-data-structures","category-c-c-plus-plus","category-csharp","category-cplusplus","category-data-structures","category-data-structures-and-algorithms","category-dsa","category-java","category-javascript","category-python"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/posts\/10146","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/users\/108"}],"replies":[{"embeddable":true,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/comments?post=10146"}],"version-history":[{"count":1,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/posts\/10146\/revisions"}],"predecessor-version":[{"id":10147,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/posts\/10146\/revisions\/10147"}],"wp:attachment":[{"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/media?parent=10146"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/categories?post=10146"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/tags?post=10146"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}