{"id":9442,"date":"2025-08-18T20:13:53","date_gmt":"2025-08-18T14:43:53","guid":{"rendered":"https:\/\/namastedev.com\/blog\/?p=9442"},"modified":"2025-10-15T16:34:39","modified_gmt":"2025-10-15T11:04:39","slug":"combination-sum-2","status":"publish","type":"post","link":"https:\/\/namastedev.com\/blog\/combination-sum-2\/","title":{"rendered":"Combination Sum ||"},"content":{"rendered":"\n<!-- Evaluate Reverse Polish Notation 10-->\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; 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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 collection of candidate numbers (<code>candidates<\/code>) and a <code>target<\/code> number (target), find all unique combinations in <code>candidates<\/code> where the candidate numbers sum to <code>target<\/code>.\n    <\/p>\n\n    <p>Each number in <code>candidates<\/code> may only be used once in the combination.<\/p>\n\n    <p><strong>Note:<\/strong> The solution set must not contain duplicate combinations.<\/p>\n    \n    <p><strong>Example 1:<\/strong><\/p>\n    <p><strong>Input:<\/strong> candidates = [10,1,2,7,6,1,5], target = 8<\/p>\n    <p><strong>Output:<\/strong><code> \n        [\n        [1,1,6],\n        [1,2,5],\n        [1,7],\n        [2,6]\n        ]\n<\/code><\/p>\n\n    <p><strong>Example 2:<\/strong><\/p>\n    <p><strong>Input:<\/strong> candidates = [2,5,2,1,2], target = 5<\/p>\n    <p><strong>Output:<\/strong><code> \n            [\n            [1,2,2],\n            [5]\n            ]\n\n    <\/code><\/p>\n\n    <h2>Constraints:<\/h2>\n    <ul>\n        <li><code>1 <= candidates.length <= 100<\/code><\/li>\n        <li><code>1 <= candidates[i] <= 50<\/code><\/li>\n        <li><code>1 <= target <= 30<\/code><\/li>\n    <\/ul>\n\n    <h2>Approach:<\/h2>\n    <ul>\n        <li><strong>Sort the array<\/strong> \u2192 ensures duplicates are adjacent and helps in skipping them.<\/li>\n        <li><strong>Use backtracking<\/strong> to explore all possible combinations:\n        <ul>\n            <li>Keep reducing the <code>remainingSum<\/code>.<\/li>\n            <li>If it <code> becomes 0<\/code>, add the current path to result.<\/li>\n            <li>If it <code>becomes negative<\/code>, stop exploring further.<\/li>\n        <\/ul>\n        <\/li>\n\n        <li><strong>Skip duplicates<\/strong> \u2192 when iterating, if the current number is the same as the previous <code>(arr[i] === arr[i-1])<\/code> and not at the starting index of the loop, continue to next iteration.<\/li>\n    \n        <li><strong>Move forward<\/strong> (use <code>i+1<\/code>) because each element can only be used once.<\/li>\n    <\/ul>\n\n    <h2>Time Complexity:<\/h2>\n    <li><p><strong>Time Complexity = O(2<sup>n<\/sup> * n)<\/strong><\/p><\/li> \n    <h2>Space Complexity:<\/h2>\n    <li><p><strong>Space Complexity =  O(2<sup>n<\/sup> * n)<\/strong> (output) + O(n) (stack)<\/p><\/li>\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\n  <p><strong>Input:<\/strong> <code>arr = [2, 3, 6, 7], target = 7<\/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);\">\nStep 0: Start Function combinationSum2([2, 3, 6, 7], 7)\nSort arr \u2192 arr = [2, 3, 6, 7]\n\nInitialize:\nresult = []\npath = []\n\nCall backtrack(7, [], 0)\n\nLoop i = 0 \u2192 arr[0] = 2\npath.push(2) \u2192 path = [2]\nCall backtrack(5, [2], 1)\n\n  Loop i = 1 \u2192 arr[1] = 3\n  path.push(3) \u2192 path = [2, 3]\n  Call backtrack(2, [2, 3], 2)\n\n    Loop i = 2 \u2192 arr[2] = 6 \u2192 exceeds remainingSum \u2192 stop\n    Loop i = 3 \u2192 arr[3] = 7 \u2192 exceeds remainingSum \u2192 stop\n\n  path.pop() \u2192 path = [2]\n\n  Loop i = 2 \u2192 arr[2] = 6\n  path.push(6) \u2192 path = [2, 6]\n  Call backtrack(-1, [2, 6], 3)\n    remainingSum < 0 \u2192 return\n  path.pop() \u2192 path = [2]\n\n  Loop i = 3 \u2192 arr[3] = 7\n  path.push(7) \u2192 path = [2, 7]\n  Call backtrack(-2, [2, 7], 4)\n    remainingSum < 0 \u2192 return\n  path.pop() \u2192 path = [2]\n\nLoop ends\npath.pop() \u2192 path = []\n\nLoop i = 1 \u2192 arr[1] = 3\npath.push(3) \u2192 path = [3]\nCall backtrack(4, [3], 2)\n\n  Loop i = 2 \u2192 arr[2] = 6\n  path.push(6) \u2192 path = [3, 6]\n  Call backtrack(-2, [3, 6], 3)\n    remainingSum < 0 \u2192 return\n  path.pop() \u2192 path = [3]\n\n  Loop i = 3 \u2192 arr[3] = 7\n  path.push(7) \u2192 path = [3, 7]\n  Call backtrack(-3, [3, 7], 4)\n    remainingSum < 0 \u2192 return\n  path.pop() \u2192 path = [3]\n\nLoop ends\npath.pop() \u2192 path = []\n\nLoop i = 2 \u2192 arr[2] = 6\npath.push(6) \u2192 path = [6]\nCall backtrack(1, [6], 3)\n\n  Loop i = 3 \u2192 arr[3] = 7 \u2192 exceeds remainingSum \u2192 stop\npath.pop() \u2192 path = []\n\nLoop i = 3 \u2192 arr[3] = 7\npath.push(7) \u2192 path = [7]\nCall backtrack(0, [7], 4)\n  remainingSum == 0 \u2192 result.push([7])\npath.pop() \u2192 path = []\n\nLoop ends\n\nStep 3: End\nReturn result = [[7]]\n  <\/pre>\n\n  <p><strong>Output:<\/strong> \n    <code>[[7]]<\/code>\n  <\/p>\n\n  <p><strong>Explanation:<\/strong> Unlike <code>combinationSum<\/code>, here each element can be used at most once (recursion advances with <code>i+1<\/code>). That\u2019s why <code>[2, 2, 3]<\/code> is <em>not<\/em> included, leaving only <code>[7]<\/code> as the valid combination.<\/p>\n  \n<\/div>\n\n<\/div>\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 combinationSum2 = function(arr, target) {\n    let result = [];\n    arr.sort((a,b) => (a-b));\n\n    let backtrack = (remainingSum, path, start) => {\n        if(remainingSum == 0){\n            result.push([...path]);\n        }\n        if(remainingSum <= 0) return;\n\n        for(let i=start; i<arr.length &#038;&#038; arr[i] <= remainingSum; i++){\n            if(i > start && arr[i-1] === arr[i])\n            continue;\n            path.push(arr[i]);\n            backtrack(remainingSum - arr[i], path, i+1);\n            path.pop();\n        }\n    }\n    backtrack(target, [], 0);\n    return result;\n};\n<\/code><\/pre>\n<\/div>\n\n<div class=\"wp_blog_code-tab-content\" data-lang=\"py\">\n<pre><code class=\"language-python\">\nfrom typing import List\n\nclass Solution:\n    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:\n        candidates.sort()\n        result = []\n\n        def backtrack(remaining, path, start):\n            if remaining == 0:\n                result.append(path[:])\n                return\n            for i in range(start, len(candidates)):\n                if candidates[i] > remaining:\n                    break\n                if i > start and candidates[i] == candidates[i-1]:\n                    continue\n                path.append(candidates[i])\n                backtrack(remaining - candidates[i], path, i+1)\n                path.pop()\n\n        backtrack(target, [], 0)\n        return result\n\n<\/code><\/pre>\n<\/div>\n\n<div class=\"wp_blog_code-tab-content\" data-lang=\"java\">\n<pre><code class=\"language-java\">\nimport java.util.*;\n\nclass Solution {\n    public List<List<Integer>> combinationSum2(int[] candidates, int target) {\n        Arrays.sort(candidates);\n        List<List<Integer>> result = new ArrayList<>();\n        backtrack(candidates, target, 0, new ArrayList<>(), result);\n        return result;\n    }\n\n    private void backtrack(int[] arr, int remaining, int start, List<Integer> path, List<List<Integer>> result) {\n        if (remaining == 0) {\n            result.add(new ArrayList<>(path));\n            return;\n        }\n        for (int i = start; i < arr.length &#038;&#038; arr[i] <= remaining; i++) {\n            if (i > start && arr[i] == arr[i-1]) continue;\n            path.add(arr[i]);\n            backtrack(arr, remaining - arr[i], i+1, path, result);\n            path.remove(path.size() - 1);\n        }\n    }\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\">\nclass Solution {\npublic:\n    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {\n        sort(candidates.begin(), candidates.end());\n        vector<vector<int>> result;\n        vector<int> path;\n        backtrack(candidates, target, 0, path, result);\n        return result;\n    }\nprivate:\n    void backtrack(vector<int>& arr, int remaining, int start, vector<int>& path, vector<vector<int>>& result) {\n        if (remaining == 0) {\n            result.push_back(path);\n            return;\n        }\n        for (int i = start; i < arr.size() &#038;&#038; arr[i] <= remaining; i++) {\n            if (i > start && arr[i] == arr[i-1]) continue;\n            path.push_back(arr[i]);\n            backtrack(arr, remaining - arr[i], i + 1, path, result);\n            path.pop_back();\n        }\n    }\n};\n<\/code><\/pre>\n<\/div>\n\n<div class=\"wp_blog_code-tab-content\" data-lang=\"c\">\n<pre><code class=\"language-c\">\nint cmp(const void* a, const void* b) {\n    return (*(int*)a - *(int*)b);\n}\n\nvoid backtrack(int* arr, int arrSize, int remaining, int start,\n               int* path, int pathSize, int** results, int* colSizes, int* returnSize) {\n    if (remaining == 0) {\n        results[*returnSize] = (int*)malloc(pathSize * sizeof(int));\n        for (int i = 0; i < pathSize; i++) results[*returnSize][i] = path[i];\n        colSizes[*returnSize] = pathSize;\n        (*returnSize)++;\n        return;\n    }\n    for (int i = start; i < arrSize &#038;&#038; arr[i] <= remaining; i++) {\n        if (i > start && arr[i] == arr[i-1]) continue;\n        path[pathSize] = arr[i];\n        backtrack(arr, arrSize, remaining - arr[i], i+1, path, pathSize+1, results, colSizes, returnSize);\n    }\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;\nusing System.Collections.Generic;\n\npublic class Solution {\n    public IList<IList<int>> CombinationSum2(int[] candidates, int target) {\n        Array.Sort(candidates);\n        IList<IList<int>> result = new List<IList<int>>();\n        Backtrack(candidates, target, 0, new List<int>(), result);\n        return result;\n    }\n\n    private void Backtrack(int[] arr, int remaining, int start, List<int> path, IList<IList<int>> result) {\n        if (remaining == 0) {\n            result.Add(new List<int>(path));\n            return;\n        }\n        for (int i = start; i < arr.Length &#038;&#038; arr[i] <= remaining; i++) {\n            if (i > start && arr[i] == arr[i-1]) continue;\n            path.Add(arr[i]);\n            Backtrack(arr, remaining - arr[i], i+1, path, result);\n            path.RemoveAt(path.Count - 1);\n        }\n    }\n}\n<\/code><\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\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","protected":false},"excerpt":{"rendered":"<p>\ud83c\udf19 Problem Statement: Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sum to target. Each number in candidates may only be used once in the combination. Note: The solution set must not contain duplicate combinations. Example 1: Input: candidates = [10,1,2,7,6,1,5],<\/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-9442","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\/9442","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=9442"}],"version-history":[{"count":2,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/posts\/9442\/revisions"}],"predecessor-version":[{"id":10331,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/posts\/9442\/revisions\/10331"}],"wp:attachment":[{"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/media?parent=9442"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/categories?post=9442"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/namastedev.com\/blog\/wp-json\/wp\/v2\/tags?post=9442"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}