{"id":934,"date":"2025-08-20T13:45:50","date_gmt":"2025-08-20T08:15:50","guid":{"rendered":"https:\/\/codeanddebug.in\/blog\/?p=934"},"modified":"2025-08-20T13:45:51","modified_gmt":"2025-08-20T08:15:51","slug":"maximum-points-you-can-obtain-from-cards","status":"publish","type":"post","link":"https:\/\/codeanddebug.in\/blog\/maximum-points-you-can-obtain-from-cards\/","title":{"rendered":"Maximum Points You Can Obtain from Cards | Leetcode 1423"},"content":{"rendered":"\n<p>The problem\u00a0<strong>\u201cMaximum Points\u00a0You Can Obtain\u00a0from Cards\u201d<\/strong>\u00a0asks\u00a0you to pick exactly\u00a0<strong>k<\/strong>\u00a0cards from\u00a0either the\u00a0<strong>start<\/strong>\u00a0or the\u00a0<strong>end<\/strong>\u00a0of the\u00a0array\u00a0<code>cardPoints<\/code>\u00a0to\u00a0maximize the\u00a0total points. The key insight\u00a0is that if you\u00a0must pick\u00a0<strong>k<\/strong>\u00a0cards from\u00a0the ends, then\u00a0you\u2019re effectively\u00a0choosing some\u00a0from the\u00a0<strong>left prefix<\/strong>\u00a0and the rest\u00a0from the\u00a0<strong>right suffix<\/strong>, and you can\u00a0try all splits\u00a0in\u00a0<strong>O(k)<\/strong>\u00a0after\u00a0a couple of passes.<\/p>\n\n\n\n<p>Here&#8217;s the [<strong><a href=\"https:\/\/leetcode.com\/problems\/maximum-points-you-can-obtain-from-cards\/description\/\" target=\"_blank\" rel=\"noreferrer noopener\"><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-purple-color\"><span style=\"text-decoration: underline;\">Problem Link<\/span><\/mark><\/a><\/strong>] to begin with.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>There are several cards&nbsp;<strong>arranged in a row<\/strong>, and each card has an associated number of points. The points are given in the integer array&nbsp;<code>cardPoints<\/code>.<\/p>\n\n\n\n<p>In one step, you can take one card from the beginning or from the end of the row. You have to take exactly&nbsp;<code>k<\/code>&nbsp;cards.<\/p>\n\n\n\n<p>Your score is the sum of the points of the cards you have taken.<\/p>\n\n\n\n<p>Given the integer array&nbsp;<code>cardPoints<\/code>&nbsp;and the integer&nbsp;<code>k<\/code>, return the&nbsp;<em>maximum score<\/em>&nbsp;you can obtain.<\/p>\n\n\n\n<p><strong>Example 1:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\"><strong>Input:<\/strong> cardPoints = [1,2,3,4,5,6,1], k = 3<br><strong>Output:<\/strong> 12<br><strong>Explanation:<\/strong> After the first step, your score will always be 1. However, choosing the rightmost card first will maximize your total score. The optimal strategy is to take the three cards on the right, giving a final score of 1 + 6 + 5 = 12.<\/pre>\n\n\n\n<p><strong>Example 2:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\"><strong>Input:<\/strong> cardPoints = [2,2,2], k = 2<br><strong>Output:<\/strong> 4<br><strong>Explanation:<\/strong> Regardless of which two cards you take, your score will always be 4.<\/pre>\n\n\n\n<p><strong>Example 3:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\"><strong>Input:<\/strong> cardPoints = [9,7,7,9,7,7,9], k = 7<br><strong>Output:<\/strong> 55<br><strong>Explanation:<\/strong> You have to take all the cards. Your score is the sum of points of all cards.<\/pre>\n\n\n\n<p><strong>Constraints:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><code>1 &lt;= cardPoints.length &lt;= 10<sup>5<\/sup><\/code><\/li>\n\n\n\n<li><code>1 &lt;= cardPoints[i] &lt;= 10<sup>4<\/sup><\/code><\/li>\n\n\n\n<li><code>1 &lt;= k &lt;= cardPoints.length<\/code><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Optimal Solution:&nbsp;<strong>Enumerate Left\/Right Split in O(k)<\/strong><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\">Why this works<\/h3>\n\n\n\n<p>Picking from ends is equivalent to leaving a&nbsp;<strong>middle subarray<\/strong>&nbsp;of length&nbsp;<code>n - k<\/code>&nbsp;unpicked. The complementary viewpoint: the maximum you can pick equals the&nbsp;<strong>total sum minus the minimum sum<\/strong>&nbsp;subarray of length&nbsp;<code>n - k<\/code>. The provided method is an equivalent and intuitive approach that directly builds left and right sums as you slide how many you take from each side.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Code <\/h3>\n\n\n\n<div class=\"wp-block-kevinbatdorf-code-block-pro cbp-has-line-numbers\" data-code-block-pro-font-family=\"Code-Pro-JetBrains-Mono\" style=\"font-size:.875rem;font-family:Code-Pro-JetBrains-Mono,ui-monospace,SFMono-Regular,Menlo,Monaco,Consolas,monospace;--cbp-line-number-color:#D4D4D4;--cbp-line-number-width:calc(2 * 0.6 * .875rem);line-height:1.5rem;--cbp-tab-width:2;tab-size:var(--cbp-tab-width, 2)\"><span style=\"display:block;padding:16px 0 0 16px;margin-bottom:-1px;width:100%;text-align:left;background-color:#1E1E1E\"><svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"54\" height=\"14\" viewBox=\"0 0 54 14\"><g fill=\"none\" fill-rule=\"evenodd\" transform=\"translate(1 1)\"><circle cx=\"6\" cy=\"6\" r=\"6\" fill=\"#FF5F56\" stroke=\"#E0443E\" stroke-width=\".5\"><\/circle><circle cx=\"26\" cy=\"6\" r=\"6\" fill=\"#FFBD2E\" stroke=\"#DEA123\" stroke-width=\".5\"><\/circle><circle cx=\"46\" cy=\"6\" r=\"6\" fill=\"#27C93F\" stroke=\"#1AAB29\" stroke-width=\".5\"><\/circle><\/g><\/svg><\/span><span role=\"button\" tabindex=\"0\" data-code=\"class Solution:\n    def maxScore(self, cardPoints: List[int], k: int) -&gt; int:\n        &quot;&quot;&quot;If k equals the length of the cardPoints,\n        the maximum points are achieved by picking all cards.\n        TC - O(N) (sum() function takes N time, where N is len cardPoints)\n        SC - O(1), No extra space used.&quot;&quot;&quot;\n        if k == len(cardPoints):\n            return sum(cardPoints)\n\n        max_sum = 0\n        left_sum = 0\n        right_sum = 0\n        n = len(cardPoints)\n        # Take all k from the left initially\n        for i in range(k):\n            left_sum += cardPoints[i]\n        max_sum = left_sum\n\n        # Slide: give back from left, take from right\n        right_index = n - 1\n        for i in range(k - 1, -1, -1):\n            left_sum -= cardPoints[i]          # remove one from the left window\n            right_sum += cardPoints[right_index]  # add one from the right tail\n            right_index -= 1\n            max_sum = max(max_sum, left_sum + right_sum)\n        return max_sum\" style=\"color:#D4D4D4;display:none\" aria-label=\"Copy\" class=\"code-block-pro-copy-button\"><svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" style=\"width:24px;height:24px\" fill=\"none\" viewBox=\"0 0 24 24\" stroke=\"currentColor\" stroke-width=\"2\"><path class=\"with-check\" stroke-linecap=\"round\" stroke-linejoin=\"round\" d=\"M9 5H7a2 2 0 00-2 2v12a2 2 0 002 2h10a2 2 0 002-2V7a2 2 0 00-2-2h-2M9 5a2 2 0 002 2h2a2 2 0 002-2M9 5a2 2 0 012-2h2a2 2 0 012 2m-6 9l2 2 4-4\"><\/path><path class=\"without-check\" stroke-linecap=\"round\" stroke-linejoin=\"round\" d=\"M9 5H7a2 2 0 00-2 2v12a2 2 0 002 2h10a2 2 0 002-2V7a2 2 0 00-2-2h-2M9 5a2 2 0 002 2h2a2 2 0 002-2M9 5a2 2 0 012-2h2a2 2 0 012 2\"><\/path><\/svg><\/span><pre class=\"shiki dark-plus\" style=\"background-color: #1E1E1E\" tabindex=\"0\"><code><span class=\"line\"><span style=\"color: #569CD6\">class<\/span><span style=\"color: #D4D4D4\"> <\/span><span style=\"color: #4EC9B0\">Solution<\/span><span style=\"color: #D4D4D4\">:<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">    <\/span><span style=\"color: #569CD6\">def<\/span><span style=\"color: #D4D4D4\"> <\/span><span style=\"color: #DCDCAA\">maxScore<\/span><span style=\"color: #D4D4D4\">(<\/span><span style=\"color: #9CDCFE\">self<\/span><span style=\"color: #D4D4D4\">, <\/span><span style=\"color: #9CDCFE\">cardPoints<\/span><span style=\"color: #D4D4D4\">: List[<\/span><span style=\"color: #4EC9B0\">int<\/span><span style=\"color: #D4D4D4\">], <\/span><span style=\"color: #9CDCFE\">k<\/span><span style=\"color: #D4D4D4\">: <\/span><span style=\"color: #4EC9B0\">int<\/span><span style=\"color: #D4D4D4\">) -&gt; <\/span><span style=\"color: #4EC9B0\">int<\/span><span style=\"color: #D4D4D4\">:<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">        <\/span><span style=\"color: #CE9178\">&quot;&quot;&quot;If k equals the length of the cardPoints,<\/span><\/span>\n<span class=\"line\"><span style=\"color: #CE9178\">        the maximum points are achieved by picking all cards.<\/span><\/span>\n<span class=\"line\"><span style=\"color: #CE9178\">        TC - O(N) (sum() function takes N time, where N is len cardPoints)<\/span><\/span>\n<span class=\"line\"><span style=\"color: #CE9178\">        SC - O(1), No extra space used.&quot;&quot;&quot;<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">        <\/span><span style=\"color: #C586C0\">if<\/span><span style=\"color: #D4D4D4\"> k == <\/span><span style=\"color: #DCDCAA\">len<\/span><span style=\"color: #D4D4D4\">(cardPoints):<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">            <\/span><span style=\"color: #C586C0\">return<\/span><span style=\"color: #D4D4D4\"> <\/span><span style=\"color: #DCDCAA\">sum<\/span><span style=\"color: #D4D4D4\">(cardPoints)<\/span><\/span>\n<span class=\"line\"><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">        max_sum = <\/span><span style=\"color: #B5CEA8\">0<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">        left_sum = <\/span><span style=\"color: #B5CEA8\">0<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">        right_sum = <\/span><span style=\"color: #B5CEA8\">0<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">        n = <\/span><span style=\"color: #DCDCAA\">len<\/span><span style=\"color: #D4D4D4\">(cardPoints)<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">        <\/span><span style=\"color: #6A9955\"># Take all k from the left initially<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">        <\/span><span style=\"color: #C586C0\">for<\/span><span style=\"color: #D4D4D4\"> i <\/span><span style=\"color: #C586C0\">in<\/span><span style=\"color: #D4D4D4\"> <\/span><span style=\"color: #DCDCAA\">range<\/span><span style=\"color: #D4D4D4\">(k):<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">            left_sum += cardPoints[i]<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">        max_sum = left_sum<\/span><\/span>\n<span class=\"line\"><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">        <\/span><span style=\"color: #6A9955\"># Slide: give back from left, take from right<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">        right_index = n - <\/span><span style=\"color: #B5CEA8\">1<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">        <\/span><span style=\"color: #C586C0\">for<\/span><span style=\"color: #D4D4D4\"> i <\/span><span style=\"color: #C586C0\">in<\/span><span style=\"color: #D4D4D4\"> <\/span><span style=\"color: #DCDCAA\">range<\/span><span style=\"color: #D4D4D4\">(k - <\/span><span style=\"color: #B5CEA8\">1<\/span><span style=\"color: #D4D4D4\">, -<\/span><span style=\"color: #B5CEA8\">1<\/span><span style=\"color: #D4D4D4\">, -<\/span><span style=\"color: #B5CEA8\">1<\/span><span style=\"color: #D4D4D4\">):<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">            left_sum -= cardPoints[i]          <\/span><span style=\"color: #6A9955\"># remove one from the left window<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">            right_sum += cardPoints[right_index]  <\/span><span style=\"color: #6A9955\"># add one from the right tail<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">            right_index -= <\/span><span style=\"color: #B5CEA8\">1<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">            max_sum = <\/span><span style=\"color: #DCDCAA\">max<\/span><span style=\"color: #D4D4D4\">(max_sum, left_sum + right_sum)<\/span><\/span>\n<span class=\"line\"><span style=\"color: #D4D4D4\">        <\/span><span style=\"color: #C586C0\">return<\/span><span style=\"color: #D4D4D4\"> max_sum<\/span><\/span><\/code><\/pre><\/div>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"step-by-step-explanation\">Step-by-Step Explanation<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Edge case:<\/strong>\u00a0If\u00a0<code>k == n<\/code>, picking all cards is optimal \u2192 return\u00a0<strong>sum(cardPoints)<\/strong>.<\/li>\n\n\n\n<li><strong>Initialize with all-left pick:<\/strong>\u00a0Sum the first\u00a0<strong>k<\/strong>\u00a0cards into\u00a0<strong>left_sum<\/strong>; set\u00a0<strong>max_sum = left_sum<\/strong>.<\/li>\n\n\n\n<li><strong>Iteratively swap picks:<\/strong>\u00a0For each iteration\u00a0<code>t<\/code>\u00a0from 1 to k:\n<ul class=\"wp-block-list\">\n<li>Remove the last included left card from\u00a0<strong>left_sum<\/strong>\u00a0(shrink left window by 1).<\/li>\n\n\n\n<li>Add one more card from the\u00a0<strong>right<\/strong>\u00a0end into\u00a0<strong>right_sum<\/strong>.<\/li>\n\n\n\n<li>Update\u00a0<strong>max_sum<\/strong>\u00a0with\u00a0<code>left_sum + right_sum<\/code>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>The loop explores all splits:\n<ul class=\"wp-block-list\">\n<li>x left and (k\u2212x) right for x = k, k\u22121, &#8230;, 0.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Return\u00a0<strong>max_sum<\/strong>.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"dry-run-quick\">Dry Run (Quick)<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>cardPoints = , k = 3<a href=\"https:\/\/leetcode.com\/problems\/maximum-points-you-can-obtain-from-cards\/description\/\" target=\"_blank\" rel=\"noreferrer noopener\"><\/a>\n<ul class=\"wp-block-list\">\n<li>Start with left 3: left_sum = 1+2+3 = 6, max_sum=6<\/li>\n\n\n\n<li>Move 1 from left to right: left_sum=6\u22123=3, right_sum=1 \u2192 3+1=4 \u2192 max=6<\/li>\n\n\n\n<li>Move another: left_sum=3\u22122=1, right_sum=1+6=7 \u2192 1+7=8 \u2192 max=8<\/li>\n\n\n\n<li>Move last: left_sum=1\u22121=0, right_sum=7+5=12 \u2192 12 \u2192 max=12<br>Answer:\u00a0<strong>12<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>This matches the known optimal choice: pick 6 and 5 from right, and 1 from left \u2192 total&nbsp;<strong>12<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"complexity\">Complexity<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Time:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Initial left sum:\u00a0<strong>O(k)<\/strong><\/li>\n\n\n\n<li>Sliding k times:\u00a0<strong>O(k)<\/strong><\/li>\n\n\n\n<li>Total:\u00a0<strong>O(k)<\/strong>\u00a0(plus O(1) checks), which is excellent when k \u2264 n.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Space:<\/strong>\u00a0<strong>O(1)<\/strong>\u00a0&#8211; only a handful of variables.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Alternative View:&nbsp;<strong>Minimum Window to Skip<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Another elegant approach: With total sum\u00a0<code>S<\/code>, the maximum pick of k cards equals\u00a0<code>S \u2212 minSumWindow(n \u2212 k)<\/code>.<\/li>\n\n\n\n<li>Use a sliding window of length\u00a0<code>n \u2212 k<\/code>\u00a0to find the\u00a0<strong>minimum<\/strong>\u00a0sum window to skip.<\/li>\n\n\n\n<li>Equivalent in result; the provided implementation is often more intuitive.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"tips-and-edge-cases\">Tips and Edge Cases<\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>k = 0<\/strong>\u00a0\u2192 result should be\u00a0<strong>0<\/strong>\u00a0(pick no cards).<\/li>\n\n\n\n<li><strong>k = n<\/strong>\u00a0\u2192 result is\u00a0<strong>sum(cardPoints)<\/strong>.<\/li>\n\n\n\n<li>Works for all\u00a0<strong>non-negative<\/strong>\u00a0card values; if negatives were allowed, the logic still holds since we control left\/right picks explicitly.<\/li>\n\n\n\n<li>Watch out for off-by-one in indices; use clear variable names like\u00a0<strong>right_index<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"takeaway\">Takeaway<\/h2>\n\n\n\n<p>This problem is a classic example of&nbsp;<strong>picking from ends<\/strong>&nbsp;with a fixed count. The optimal strategy enumerates how many to take from the&nbsp;<strong>left<\/strong>&nbsp;versus the&nbsp;<strong>right<\/strong>&nbsp;in&nbsp;<strong>O(k)<\/strong>&nbsp;and keeps a running best. It\u2019s clean,&nbsp;<strong>efficient<\/strong>, and extremely&nbsp;<strong>interview-friendly<\/strong>.<\/p>\n\n\n\n<div class=\"wp-block-buttons is-content-justification-center is-layout-flex wp-container-core-buttons-is-layout-16018d1d wp-block-buttons-is-layout-flex\">\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link wp-element-button\" href=\"https:\/\/codeanddebug.in\/course\/zero-to-hero-python-dsa\" target=\"_blank\" rel=\"noreferrer noopener\">Join our Advance DSA COURSE<\/a><\/div>\n<\/div>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><em>For any changes to the article, kindly email at <a href=\"mailto:code@codeanddebug.in\" target=\"_blank\" rel=\"noreferrer noopener\">code@codeanddebug.in<\/a> or contact us at <a href=\"tel:+91-9712928220\" target=\"_blank\" rel=\"noreferrer noopener\">+91-9712928220<\/a>.<\/em><\/p>\n","protected":false},"excerpt":{"rendered":"<p>The problem\u00a0\u201cMaximum Points\u00a0You Can Obtain\u00a0from Cards\u201d\u00a0asks\u00a0you to pick exactly\u00a0k\u00a0cards from\u00a0either the\u00a0start\u00a0or the\u00a0end\u00a0of the\u00a0array\u00a0cardPoints\u00a0to\u00a0maximize the\u00a0total points. The key insight\u00a0is that if you\u00a0must pick\u00a0k\u00a0cards from\u00a0the ends, then\u00a0you\u2019re effectively\u00a0choosing some\u00a0from the\u00a0left prefix\u00a0and the rest\u00a0from the\u00a0right suffix, and you can\u00a0try all splits\u00a0in\u00a0O(k)\u00a0after\u00a0a couple of passes. Here&#8217;s the [Problem Link] to begin with. There are several cards&nbsp;arranged in a row,<\/p>\n","protected":false},"author":1,"featured_media":936,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3,6],"tags":[19,40],"class_list":{"0":"post-934","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-data-structures-and-algorithm","8":"category-intermediate","9":"tag-medium","10":"tag-sliding-window-and-two-pointers"},"featured_image_src":"https:\/\/codeanddebug.in\/blog\/wp-content\/uploads\/2025\/08\/maximum-points-you-can-obtain-from-cards-featured-image.png","author_info":{"display_name":"codeanddebug","author_link":"https:\/\/codeanddebug.in\/blog\/author\/codeanddebug\/"},"_links":{"self":[{"href":"https:\/\/codeanddebug.in\/blog\/wp-json\/wp\/v2\/posts\/934","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/codeanddebug.in\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/codeanddebug.in\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/codeanddebug.in\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/codeanddebug.in\/blog\/wp-json\/wp\/v2\/comments?post=934"}],"version-history":[{"count":2,"href":"https:\/\/codeanddebug.in\/blog\/wp-json\/wp\/v2\/posts\/934\/revisions"}],"predecessor-version":[{"id":940,"href":"https:\/\/codeanddebug.in\/blog\/wp-json\/wp\/v2\/posts\/934\/revisions\/940"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/codeanddebug.in\/blog\/wp-json\/wp\/v2\/media\/936"}],"wp:attachment":[{"href":"https:\/\/codeanddebug.in\/blog\/wp-json\/wp\/v2\/media?parent=934"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/codeanddebug.in\/blog\/wp-json\/wp\/v2\/categories?post=934"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/codeanddebug.in\/blog\/wp-json\/wp\/v2\/tags?post=934"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}