{"id":11775,"date":"2026-03-14T19:01:40","date_gmt":"2026-03-14T13:31:40","guid":{"rendered":"https:\/\/namastedev.com\/blog\/?p=11775"},"modified":"2026-03-14T19:23:25","modified_gmt":"2026-03-14T13:53:25","slug":"min-cost-to-cut-a-stick-rod-cutting-problem","status":"publish","type":"post","link":"https:\/\/namastedev.com\/blog\/min-cost-to-cut-a-stick-rod-cutting-problem\/","title":{"rendered":"Min Cost to Cut a Stick (Rod Cutting Problem)"},"content":{"rendered":"\n<link\n    href=\"https:\/\/cdn.jsdelivr.net\/npm\/prismjs@1.29.0\/themes\/prism-tomorrow.min.css\"\n    rel=\"stylesheet\"\n\/>\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;   \/* light bg instead of dark blue *\/\n  color: #E58C32;        \/* brand orange *\/\n  padding: 3px 6px;\n  border-radius: 4px;\n  font-family: 'Courier New', monospace;\n  font-weight: 600;      \/* optional, makes it pop *\/\n}\n\n.wp_blog_explanation img {\n  max-width: 100%;\n  border-radius: var(--tab-radius);\n  margin-top: 1rem;\n  box-shadow: 0 2px 12px rgba(0, 0, 0, 0.06);\n}\n\n\/* Tab Buttons *\/\n.wp_blog_code-tabs-header {\n  display: flex;\n  flex-wrap: wrap;\n  gap: 0.5rem;\n  margin-bottom: 1rem;\n}\n\n.wp_blog_code-tab-button {\n  padding: 0.6rem 1.2rem;\n  border: 1px solid var(--primary);\n  background: white;\n  color: var(--primary);\n  border-radius: 50px;\n  font-weight: 600;\n  cursor: pointer;\n  transition: all 0.3s ease;\n}\n\n.wp_blog_code-tab-button:hover {\n  background: var(--secondary);\n}\n\n.wp_blog_code-tab-button.active {\n  background: var(--primary);\n  color: white;\n}\n\n\/* Code Content *\/\n.wp_blog_code-tab-content {\n  display: none;\n  background: var(--code-bg);\n  border-radius: var(--tab-radius);\n}\n\n.wp_blog_code-tab-content.active {\n  display: block;\n}\n\n.wp_blog_code-tab-content pre {\n  margin: 0;\n  padding: 1.5rem;\n  font-size: 1rem;\n  overflow-x: auto;\n  background: var(--code-bg);\n  border-radius: var(--tab-radius);\n  color: var(--code-text);\n}\n\n\/* Dark mode variables *\/\n.wp_blog_theme.dark-mode {\n  --light-bg: #121212;\n  --text-dark: #f5f5f5;\n  --shadow: 0 4px 12px rgba(255, 255, 255, 0.08);\n  --code-bg: #1e1e1e;\n  --code-text: #c5f0ff;\n}\n\n.wp_blog_theme.dark-mode .wp_blog_explanation {\n  background: #1e1e1e;\n}\n\n\/* Dark mode code highlight *\/\n.wp_blog_theme.dark-mode .wp_blog_explanation code {\n  background: #333;\n  color: #ffd27f;\n}\n\n.wp_blog_theme {\n  position: relative; \/* 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    <div class=\"wp_blog_explanation\">\n        <h2>Problem Statement:<\/h2>\n        <p>\n            Given a wooden stick of length n units. The stick is labelled from 0 to n. For example, a stick of length 6 is labelled as follows:\n        <\/p>\n\n        <img decoding=\"async\" src=\"https:\/\/namastedev.com\/blog\/wp-content\/uploads\/2026\/03\/Screenshot-2026-03-14-at-6.43.13-PM.png\" alt=\"\">\n\n        <p>\n            Given an integer array cuts where cuts[i] denotes a position you should perform a cut at.\n        <\/p>\n\n        <p>\n            You should perform the cuts in order, you can change the order of the cuts as you wish.\n        <\/p>\n\n        <p>\n            The cost of one cut is the length of the stick to be cut, the total cost is the sum of costs of all cuts. When you cut a stick, it will be split into two smaller sticks (i.e. the sum of their lengths is the length of the stick before the cut). Please refer to the first example for a better explanation.\n        <\/p>\n\n        <p>\n            Return the minimum total cost of the cuts.\n        <\/p>\n\n                <h2>Examples:<\/h2>\n\n                <h3>Example 1:<\/h3>\n                <img decoding=\"async\" src=\"https:\/\/namastedev.com\/blog\/wp-content\/uploads\/2026\/03\/Screenshot-2026-03-14-at-6.43.22-PM.png\" alt=\"\">\n                <p><strong>Input:<\/strong> n = 7, cuts = [1,3,4,5]\n                <\/p>\n                <p><strong>Output:<\/strong> 16<\/p>\n                <p>\n                  <code>Explanation:<\/code>\n                  Using cuts order = [1, 3, 4, 5] as in the input leads to the following scenario:\n                  <img decoding=\"async\" src=\"https:\/\/namastedev.com\/blog\/wp-content\/uploads\/2026\/03\/Screenshot-2026-03-14-at-6.43.28-PM.png\" alt=\"\">\n                <\/p>\n                <p>\n                    The first cut is done to a rod of length 7 so the cost is 7. The second cut is done to a rod of length 6 (i.e. the second part of the first cut), the third is done to a rod of length 4 and the last cut is to a rod of length 3. The total cost is 7 + 6 + 4 + 3 = 20.\nRearranging the cuts to be [3, 5, 1, 4] for example will lead to a scenario with total cost = 16 (as shown in the example photo 7 + 4 + 3 + 2 = 16).\n                <\/p>\n\n                <h3>Example 2:<\/h3>\n                <p><strong>Input:<\/strong> n = 9, cuts = [5,6,1,4,2]<\/p>\n\n                <p><strong>Output:<\/strong> 22<\/p>\n                  <p>\n                  <code>Explanation:<\/code>\n                    If you try the given cuts ordering the cost will be 25.\nThere are much ordering with total cost <= 25, for example, the order [4, 6, 5, 2, 1] has total cost = 22 which is the minimum possible.\n                <\/p>\n\n                <!-- <h5>Example 3:<\/h5>\n                <p><strong>Input:<\/strong>\n                <\/p>\n                <p>\n                    nums1 = [0], m = 0\n                    nums2 = [1], n = 1\n                <\/p>    \n                <p><strong>Output:<\/strong>[1]<\/p>\n                  <p>\n                  <code>Explanation:<\/code>\n                 The arrays we are merging are [] and [1].\n                    The result of the merge is [1].\n                    Note that because m = 0, there are no elements in nums1. \n                    The 0 is only there to ensure the merge result can fit in nums1.\n                <\/p> -->\n\n                <h2>Constraints<\/h2>\n                <p><code>\n                  2 <= n <= 10<sup>6<\/sup>\n                <\/code><\/p>\n\n                <p>\n                  <code>1 <= cuts.length <= min(n - 1, 100)<\/code>\n                <\/p>\n\n                <p><code>1 <= cuts[i] <= n - 1<\/code><\/p>\n                <p>All the integers in <code>cuts<\/code> array are <strong>distinct<\/strong>.<\/p>\n                    \n                    <h2>Time Complexity:<\/h2>\n                    <li>\n                      <p><strong>Time Complexity = O(m * n<sup>2<\/sup>)<\/strong>\n                      <\/li>\n                    <h2>Space Complexity:<\/h2>\n                    <li>\n                      <p><strong>Space Complexity = O(n<sup>2<\/sup>)<\/strong><\/p>\n                    <\/li>\n\n                <h2>Approach<\/h2>\n                <ul>\n                  <li>\n                    We want to find the <strong>minimum cost to cut<\/strong> a stick of length n at given positions in cuts.\n                  <\/li>\n                  <li>Each <code>time we make a cut inside a segment (start, end)<\/code>, the cost of that cut is the current segment length (end - start).<\/li>\n                  <li>After cutting at <strong>position c<\/strong>, the stick splits into two smaller segments:\n                    <ul>\n                        <li><strong>Left segment<\/strong> \u2192 <code>(start, c)<\/code><\/li>\n                        <li><strong>Right segment <\/strong>\u2192 <code>(c, end)<\/code><\/li>\n                    <\/ul>\n                  <\/li>\n                  <li>Use <strong>DFS recursion<\/strong> to try every possible cut c that lies between start and end.<\/li>\n                  <li>For each valid cut:\n                    <ul>\n                        <li>Compute <code>currCost = (end - start) + dfs(start, c) + dfs(c, end)<\/code><\/li>\n                        <li>Keep the <code>minimum cost <\/code>among all choices.<\/li>\n                    <\/ul>\n                  <\/li>\n                  <li>Use a <strong>Map (dp)<\/strong> for memoization with key <code>\"start_end\"<\/code> to store the minimum cost for each segment so that the same subproblem is not recomputed.<\/li>\n                  <li>If no cut exists inside the <strong>segment<\/strong>, the cost is <code>0<\/code>.<\/li>\n                  <li>Start the recursion with the whole stick: <code>dfs(0, n)<\/code>.<\/li>\n                <\/ul>\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>n = 7, cuts = [1, 3, 4, 5]<\/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);\">\nGoal:\nWe want the minimum cost to cut a stick of length 7.\nEach cut costs the current stick length.\n\nCall dfs(0, 7)\n\nStep 1: dfs(0,7)\nPossible cuts between 0 and 7 \u2192 [1,3,4,5]\n\nTry cut = 1\nCost = (7 - 0) + dfs(0,1) + dfs(1,7)\n\n   dfs(0,1)\n   No cuts between 0 and 1\n   \u2192 return 0\n\n   dfs(1,7)\n   Possible cuts \u2192 [3,4,5]\n\n   Try cut = 3\n   Cost = (7 - 1) + dfs(1,3) + dfs(3,7)\n\n      dfs(1,3)\n      No cuts inside\n      \u2192 return 0\n\n      dfs(3,7)\n      Possible cuts \u2192 [4,5]\n\n      Try cut = 4\n      Cost = (7 - 3) + dfs(3,4) + dfs(4,7)\n\n         dfs(3,4) \u2192 0\n         dfs(4,7)\n         Possible cut \u2192 [5]\n\n         Try cut = 5\n         Cost = (7 - 4) + dfs(4,5) + dfs(5,7)\n              = 3 + 0 + 0\n              = 3\n\n         \u2192 dfs(4,7) = 3\n\n      Total cost = 4 + 0 + 3 = 7\n\n      Try cut = 5\n      Cost = (7 - 3) + dfs(3,5) + dfs(5,7)\n\n         dfs(3,5)\n         Possible cut \u2192 [4]\n\n         Cost = (5 - 3) + dfs(3,4) + dfs(4,5)\n              = 2 + 0 + 0\n              = 2\n\n         dfs(5,7) \u2192 0\n\n      Total cost = 4 + 2 + 0 = 6\n\n      Minimum = 6\n      \u2192 dfs(3,7) = 6\n\n   Total cost for cut=3\n   = 6 + 0 + 6\n   = 12\n\nContinue checking other first cuts (3,4,5) similarly.\n\nMinimum cost among all choices becomes:\n\u2192 16\n\nMemoization stores results for intervals like:\n(0,7), (1,7), (3,7), (4,7), (3,5) etc.\n  <\/pre> \n  \n  <p><strong>Output:<\/strong> <code>16<\/code><\/p> \n<\/div>\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\">\n            JavaScript\n        <\/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 minCost = function(n, cuts) {\n    let dp = new Map();\n\n    let dfs = (start, end) => {\n        if(start >= end) return 0;\n\n        let key = start + \"_\" + end;\n        if(dp.has(key)) return dp.get(key);\n\n        let minCost = Infinity;\n        for(let c of cuts) {\n            if(c > start && c < end) {\n                let currCost = (end - start) + \n                dfs(start, c) + dfs(c, end);\n                minCost = Math.min(minCost, currCost)\n            }\n        }\n\n        minCost = minCost == Infinity ? 0 : minCost;\n\n        dp.set(key, minCost);\n        return minCost;\n    }\n    return dfs(0, n);\n};\n    <\/code><\/pre>\n    <\/div>\n    <div class=\"wp_blog_code-tab-content\" data-lang=\"py\">\n        <pre><code class=\"language-python\">\ndef minCost(n, cuts):\n    dp = {}\n\n    def dfs(start, end):\n        if start >= end:\n            return 0\n\n        key = (start, end)\n        if key in dp:\n            return dp[key]\n\n        min_cost = float('inf')\n\n        for c in cuts:\n            if start < c < end:\n                curr_cost = (end - start) + dfs(start, c) + dfs(c, end)\n                min_cost = min(min_cost, curr_cost)\n\n        if min_cost == float('inf'):\n            min_cost = 0\n\n        dp[key] = min_cost\n        return min_cost\n\n    return dfs(0, n)\n    <\/code><\/pre>\n    <\/div>\n    <div class=\"wp_blog_code-tab-content\" data-lang=\"java\">\n        <pre><code class=\"language-java\">\nimport java.util.*;\n\nclass Solution {\n    Map<String, Integer> dp = new HashMap<>();\n\n    public int minCost(int n, int[] cuts) {\n        return dfs(0, n, cuts);\n    }\n\n    private int dfs(int start, int end, int[] cuts) {\n        if(start >= end) return 0;\n\n        String key = start + \"_\" + end;\n        if(dp.containsKey(key)) return dp.get(key);\n\n        int minCost = Integer.MAX_VALUE;\n\n        for(int c : cuts) {\n            if(c > start && c < end) {\n                int currCost = (end - start) + dfs(start, c, cuts) + dfs(c, end, cuts);\n                minCost = Math.min(minCost, currCost);\n            }\n        }\n\n        if(minCost == Integer.MAX_VALUE) minCost = 0;\n\n        dp.put(key, minCost);\n        return minCost;\n    }\n}\n<\/code><\/pre>\n    <\/div>\n    <div class=\"wp_blog_code-tab-content\" data-lang=\"cpp\">\n        <pre><code class=\"language-cpp\">\n#include &lt;bits\/stdc++.h&gt;\nusing namespace std;\n\nclass Solution {\npublic:\n    unordered_map<string, int> dp;\n\n    int dfs(int start, int end, vector<int>& cuts) {\n        if(start >= end) return 0;\n\n        string key = to_string(start) + \"_\" + to_string(end);\n        if(dp.count(key)) return dp[key];\n\n        int minCost = INT_MAX;\n\n        for(int c : cuts) {\n            if(c > start && c < end) {\n                int currCost = (end - start) + dfs(start, c, cuts) + dfs(c, end, cuts);\n                minCost = min(minCost, currCost);\n            }\n        }\n\n        if(minCost == INT_MAX) minCost = 0;\n\n        return dp[key] = minCost;\n    }\n\n    int minCost(int n, vector<int>& cuts) {\n        return dfs(0, n, cuts);\n    }\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;stdio.h&gt;\n#include &lt;limits.h&gt;\n#include &lt;string.h&gt;\n\nint dp[101][101];\n\nint dfs(int start, int end, int cuts[], int size) {\n    if(start >= end) return 0;\n\n    if(dp[start][end] != -1) return dp[start][end];\n\n    int minCost = INT_MAX;\n\n    for(int i = 0; i < size; i++) {\n        int c = cuts[i];\n        if(c > start && c < end) {\n            int currCost = (end - start) +\n                           dfs(start, c, cuts, size) +\n                           dfs(c, end, cuts, size);\n\n            if(currCost < minCost)\n                minCost = currCost;\n        }\n    }\n\n    if(minCost == INT_MAX) minCost = 0;\n\n    return dp[start][end] = minCost;\n}\n\nint minCost(int n, int cuts[], int size) {\n    memset(dp, -1, sizeof(dp));\n    return dfs(0, n, cuts, size);\n}\n   <\/code><\/pre>\n    <\/div>\n    <div class=\"wp_blog_code-tab-content\" data-lang=\"cs\">\n        <pre><code class=\"language-csharp\">\nusing System;\nusing System.Collections.Generic;\n\npublic class Solution {\n    Dictionary<string, int> dp = new Dictionary<string, int>();\n\n    public int MinCost(int n, int[] cuts) {\n        return dfs(0, n, cuts);\n    }\n\n    private int dfs(int start, int end, int[] cuts) {\n        if(start >= end) return 0;\n\n        string key = start + \"_\" + end;\n        if(dp.ContainsKey(key)) return dp[key];\n\n        int minCost = int.MaxValue;\n\n        foreach(int c in cuts) {\n            if(c > start && c < end) {\n                int currCost = (end - start) + dfs(start, c, cuts) + dfs(c, end, cuts);\n                minCost = Math.Min(minCost, currCost);\n            }\n        }\n\n        if(minCost == int.MaxValue) minCost = 0;\n\n        dp[key] = minCost;\n        return minCost;\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 wooden stick of length n units. The stick is labelled from 0 to n. For example, a stick of length 6 is labelled as follows: Given an integer array cuts where cuts[i] denotes a position you should perform a cut at. You should perform the cuts in order, you can<\/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":[322,260,176,175,211,811,810,174,172,173],"tags":[],"class_list":{"0":"post-11775","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-algorithms-and-data-structures","7":"category-c-c-plus-plus","8":"category-csharp","9":"category-cplusplus","10":"category-data-structures","11":"category-data-structures-and-algorithms","12":"category-dsa","13":"category-java","14":"category-javascript","15":"category-python"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/posts\/11775","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=11775"}],"version-history":[{"count":1,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/posts\/11775\/revisions"}],"predecessor-version":[{"id":11776,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/posts\/11775\/revisions\/11776"}],"wp:attachment":[{"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/media?parent=11775"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/categories?post=11775"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/tags?post=11775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}