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reorder-routes-to-make-all-paths-lead-to-the-city-zero | Java (Using Queue | simple dfs) | java-using-queue-simple-dfs-by-dhruvpanw-ennc | First approach:(Using a queue)\n1. Make a boolean array hasReached[] where hasReached[i] means ith vertex can now reach 0.\n2. Push all edges in a queue\n3. Pop | dhruvpanwar31 | NORMAL | 2021-08-13T21:32:26.651543+00:00 | 2021-08-13T21:32:26.651585+00:00 | 458 | false | First approach:(Using a queue)\n1. Make a boolean array hasReached[] where hasReached[i] means ith vertex can now reach 0.\n2. Push all edges in a queue\n3. Pop queue, see if in the edge (u,v) we have a node that reaches 0.( hasReached[u] or hasReached[v] ) \n4. Let the node that reaches 0 be \'u\'.\n5. So now for \'v\... | 4 | 0 | ['Depth-First Search', 'Queue', 'Java'] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | C++ | DFS with explanation | c-dfs-with-explanation-by-xpharry-38vt | Assume each city can reach the city 0, then from the city 0, it can always reach all the cities if the connections are all reversed. Then we can traverse this r | xpharry | NORMAL | 2021-03-09T23:15:53.069256+00:00 | 2021-03-09T23:15:53.069289+00:00 | 331 | false | Assume each city can reach the city 0, then from the city 0, it can always reach all the cities if the connections are all reversed. Then we can traverse this reversed directed-graph and check each connection without repeatation, we can get the minimum number of edges to change by counting the connection that can be fo... | 4 | 1 | [] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Java || BFS || O(n) | java-bfs-on-by-legendarycoder-8hkb | \n public int minReorder(int n, int[][] connections) {\n int count = 0;\n\t\tList[] graph = new ArrayList[n];\n\t\tfor (int i = 0; i < n; i++)\n\t\t\t | LegendaryCoder | NORMAL | 2021-02-20T06:48:17.723022+00:00 | 2021-02-20T06:48:17.723051+00:00 | 285 | false | \n public int minReorder(int n, int[][] connections) {\n int count = 0;\n\t\tList<Integer>[] graph = new ArrayList[n];\n\t\tfor (int i = 0; i < n; i++)\n\t\t\tgraph[i] = new ArrayList<Integer>();\n\n\t\tfor (int[] connection : connections) {\n\t\t\tgraph[connection[0]].add(connection[1]);\n\t\t\tgraph[connect... | 4 | 0 | [] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Java 100%, simplified topological sort, no tree/graph needed | java-100-simplified-topological-sort-no-fa2wg | Simplified topological sort, keep iterate cities that can be reached by flipping direction until all cities can be reached.\n\nclass Solution {\n public int | tao68 | NORMAL | 2021-01-04T22:41:43.470108+00:00 | 2021-01-04T22:41:43.470146+00:00 | 302 | false | Simplified topological sort, keep iterate cities that can be reached by flipping direction until all cities can be reached.\n```\nclass Solution {\n public int minReorder(int n, int[][] connections) {\n boolean[] reach = new boolean[n]; // this array tracks cities reachable so far\n reach[0] = true; //... | 4 | 0 | [] | 2 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Javascript BFS, easy to understand. | javascript-bfs-easy-to-understand-by-and-4jd9 | Thinking process:\nSince the target is to minimize the number of road to reverse so that all cities can reach city 0, then for those directly connected to 0, sa | andyoung | NORMAL | 2020-10-03T22:28:30.287882+00:00 | 2020-10-03T22:58:45.890791+00:00 | 721 | false | **Thinking process:**\nSince the target is to minimize the number of road to reverse so that all cities can reach city `0`, then for those directly connected to `0`, say `nodes` , if it\'s already in the right direction, no change; if otherwise, reverse it. Same idea applies to the remaining cities that connected to `n... | 4 | 0 | ['Breadth-First Search', 'JavaScript'] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | [Python] 7-line DFS solution | python-7-line-dfs-solution-by-m-just-2n03 | python\ndef minReorder(self, n: int, connections: List[List[int]]) -> int:\n edges = collections.defaultdict(list)\n for a, b in connections:\n edg | m-just | NORMAL | 2020-09-21T19:51:19.182805+00:00 | 2020-09-21T19:51:19.182837+00:00 | 425 | false | ```python\ndef minReorder(self, n: int, connections: List[List[int]]) -> int:\n edges = collections.defaultdict(list)\n for a, b in connections:\n edges[a].append((b, 1))\n edges[b].append((a, 0))\n\n def dfs(i, p):\n return sum(d + dfs(j, i) for j, d in edges[i] if j != p)\n\n return d... | 4 | 0 | ['Depth-First Search', 'Python'] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Python Beats 100% with Explanation (Better than Editorial) | python-beats-100-with-explanation-better-p63j | Approach: BFS/Greedy Search with a Seen SetProblem Breakdown
You are givenncities labeled from0ton-1and a list ofconnections, whereconnections[i] = [a, b]repres | jxmils | NORMAL | 2025-02-19T14:20:00.915629+00:00 | 2025-02-19T14:20:00.915629+00:00 | 744 | false | ### `Approach: BFS/Greedy Search with a Seen Set`
#### `Problem Breakdown`
- You are given `n` cities labeled from `0` to `n-1` and a list of `connections`, where `connections[i] = [a, b]` represents a directed road from city `a` to city `b`.
- You must reorder some roads so that **every city can reach city `0`**.
- R... | 3 | 0 | ['Python3'] | 3 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Python, beats 87%, clean solution with recursion | python-beats-87-clean-solution-with-recu-wsot | IntuitionDespite directioned edge, we should setup graph with biderectional edges, with marking current direction.
Setup recursion, which will consider conectio | dev_lvl_80 | NORMAL | 2025-02-03T06:43:51.289009+00:00 | 2025-02-03T06:43:51.289009+00:00 | 488 | false | # Intuition
Despite directioned edge, we should setup graph with biderectional edges, with marking current direction.
Setup recursion, which will consider conection existance and current direction
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
O(N)
- Space complex... | 3 | 0 | ['Python3'] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Easy to Understand simple Bfs beats 100% O(n) | easy-to-understand-simple-bfs-beats-100-en6f0 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | hemanth2-ko | NORMAL | 2024-10-23T17:02:25.007594+00:00 | 2024-10-23T17:02:25.007631+00:00 | 240 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:... | 3 | 0 | ['C++'] | 3 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Solution By Dare2Solve | Detailed Explanation | Clean Code | solution-by-dare2solve-detailed-explanat-59kg | Explanation []\nauthorslog.com/blog/a89c2S0zHB\n\n# Code\n\ncpp []\nclass Solution {\npublic:\n int minReorder(int n, vector<vector<int>>& connections) {\n | Dare2Solve | NORMAL | 2024-08-22T08:44:24.649503+00:00 | 2024-08-22T08:44:24.649532+00:00 | 1,265 | false | ```Explanation []\nauthorslog.com/blog/a89c2S0zHB\n```\n# Code\n\n```cpp []\nclass Solution {\npublic:\n int minReorder(int n, vector<vector<int>>& connections) {\n vector<vector<int>> grid(n);\n vector<int> vis(n, 0);\n\n for (const auto& conn : connections) {\n int src = conn[0];\n ... | 3 | 1 | ['Breadth-First Search', 'Graph', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Java Level-Order Traversal Solution Using BFS with Detailed Explanations | java-level-order-traversal-solution-usin-bxqr | Intuition\nWhen should we change the direction of a route? If the route starts from a city close to city 0 and ends at a city far from city 0, we need to change | Michael-Qu | NORMAL | 2024-06-17T22:38:51.968529+00:00 | 2024-06-17T22:38:51.968550+00:00 | 287 | false | # Intuition\nWhen should we change the direction of a route? If the route starts from a city close to city 0 and ends at a city far from city 0, we need to change its direction since it brings people "away" from city 0. Otherwise we don\'t need to change its direction since it already meets our need. In this case, once... | 3 | 0 | ['Tree', 'Breadth-First Search', 'Graph', 'Queue', 'Java'] | 1 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Python | BFS | DFS | Graphs | Easy to Understand | python-bfs-dfs-graphs-easy-to-understand-zlw9 | Approach: Create a graph with both original edges and their reversed counterparts (with scores 1 and 0, respectively). Traverse the graph starting from node \' | Coder1918 | NORMAL | 2024-04-01T20:52:22.027053+00:00 | 2024-04-01T21:37:17.015638+00:00 | 933 | false | **Approach:** Create a graph with both original edges and their reversed counterparts (with scores 1 and 0, respectively). Traverse the graph starting from node \'0\', incrementing a count for each original edge that points away from node \'0\'. This count represents the necessary reorientations.\nBelow code provides ... | 3 | 0 | ['Depth-First Search', 'Breadth-First Search', 'Graph', 'Python3'] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Breadth-first search | breadth-first-search-by-burkh4rt-dnfi | Approach\nFirst create a lookup dictionary adj that maps vertices to adjacent/contiguous edges. Initialize 0 as the current vertex set curr and the set seen of | burkh4rt | NORMAL | 2024-01-28T15:02:28.931059+00:00 | 2024-01-28T15:02:28.931094+00:00 | 911 | false | # Approach\nFirst create a lookup dictionary `adj` that maps vertices to adjacent/contiguous edges. Initialize 0 as the current vertex set `curr` and the set `seen` of vertices that have been visited. Iteratively update `curr` by adding vertices adjacent to the current set if they have not already been seen. Edges `e` ... | 3 | 0 | ['Breadth-First Search', 'Python3'] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Reorder Nodes | reorder-nodes-by-noob_1012-q88m | \n/**\n * @param {number} n\n * @param {number[][]} connections\n * @return {number}\n */\nvar minReorder = function(n, connections) {\n let visitSet = new S | noob_1012 | NORMAL | 2024-01-08T14:13:58.358096+00:00 | 2024-01-08T14:14:21.137845+00:00 | 151 | false | ```\n/**\n * @param {number} n\n * @param {number[][]} connections\n * @return {number}\n */\nvar minReorder = function(n, connections) {\n let visitSet = new Set();\n let count = 0;\n let neighbours = new Map();\n let edges = new Set();\n for(let i=0;i<n;i++){\n neighbours[i] = [];\n }\n \n... | 3 | 0 | ['Depth-First Search', 'JavaScript'] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | [C] DFS, graph | c-dfs-graph-by-leetcodebug-im4d | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | leetcodebug | NORMAL | 2023-07-17T15:36:33.281927+00:00 | 2023-07-17T15:36:33.281977+00:00 | 24 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here... | 3 | 0 | ['Depth-First Search', 'Graph', 'C'] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Java easy with bfs | java-easy-with-bfs-by-mohd_danish_khan-g019 | Approach\ncreate a map {city : neighbours} and set of all the edges\n\nnow do a bfs and check on every city if there is an edge (neigh, city) in our edge set, i | Mohd_Danish_Khan | NORMAL | 2023-07-13T21:53:39.227780+00:00 | 2023-07-13T21:53:39.227797+00:00 | 632 | false | # Approach\ncreate a map {city : neighbours} and set of all the edges\n\nnow do a bfs and check on every city if there is an edge (neigh, city) in our edge set, if its not their increase the changes count. and move on to next city. keep visited set to avoid checking the already checked node.\n\n# Code\n```\nclass Solut... | 3 | 0 | ['Java'] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Solution in C++ | solution-in-c-by-ashish_madhup-x8ch | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ashish_madhup | NORMAL | 2023-03-24T15:16:22.745508+00:00 | 2023-03-24T15:16:22.745532+00:00 | 203 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 0 | ['C++'] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | reorder routes to make all paths lead to the city zero | reorder-routes-to-make-all-paths-lead-to-e7eu | Intuition\nstore all given connections in a adjaceny list smartly (in pair form) such that given connections can be checked and created connection can be checke | nobita_n0bi | NORMAL | 2023-03-24T13:56:23.061090+00:00 | 2023-03-24T13:56:23.061127+00:00 | 340 | false | # Intuition\nstore all given connections in a adjaceny list smartly (in pair form) such that given connections can be checked and created connection can be checked at the time of dfs function working\n\n# Approach\nnow let\'s talk about complete approach i stored all given connection in a adjaency list of pairs such th... | 3 | 0 | ['Greedy', 'Graph', 'C++'] | 2 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Easy understand graph problems(C++) | easy-understand-graph-problemsc-by-vikas-z7nr | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | vikash_2003 | NORMAL | 2023-03-24T12:35:21.127995+00:00 | 2023-03-24T12:35:21.128048+00:00 | 6,118 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 0 | ['C++'] | 1 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | Using BFS C++ Easy Solution | using-bfs-c-easy-solution-by-kisna-i69m | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | kisna | NORMAL | 2023-03-24T12:01:53.872379+00:00 | 2023-03-24T12:01:53.872410+00:00 | 888 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 0 | ['Breadth-First Search', 'C++'] | 0 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | C# DFS | c-dfs-by-dmitriy-maksimov-13sq | Approach\nTreat the graph as undirected. Start a dfs from the root, if you come across an edge in the forward direction, you need to reverse the edge.\n\n# Comp | dmitriy-maksimov | NORMAL | 2023-03-24T10:39:52.431748+00:00 | 2023-03-24T10:39:52.431779+00:00 | 667 | false | # Approach\nTreat the graph as undirected. Start a dfs from the root, if you come across an edge in the forward direction, you need to reverse the edge.\n\n# Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n```\npublic class Solution\n{\n private readonly Dictionary<int, List<(int cit... | 3 | 0 | ['C#'] | 2 |
reorder-routes-to-make-all-paths-lead-to-the-city-zero | dfs | dfs-by-mr_stark-3nhb | \nclass Solution {\npublic:\n \n void dfs(vector<vector<pair<int,int>>> &g,int s, int &cnt,vector<bool> &v)\n {\n if(v[s]){\n return | mr_stark | NORMAL | 2023-03-24T08:46:27.338388+00:00 | 2023-03-24T08:50:22.777067+00:00 | 275 | false | ```\nclass Solution {\npublic:\n \n void dfs(vector<vector<pair<int,int>>> &g,int s, int &cnt,vector<bool> &v)\n {\n if(v[s]){\n return ;\n }\n v[s] = true;\n \n for(auto &i: g[s])\n { \n if(!v[i.first]){\n cnt+=i.second;\n ... | 3 | 0 | ['C'] | 0 |
smallest-string-with-a-given-numeric-value | [Java/C++] Easiest Possible Exaplained!! | javac-easiest-possible-exaplained-by-hi-4imxc | How\'s going Ladies - n - Gentlemen, today we are going to solve another coolest problem i.e. Smallest String With A Given Numeric Value\n\nWhat the problem is | hi-malik | NORMAL | 2022-03-22T00:54:32.589593+00:00 | 2022-03-22T01:01:25.144052+00:00 | 7,304 | false | How\'s going Ladies - n - Gentlemen, today we are going to solve another coolest problem i.e. **Smallest String With A Given Numeric Value**\n\nWhat the problem is saying, we have to generate the **string** with **numeric sum** equals to `k` & have the `n` characters\n\nLet\'s understand with an example,\n\n**Input**: ... | 220 | 7 | [] | 30 |
smallest-string-with-a-given-numeric-value | [C++] Simple O(N) with explanation | c-simple-on-with-explanation-by-cgsgak1-rjhq | The idea is simple:\n1) Build the string of length k, which consists of letter \'a\' (lexicographically smallest string).\n2) Increment string from right to lef | cgsgak1 | NORMAL | 2020-11-22T04:01:59.788252+00:00 | 2021-03-18T20:13:00.454231+00:00 | 6,930 | false | The idea is simple:\n1) Build the string of length k, which consists of letter \'a\' (lexicographically smallest string).\n2) Increment string from right to left until it\'s value won\'t reach the target.\n\nTime complexity O(n), space complexity O(1) - no additional space used except for result string.\n```\nclass Sol... | 131 | 10 | ['C', 'C++'] | 13 |
smallest-string-with-a-given-numeric-value | 🌻|| C++|| Easy Approach || PROPER EXPLANATION | c-easy-approach-proper-explanation-by-ab-kik0 | ALGORITHM\n\nStep - 1 Initialize a string s with n number of \'a\' to make it as a smallest possible lexiographic.\nStep - 2 Count of \'a = 1\' , so if i have | Abhay_Rautela | NORMAL | 2022-03-22T01:24:48.597520+00:00 | 2022-03-22T01:24:48.597550+00:00 | 7,437 | false | ### ALGORITHM\n```\nStep - 1 Initialize a string s with n number of \'a\' to make it as a smallest possible lexiographic.\nStep - 2 Count of \'a = 1\' , so if i have n number of \'a\' it means it sum is also \'n\'.\nStep - 3 Reduce the value of k by n (i.e k=k-n);\nStep - 4 Start traversing the string from the end bec... | 88 | 1 | ['C'] | 15 |
smallest-string-with-a-given-numeric-value | C++ Reverse Fill | c-reverse-fill-by-votrubac-iazt | Generate the initial string with all \'a\' characters. This will reduce k by n. \n \nThen, turn rightmost \'a\' into \'z\' (\'a\' + 25, or \'a\' + k) while k | votrubac | NORMAL | 2020-11-22T04:13:25.815258+00:00 | 2020-11-22T06:21:44.391096+00:00 | 3,634 | false | Generate the initial string with all \'a\' characters. This will reduce `k` by `n`. \n \nThen, turn rightmost \'a\' into \'z\' (\'a\' + 25, or \'a\' + k) while `k` is positive.\n\n```cpp\nstring getSmallestString(int n, int k) {\n string res = string(n, \'a\');\n k -= n;\n while (k > 0) {\n res[--n] ... | 85 | 7 | [] | 10 |
smallest-string-with-a-given-numeric-value | ✔️ [Python3] GREEDY FILLING (🌸¬‿¬), Explained | python3-greedy-filling-_-explained-by-ar-itri | UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.\n\nSince we are forming the lexicographically smallest string, | artod | NORMAL | 2022-03-22T02:12:40.544744+00:00 | 2022-03-22T03:22:21.950923+00:00 | 4,360 | false | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nSince we are forming the lexicographically smallest string, we just simply fill our result with `a`s ( o\u02D8\u25E1\u02D8o). But hold on, that result will not necessarily have the required score c(\uFF9F.\uFF9F*) U... | 66 | 2 | ['Python3'] | 7 |
smallest-string-with-a-given-numeric-value | ✅ C++ || Dry Run || Easy || Simple & Short || O(N) | c-dry-run-easy-simple-short-on-by-knockc-vmd3 | 1663. Smallest String With A Given Numeric Value\nKNOCKCAT\n\n\n1. Easy C++\n2. Line by Line Explanation with Comments.\n3. Detailed Explanation \u2705\n4. Lexi | knockcat | NORMAL | 2022-03-22T00:47:14.032205+00:00 | 2022-12-31T07:04:57.128016+00:00 | 2,027 | false | # 1663. Smallest String With A Given Numeric Value\n**KNOCKCAT**\n\n```\n1. Easy C++\n2. Line by Line Explanation with Comments.\n3. Detailed Explanation \u2705\n4. Lexicographically smallest string of length n and sum k with Dry Run.\n5. Please Upvote if it helps\u2B06\uFE0F\n```\n``` ```\n[LeetCode](http://github.com... | 65 | 2 | ['String', 'C'] | 9 |
smallest-string-with-a-given-numeric-value | [Java/Python 3] Two O(n) codes w/ brief explanation and analysis. | javapython-3-two-on-codes-w-brief-explan-v67l | Mehtod 1: Greedily reversely place values\n\n1. Make sure each value of the n characters is at least 1: initialized all as \'a\';\n2. Put as more value at the | rock | NORMAL | 2020-11-22T04:25:28.789229+00:00 | 2022-03-22T12:50:20.922441+00:00 | 4,650 | false | **Mehtod 1: Greedily reversely place values**\n\n1. Make sure each value of the n characters is at least `1`: initialized all as `\'a\'`;\n2. Put as more value at the end of the String as possible.\n```java\n public String getSmallestString(int n, int k) {\n k -= n;\n char[] ans = new char[n];\n ... | 65 | 1 | ['Java', 'Python3'] | 9 |
smallest-string-with-a-given-numeric-value | [Python] O(n) math solution, explained | python-on-math-solution-explained-by-dba-yiis | In this problem we need to construct lexicographically smallest string with given properties, and here we can use greedy strategy: try to take as small symbol a | dbabichev | NORMAL | 2021-01-28T08:59:35.172473+00:00 | 2021-01-28T08:59:35.172508+00:00 | 2,378 | false | In this problem we need to construct **lexicographically smallest string** with given properties, and here we can use **greedy** strategy: try to take as small symbol as possible until we can. Smallest letter means `a`, so our string in general case will look like this:\n\n`aa...aa?zz...zz`,\n\nwhere some of the groups... | 59 | 8 | ['Greedy'] | 5 |
smallest-string-with-a-given-numeric-value | ✅ Explained with Comments | Easy | 90% | O(n) | JS | explained-with-comments-easy-90-on-js-by-uetk | \n/* \nIdea is to build the string from the end.\n1. We create an array (numericArr) to store string (1 indexed) corresponding to it\'s \nnumeric value as given | _kapi1 | NORMAL | 2022-03-22T02:09:30.664841+00:00 | 2022-03-22T02:09:30.664883+00:00 | 6,684 | false | ```\n/* \nIdea is to build the string from the end.\n1. We create an array (numericArr) to store string (1 indexed) corresponding to it\'s \nnumeric value as given i.e [undefined,a,b,c,......,z] \n\nNow we have to fill n places i.e _ _ _ _ _ _ (Think it as fill in the blanks)\n\nWe will fill this from the end. The hig... | 26 | 1 | ['JavaScript'] | 11 |
smallest-string-with-a-given-numeric-value | C++ Simple and Short O(n) Solution, faster than 97% | c-simple-and-short-on-solution-faster-th-pu91 | At first, we will initialize the resulting string in the target size with \'a\'s.\nSo we used \'n\' out of the target \'k\' - we can reduce n from k.\nThen, we | yehudisk | NORMAL | 2021-01-28T09:05:10.113896+00:00 | 2021-01-28T15:58:12.696838+00:00 | 1,469 | false | At first, we will initialize the resulting string in the target size with \'a\'s.\nSo we used \'n\' out of the target \'k\' - we can reduce n from k.\nThen, we go from left to right and while k > 0 we turn the \'a\'s to \'z\'s, then there might be a middle character with some letter that equals to the leftover of k.\nO... | 26 | 4 | ['C'] | 4 |
smallest-string-with-a-given-numeric-value | Python without any loop | python-without-any-loop-by-jaguary-wskm | \n num2 = (k-n) // 25\n if num2 == n: return \'z\' * n\n\n num1 = n - num2 - 1\n num = k - (num1 + num2 * 26)\n return \'a\' | JaguarY | NORMAL | 2020-11-22T04:11:52.431191+00:00 | 2020-11-25T08:04:15.130111+00:00 | 1,616 | false | ```\n num2 = (k-n) // 25\n if num2 == n: return \'z\' * n\n\n num1 = n - num2 - 1\n num = k - (num1 + num2 * 26)\n return \'a\' * num1 + chr(num+96) + \'z\' * num2\n```\nBasically, this result will start with a bunch of "a"(num1) and end with a bunch of "z"(num2), with at most one oth... | 24 | 4 | [] | 7 |
smallest-string-with-a-given-numeric-value | [Python] Simple solution - Greedy - O(N) [Explanation + Code + Comment] | python-simple-solution-greedy-on-explana-qmjr | The lexicographically smallest string of length n could be \'aaa\'.. of length n\nThe constraint is that the numeric value must be k.\nTo satisfy this constrain | mihirrane | NORMAL | 2020-11-22T04:17:33.309872+00:00 | 2021-01-29T00:24:38.358491+00:00 | 1,504 | false | The lexicographically smallest string of length n could be \'aaa\'.. of length n\nThe constraint is that the numeric value must be k.\nTo satisfy this constraint, we will greedily add either \'z\' or character form of (k - value) to the string.\n\nFor eg. \nn = 3, k = 32\nInitially, ans = \'aaa\', val = 3, i =2\nSince ... | 19 | 0 | ['Greedy', 'Python', 'Python3'] | 2 |
smallest-string-with-a-given-numeric-value | [C++, Python, Javascript] Easy Solution w/ Explanation | beats 100% / 100% | c-python-javascript-easy-solution-w-expl-74zj | (Note: This is part of a series of Leetcode solution explanations (index). If you like this solution or find it useful, please upvote this post.)\n\n---\n\nIdea | sgallivan | NORMAL | 2021-01-28T09:33:38.406672+00:00 | 2021-02-03T05:07:07.182952+00:00 | 882 | false | *(Note: This is part of a series of Leetcode solution explanations ([**index**](https://dev.to/seanpgallivan/leetcode-solutions-index-57fl)). If you like this solution or find it useful,* ***please upvote*** *this post.)*\n\n---\n\n***Idea:***\n\nThe basic idea is simple: in order for the string to be as lexigraphicall... | 16 | 0 | ['C', 'Python', 'JavaScript'] | 3 |
smallest-string-with-a-given-numeric-value | Java Simple Solution with explanation | java-simple-solution-with-explanation-by-4eaa | Imaging an array with length n and value between 1-26 representing the character values\nFill all slots with minimumr equirement a, then we have value k-a left\ | htttth | NORMAL | 2020-11-22T04:02:26.863151+00:00 | 2020-11-22T18:26:00.795485+00:00 | 1,164 | false | Imaging an array with length n and value between 1-26 representing the character values\nFill all slots with minimumr equirement a, then we have value k-a left\nstart filling slots up to 26 (\'z\') fromt he end of the array, we can calculate with:\nnumber of \'z\' = k%25\na single character with remaining value of k%2... | 14 | 5 | [] | 7 |
smallest-string-with-a-given-numeric-value | ✅[Python] Only 2 Lines Solution || O(1) || Greedy | python-only-2-lines-solution-o1-greedy-b-0rrv | \nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n z, alp = divmod(k-n-1, 25)\n return (n-z-1)*\'a\'+chr(ord(\'a\')+alp+ | subinium | NORMAL | 2022-03-22T00:47:14.219204+00:00 | 2022-03-22T00:49:50.425814+00:00 | 1,248 | false | ```\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n z, alp = divmod(k-n-1, 25)\n return (n-z-1)*\'a\'+chr(ord(\'a\')+alp+1)+z*\'z\'\n``` | 12 | 0 | ['Greedy', 'Python'] | 2 |
smallest-string-with-a-given-numeric-value | [Python3] Runtime: 24 ms, faster than 99.60% | Memory: 14.8 MB, less than 95.18% | python3-runtime-24-ms-faster-than-9960-m-z0pe | \nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n diff = k - n\n q = diff // 25\n r = diff % 25\n ans = " | anubhabishere | NORMAL | 2022-03-22T05:18:17.558364+00:00 | 2022-03-22T05:18:17.558390+00:00 | 737 | false | ```\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n diff = k - n\n q = diff // 25\n r = diff % 25\n ans = "a"*(n-q-1) + chr(97+r) + "z"*qt if r else "a"*(n-q)+ "z"*q\n return ans\n``` | 9 | 0 | ['Math', 'Greedy', 'Python'] | 0 |
smallest-string-with-a-given-numeric-value | Python - very easy to follow - O(N) - Char Dict | python-very-easy-to-follow-on-char-dict-6dxao | \n\ndef getSmallestString(self, n: int, k: int) -> str:\n\tc_dict = {i+1: chr(ord(\'a\') + i) for i in range(26)} # {1:\'a\', 2:\'b\' ... 26:\'z\'}\n\n\tlst = | vvvirenyu | NORMAL | 2021-01-28T09:05:03.384979+00:00 | 2021-01-28T09:12:53.323592+00:00 | 403 | false | \n```\ndef getSmallestString(self, n: int, k: int) -> str:\n\tc_dict = {i+1: chr(ord(\'a\') + i) for i in range(26)} # {1:\'a\', 2:\'b\' ... 26:\'z\'}\n\n\tlst = [\'\']*n\n \n\tfor i in range(n-1, -1, -1): # Reverse fill\n\t\tc = min(26, k-i) # if k-i > 26, character is \'z\' else c_dict[k-i]\n\t\tlst[i... | 9 | 0 | [] | 2 |
smallest-string-with-a-given-numeric-value | Easy code with comments and key idea | easy-code-with-comments-and-key-idea-by-2jxx8 | \n// Key idea is to have as many "a"s at the beginning as possible, so that the \n// string will be lexicographically smallest.\n// Similarly, maximize "z"s at | interviewrecipes | NORMAL | 2020-11-22T04:02:25.869336+00:00 | 2020-11-22T04:02:55.717266+00:00 | 712 | false | ```\n// Key idea is to have as many "a"s at the beginning as possible, so that the \n// string will be lexicographically smallest.\n// Similarly, maximize "z"s at the end so that the character will be between \n// "a"s and "z"s will be as small as possible.\n// Say, n = 6 and k is 57, then we can\'t have more than 3 "a... | 9 | 2 | [] | 2 |
smallest-string-with-a-given-numeric-value | Simple java solution with explaination| o(n) time,o(n) space | simple-java-solution-with-explaination-o-mcy2 | we want String of length n and lexographic shortest string as possible.\n2. so take an character array of size n and initialize every index value with \'a\'. an | kushguptacse | NORMAL | 2021-01-29T06:33:04.067322+00:00 | 2022-03-22T06:51:55.405084+00:00 | 613 | false | 1. we want String of length n and lexographic shortest string as possible.\n2. so take an character array of size n and initialize every index value with \'a\'. and decrement k.\n3. by doing above step if k reaches 0 it means it is the desired answer. why? becuase all a together are enough to make k sum. \n4. run repea... | 8 | 1 | ['Greedy', 'Java'] | 1 |
smallest-string-with-a-given-numeric-value | simple java solution | simple-java-solution-by-manishkumarsah-3rq7 | \nclass Solution {\n public String getSmallestString(int n, int k) {\n char[] res = new char[n];\n Arrays.fill(res,\'a\'); // filling the whole | manishkumarsah | NORMAL | 2021-01-28T08:43:57.422720+00:00 | 2021-01-28T08:44:47.349626+00:00 | 335 | false | ```\nclass Solution {\n public String getSmallestString(int n, int k) {\n char[] res = new char[n];\n Arrays.fill(res,\'a\'); // filling the whole array with starting alphabet a\n \n // now update the k as we have fill the array with character a\n k = k-n;\n \n while... | 7 | 0 | [] | 0 |
smallest-string-with-a-given-numeric-value | C++ O(N) Time Greedy | c-on-time-greedy-by-lzl124631x-0874 | See my latest update in repo LeetCode\n\n## Solution 1. Greedy\n\nIntuition: We can do it greedily:\n\n If picking a won\'t result in unsolvable problem, we pre | lzl124631x | NORMAL | 2020-11-22T04:03:04.578325+00:00 | 2020-11-22T07:00:15.137565+00:00 | 602 | false | See my latest update in repo [LeetCode](https://github.com/lzl124631x/LeetCode)\n\n## Solution 1. Greedy\n\n**Intuition**: We can do it greedily:\n\n* If picking `a` won\'t result in unsolvable problem, we prepend `a` to the start of the string.\n* Otherwise, if picking `z` won\'t result in unsolvable problem, we appen... | 7 | 1 | [] | 0 |
smallest-string-with-a-given-numeric-value | java || Simple solution || O(n) | java-simple-solution-on-by-funingteng-cca1 | Assume we have n length of array with all \'a\', e.g. [\'a\', \'a\', \'a\'] for n = 3, k = 27. The total remaing is 27 - 3 = 24 now. Now we just need to allocat | funingteng | NORMAL | 2022-03-24T18:45:25.700945+00:00 | 2022-03-28T19:44:01.140198+00:00 | 419 | false | Assume we have n length of array with all \'a\', e.g. [\'a\', \'a\', \'a\'] for n = 3, k = 27. The total remaing is 27 - 3 = 24 now. Now we just need to allocate the remaining numbers from right to left. Then iterate the array in reverse order, if larger than 25 (26 - 1), we should allocate \'z\', otherwise we should ... | 6 | 0 | ['Java'] | 0 |
smallest-string-with-a-given-numeric-value | Java Code | java-code-by-rizon__kumar-iezr | \nclass Solution {\n public String getSmallestString(int n, int k) {\n char[] c = new char[n];\n \n for(int i = n-1; i>=0; i--){\n | rizon__kumar | NORMAL | 2022-03-22T02:50:56.101607+00:00 | 2022-03-22T02:50:56.101638+00:00 | 174 | false | ```\nclass Solution {\n public String getSmallestString(int n, int k) {\n char[] c = new char[n];\n \n for(int i = n-1; i>=0; i--){\n int val = Math.min(26,k - i);\n c[i] = (char)(\'a\'+val-1); //using (char) to convert ASCII to respective character\n k=k-val;\n ... | 6 | 0 | ['Greedy'] | 1 |
smallest-string-with-a-given-numeric-value | [Java] Intuitive solution. | java-intuitive-solution-by-sadriddin17-dkjj | The idea is to initialize char array with all value with \'a\' then to reach k we\'ll increase array elements to max from the end to the beginning.\n\n\t\tchar | sadriddin17 | NORMAL | 2022-03-22T01:42:14.977519+00:00 | 2022-03-22T01:43:29.058685+00:00 | 756 | false | The idea is to initialize char array with all value with \'a\' then to reach k we\'ll increase array elements to max from the end to the beginning.\n```\n\t\tchar[] result = new char[n];\n for (int i = 0; i < n; i++) {\n result[i] = \'a\';//initialize all array element with \'a\'\n k--;\n ... | 6 | 0 | ['Java'] | 2 |
smallest-string-with-a-given-numeric-value | Correct simple Python solution without loop | correct-simple-python-solution-without-l-fjgg | \nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n q, r = divmod(k - n, 25)\n return (\'a\'*(n - q - 1) + chr(ord(\'a\') | dpustovarov | NORMAL | 2020-11-23T15:58:22.025745+00:00 | 2020-11-23T16:23:25.074807+00:00 | 441 | false | ```\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n q, r = divmod(k - n, 25)\n return (\'a\'*(n - q - 1) + chr(ord(\'a\') + r) if q < n else \'\') + \'z\'*q\n```\n\n* the smallest string is a string like aaa...aszzz...z\n-- a - smallest char (0 to n length)\n-- s - char on the ... | 6 | 0 | ['Math', 'Python'] | 0 |
smallest-string-with-a-given-numeric-value | c++ || very simple solution || O(N) | c-very-simple-solution-on-by-shubhangnau-v3qy | PLEASE UPVOTE IF IT HELPS YOU\n\nclass Solution {\npublic:\n string getSmallestString(int n, int k) {\n { \n string arr = ""; \n \n for(int i = | shubhangnautiyal | NORMAL | 2022-03-22T04:57:21.304686+00:00 | 2022-03-22T04:57:21.304726+00:00 | 420 | false | **PLEASE UPVOTE IF IT HELPS YOU**\n```\nclass Solution {\npublic:\n string getSmallestString(int n, int k) {\n { \n string arr = ""; \n \n for(int i = 0; i < n; i++) \n arr += \'a\'; \n for (int i = n - 1; i >= 0; i--) \n { \n k -= i; \n if (k >= 0) \n { \n ... | 5 | 0 | ['String', 'C'] | 1 |
smallest-string-with-a-given-numeric-value | ✔️ C++ concise solution | Explanation | minimum time and space complexity | c-concise-solution-explanation-minimum-t-gcd5 | Intuition:\nAs we have to create a lexicographically smallest string with a length of n. We have to place as many as the largest number back of the output strin | foysol_ahmed | NORMAL | 2022-03-22T04:47:06.673943+00:00 | 2022-03-22T16:26:36.190312+00:00 | 252 | false | **Intuition:**\nAs we have to create ***a lexicographically smallest string with a length of n***. We have to place as many as the largest number back of the output string. To be more precise, for every iteration, our target is to place ***(i-1) a*** at the beginning, and for the last index, we just put the minimum num... | 5 | 0 | ['Greedy'] | 1 |
smallest-string-with-a-given-numeric-value | Python. 3 lines. faster than 100.00%. cool & clear solution. | python-3-lines-faster-than-10000-cool-cl-kjzn | \tclass Solution:\n\t\tdef getSmallestString(self, n: int, k: int) -> str:\n\t\t\tz = (k - n) // 25\n\t\t\tunique = chr(k - z * 25 - n + 97) if n - z else ""\n\ | m-d-f | NORMAL | 2021-01-28T12:14:12.458078+00:00 | 2021-01-28T12:15:25.030201+00:00 | 472 | false | \tclass Solution:\n\t\tdef getSmallestString(self, n: int, k: int) -> str:\n\t\t\tz = (k - n) // 25\n\t\t\tunique = chr(k - z * 25 - n + 97) if n - z else ""\n\t\t\treturn "a"*(n-z-1) + unique + "z"*z | 5 | 1 | ['Python', 'Python3'] | 0 |
smallest-string-with-a-given-numeric-value | Python Math | python-math-by-twerpapple-0vhk | \n"""\nfind the greatest x such that \n 1*x + 26*(n-x) >= k\n 26n - 25x >= k\n 26n - k >= 25x\n (26n-k)/25 >= x\n \n x = floor((26n-k)/25)\n | twerpapple | NORMAL | 2020-11-29T17:38:53.097236+00:00 | 2020-11-29T17:39:28.743732+00:00 | 136 | false | ```\n"""\nfind the greatest x such that \n 1*x + 26*(n-x) >= k\n 26n - 25x >= k\n 26n - k >= 25x\n (26n-k)/25 >= x\n \n x = floor((26n-k)/25)\n \n numeric_val = 1*x + 26*(n-x)\n string = [1]*x + [26-(numeric_val-k)] + [26]*(n-x-1)\n"""\n\n\nclass Solution:\n def getSmallestString(self, n, ... | 5 | 0 | [] | 0 |
smallest-string-with-a-given-numeric-value | Simple Basic C++ Soln. Only Loop | simple-basic-c-soln-only-loop-by-vardaan-pd93 | Intuition\nWe need the smallest string therefore I followed these simple steps to get my ans\n1--> add \'a\' till possible \n2--> add a char \n3--> now add \'z\ | vardaanpahwa02 | NORMAL | 2023-10-27T13:00:18.711045+00:00 | 2023-10-27T13:00:18.711070+00:00 | 342 | false | # Intuition\nWe need the smallest string therefore I followed these simple steps to get my ans\n1--> add \'a\' till possible \n2--> add a char \n3--> now add \'z\'\n\n# Approach\nNow the question arrises add \'a\' but till when . \nwe will use a number for example let n=3 and k=30.\n1--> add \'a\' then n=2 and k=29\n2-... | 4 | 0 | ['C++'] | 2 |
smallest-string-with-a-given-numeric-value | Javascript | Recursive | javascript-recursive-by-vihangpatel-h9ty | Here we try to find out how many \'z\'s can be placed at the end of the desired output \n Let\'s say n = 3, k = 3; n <= k <= 78 -> All three places are required | vihangpatel | NORMAL | 2022-03-23T05:44:29.775750+00:00 | 2022-03-23T05:44:29.775778+00:00 | 178 | false | Here we try to find out how many \'z\'s can be placed at the end of the desired output \n Let\'s say n = 3, k = 3; n <= k <= 78 -> All three places are required to be filled\n\n 3 -> aaa\n 4 -> aab\n .\n .\n 27 -> aay\n 28 -> aaz\n 29 -> abz\n 30 -> acz\n \nSo, with n = 3, k = 3, lexicographi... | 4 | 0 | ['JavaScript'] | 0 |
smallest-string-with-a-given-numeric-value | Easy Concise C++ O(n) Solution(Faster than 99%) | easy-concise-c-on-solutionfaster-than-99-ooht | class Solution {\npublic:\n string getSmallestString(int n, int k) {\n int curr=1;\n string ans=""; \n for(int i=1;i<=n;i++){\n if( | kartiksaxenaabcd | NORMAL | 2022-03-22T11:30:28.243614+00:00 | 2022-03-22T11:34:12.129254+00:00 | 142 | false | class Solution {\npublic:\n string getSmallestString(int n, int k) {\n int curr=1;\n string ans=""; \n for(int i=1;i<=n;i++){\n if((k-curr)< 26*(n-i)){\n ans+=char(\'a\'+curr-1);\n k-=curr;\n }\n else{\n while((k-curr)>26*(n-i))\n ... | 4 | 0 | ['Greedy', 'C'] | 0 |
smallest-string-with-a-given-numeric-value | Backward-filling with an early exit [bad code!] | backward-filling-with-an-early-exit-bad-s1iya | There is one optimization in the code below that is helpful for large n: As soon as k == n, we know that the rest of the sought string is \'a\' (s. && k > n in | anikit | NORMAL | 2022-03-22T10:24:23.935345+00:00 | 2023-01-03T23:08:09.311835+00:00 | 101 | false | There is one optimization in the code below that is helpful for large `n`: As soon as `k == n`, we know that the rest of the sought string is `\'a\'` (s. `&& k > n` in the `while` condition).\n\nWhile this code works, there are a few bad coding practices here which should be avoided:\n1. Re-use of variables: the argume... | 4 | 0 | ['C#'] | 0 |
smallest-string-with-a-given-numeric-value | C++ two approaches || commented | c-two-approaches-commented-by-suryapsg-wxh4 | Approach 1:\n\nstring getSmallestString(int n, int k) {\n\tstring res;\n\tint c;\n\tfor(int i=0;i<n;i++){\n\t\t//start with \'a\'\n\t\tc=1;\n\t\t//If current in | SuryaPSG | NORMAL | 2022-03-22T03:06:46.622178+00:00 | 2022-03-22T04:16:55.727661+00:00 | 305 | false | **Approach 1:**\n```\nstring getSmallestString(int n, int k) {\n\tstring res;\n\tint c;\n\tfor(int i=0;i<n;i++){\n\t\t//start with \'a\'\n\t\tc=1;\n\t\t//If current index holds \'a\' then remaining n-i-1 length should be enough to hold atleast \'z\' in every index\n\t\t//if it is not enough we make the current char b,c... | 4 | 0 | ['C'] | 1 |
smallest-string-with-a-given-numeric-value | [Simple and readable] - O(N) TypeScript/JavaScript | simple-and-readable-on-typescriptjavascr-aq9w | \nfunction getSmallestString(n: number, k: number): string {\n const alphabet = [\'\',\'a\',\'b\',\'c\',\'d\',\'e\',\'f\',\'g\',\'h\',\'i\',\'j\',\'k\',\'l\' | dustinkiselbach | NORMAL | 2021-01-28T15:33:47.613795+00:00 | 2021-01-28T15:33:47.613836+00:00 | 275 | false | ```\nfunction getSmallestString(n: number, k: number): string {\n const alphabet = [\'\',\'a\',\'b\',\'c\',\'d\',\'e\',\'f\',\'g\',\'h\',\'i\',\'j\',\'k\',\'l\',\'m\',\'n\',\'o\',\'p\',\'q\',\'r\',\'s\',\'t\',\'u\',\'v\',\'w\',\'x\',\'y\',\'z\']\n let sum = 0\n let s = \'\'\n \n for (let i = n - 1; i >= ... | 4 | 0 | ['TypeScript', 'JavaScript'] | 1 |
smallest-string-with-a-given-numeric-value | Smallest String With A Given Numeric Value: Python, 3 LoC | smallest-string-with-a-given-numeric-val-3qmq | python\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n if k == 26*n: return \'z\' * n\n nz, nshift = divmod(k - n, 25) | sjoin | NORMAL | 2021-01-28T15:18:45.019804+00:00 | 2021-01-28T15:18:45.019838+00:00 | 185 | false | ```python\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n if k == 26*n: return \'z\' * n\n nz, nshift = divmod(k - n, 25)\n return \'a\' * (n - nz - 1) + chr(ord(\'a\') + nshift) + \'z\' * nz\n```\n\n---\nIf you find this helpful, please **upvote**! Thank you! :-) | 4 | 0 | ['Math', 'Python'] | 0 |
smallest-string-with-a-given-numeric-value | O(n) Better Solution Greedy [Cpp] | on-better-solution-greedy-cpp-by-anil111-54jb | \nclass Solution {\npublic:\n string getSmallestString(int n, int k) \n {\n string str(n,\'a\');\n k=k-n;\n int ri=n-1;\n whil | anil111 | NORMAL | 2021-01-28T08:55:35.295039+00:00 | 2021-01-28T09:05:27.751520+00:00 | 276 | false | ```\nclass Solution {\npublic:\n string getSmallestString(int n, int k) \n {\n string str(n,\'a\');\n k=k-n;\n int ri=n-1;\n while(k>0)\n {\n str[ri]=str[ri]+min(25,k);\n ri--;\n k=k-25;\n }\n return str;\n }\n};\n``` | 4 | 0 | ['Greedy', 'C++'] | 0 |
smallest-string-with-a-given-numeric-value | Python - !beautiful && !concise && !short | python-beautiful-concise-short-by-kafola-h41x | Start with all "a" then from the end bump it up as much as posible (25 max to turn "a" into "z")\n\n\ndef getSmallestString(self, n: int, k: int) -> str:\n\t\ta | kafola | NORMAL | 2020-11-22T04:04:18.287214+00:00 | 2020-11-22T04:04:18.287246+00:00 | 313 | false | Start with all "a" then from the end bump it up as much as posible (25 max to turn "a" into "z")\n\n```\ndef getSmallestString(self, n: int, k: int) -> str:\n\t\tarr = [97] * n\n k -= n\n \n i = n - 1\n while k > 0 and i > -1:\n arr[i] += min(k, 25)\n k -= min(k, 25)\n ... | 4 | 0 | [] | 2 |
smallest-string-with-a-given-numeric-value | C++ Easy Code || Fully Explained in Simple Steps || Clean and Clear code + Explanation | c-easy-code-fully-explained-in-simple-st-ej36 | \nThe most clear explanation of the algorithm is as:\n\nStep - 1 Initialize a string s with n number of \'a\' to make it as a smallest possible lexiographic.\n | dee_stroyer | NORMAL | 2022-03-23T19:34:14.637065+00:00 | 2022-03-23T19:34:35.067496+00:00 | 140 | false | \n**The most clear explanation of the algorithm is as:**\n\n*Step - 1 Initialize a string s with n number of \'a\' to make it as a smallest possible lexiographic.\nStep - 2 Count of \'a = 1\' , so if i have n number of \'a\' it means it sum is also \'n\'.\nStep - 3 Reduce the value of k by n (i.e k=k-n);\nStep - 4 Sta... | 3 | 0 | ['String', 'Greedy', 'C', 'C++'] | 0 |
smallest-string-with-a-given-numeric-value | 1663 | C++ | Easy O(N) solution with explanation | 1663-c-easy-on-solution-with-explanation-9s7f | Please upvote if you like this solution :)\n\nApproach:\n For getting smallest lexicographical order of string, Initially we define n size string and put \'a\'s | Yash2arma | NORMAL | 2022-03-22T18:33:21.921805+00:00 | 2022-03-22T18:33:21.921875+00:00 | 72 | false | **Please upvote if you like this solution :)**\n\n**Approach:**\n* For getting smallest lexicographical order of string, Initially we define n size string and put \'a\'s into it.\n* Since string only contains \'a\'s we subtract n from k i.e. k=k-n;\n* Now, we traverse from the end of the string because we want smallest... | 3 | 0 | ['Greedy', 'C'] | 0 |
smallest-string-with-a-given-numeric-value | Simple java solution | simple-java-solution-by-siddhant_1602-4rwo | class Solution {\n\n public String getSmallestString(int n, int k) {\n char c[]=new char[n];\n Arrays.fill(c,\'a\');\n int i=n-1;\n | Siddhant_1602 | NORMAL | 2022-03-22T17:49:02.895517+00:00 | 2022-03-22T17:49:02.895562+00:00 | 203 | false | class Solution {\n\n public String getSmallestString(int n, int k) {\n char c[]=new char[n];\n Arrays.fill(c,\'a\');\n int i=n-1;\n k-=n;\n while(k>0)\n {\n if(k>25)\n {\n c[i--]=\'z\';\n k-=25;\n }\n ... | 3 | 0 | ['Java'] | 0 |
smallest-string-with-a-given-numeric-value | O(n) C++ solution with explanation and dry run | on-c-solution-with-explanation-and-dry-r-g7gd | We have to find the smallest lexiographical string with size of n and numeric value of k.\n\nFor a string to be lexiographically small, it\'s beginning charact | shivansh961 | NORMAL | 2022-03-22T13:23:20.176319+00:00 | 2022-03-22T13:52:57.518385+00:00 | 135 | false | We have to find the smallest lexiographical string with size of n and numeric value of k.\n\n**For a string to be lexiographically small, it\'s beginning characters must be smallest.**\n\nWe initialise a string as **ans** with size n, having each character as **\'a\'**. \n\nLet\'s take an example for the better unders... | 3 | 0 | [] | 2 |
smallest-string-with-a-given-numeric-value | ✅ Python Solution | python-solution-by-dhananjay79-qk56 | \nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n ans = \'\'\n while (n - 1) * 26 >= k:\n ans += \'a\'\n | dhananjay79 | NORMAL | 2022-03-22T12:40:26.217732+00:00 | 2022-03-22T12:40:26.217764+00:00 | 309 | false | ```\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n ans = \'\'\n while (n - 1) * 26 >= k:\n ans += \'a\'\n n -= 1; k -= 1\n ans += chr(ord(\'a\') + (k % 26 or 26) - 1)\n ans += \'z\' * (n - 1)\n return ans\n``` | 3 | 0 | ['Python', 'Python3'] | 0 |
smallest-string-with-a-given-numeric-value | Simple Java Code | simple-java-code-by-sai_prashant-z8ia | Idea is simple, we will make the string from end greedily i.e, putting the lexographically biggest possible character. Here is my code \n\n\nclass Solution {\n | sai_prashant | NORMAL | 2022-03-22T06:04:36.355636+00:00 | 2022-03-22T06:04:36.355687+00:00 | 169 | false | Idea is simple, we will make the string from end greedily i.e, putting the lexographically biggest possible character. Here is my code \n\n```\nclass Solution {\n public String getSmallestString(int n, int k) {\n StringBuilder sb=new StringBuilder("");\n for(int i=0;i<n;i++){\n int val=k-(n-... | 3 | 0 | ['Greedy'] | 0 |
smallest-string-with-a-given-numeric-value | 1. Failed DP, 2. Greedy | 1-failed-dp-2-greedy-by-akshay3213-5qsb | \n\n\n# Using dynamic programming - Gave me memory limit exceeding\n# \n\n\nclass Solution {\n\t//O(n * k)\n public String getSmallestString(int n, int k) {\ | akshay3213 | NORMAL | 2022-03-22T03:53:58.020877+00:00 | 2022-03-22T03:59:52.624913+00:00 | 144 | false | \n\n\n# Using dynamic programming - Gave me memory limit exceeding\n# \n\n```\nclass Solution {\n\t/... | 3 | 0 | ['Dynamic Programming', 'Greedy'] | 0 |
smallest-string-with-a-given-numeric-value | [Python3] O(n) time and O(1) space complexity solution | python3-on-time-and-o1-space-complexity-cuqu1 | python3\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n result =[0]*n\n for position in range(n-1,-1,-1):\n | bhushandhumal | NORMAL | 2022-03-22T03:20:49.899761+00:00 | 2022-03-22T04:12:13.413894+00:00 | 360 | false | ``` python3\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n result =[0]*n\n for position in range(n-1,-1,-1):\n add = min(k -position,26)\n result[position] = chr(ord("a")+add -1)\n k-=add\n \n return "".join(result)\n```\n\nCo... | 3 | 0 | ['Python', 'Python3'] | 1 |
smallest-string-with-a-given-numeric-value | python 3 || O(n) | python-3-on-by-derek-y-3xzb | ```\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n res = \'\'\n for i in range(n):\n q, r = divmod(k, 26)\ | derek-y | NORMAL | 2022-03-22T01:36:15.876719+00:00 | 2022-03-24T03:38:39.913835+00:00 | 194 | false | ```\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n res = \'\'\n for i in range(n):\n q, r = divmod(k, 26)\n if r == 0:\n q -= 1\n r = 26\n\n if n - q - i >= 2:\n res += \'a\'\n k -= 1\... | 3 | 0 | ['Greedy', 'Python', 'Python3'] | 0 |
smallest-string-with-a-given-numeric-value | C++ || Easy Explanation || With Comments || TC - O(N) | c-easy-explanation-with-comments-tc-on-b-hml3 | \n\n1. Create a string of size n with all \'a\'s in it\n2. Start from end of the created string \n3. Create a variable \'remaining\' and put \'z\' at index i if | krishna__1902 | NORMAL | 2022-03-22T01:12:48.084374+00:00 | 2022-03-22T01:16:06.898575+00:00 | 59 | false | \n\n**1. Create a string of size n with all \'a\'s in it\n2. Start from end of the created string \n3. Create a variable \'remaining\' and put \'z\' at index i if remaining>25 (i.e, \'a\'+25 , as a is already there so \'a\'+25 = \'z\'\n4. If remaining becomes less than or equal to 25, put \'a\' + remaining value at tha... | 3 | 0 | ['Greedy', 'C'] | 0 |
smallest-string-with-a-given-numeric-value | [Python] O(1) with explanation | python-o1-with-explanation-by-shoomoon-dudy | Accoding to the defination, the smallest lexicographically string must have the patten as \n\n\t\ts = \'a\' * p + t * q + \'z\' * r\n\t\twhere p, r >= 0, \'a\' | shoomoon | NORMAL | 2021-02-25T19:59:43.811115+00:00 | 2021-02-25T21:45:42.951225+00:00 | 139 | false | Accoding to the defination, the smallest lexicographically string must have the patten as \n\n\t\ts = \'a\' * p + t * q + \'z\' * r\n\t\twhere p, r >= 0, \'a\' < t < \'z\', q = 0 or 1\n\nThus, we have:\n```\n1. k = 1 * p + val_t * q + 26 * r\n2. p + q + r = n\n```\n\nThen k = (n - q - r) + val_t * q + 26 * r = n + q * ... | 3 | 0 | [] | 0 |
smallest-string-with-a-given-numeric-value | C# - O(N) - Fill array from end | c-on-fill-array-from-end-by-christris-wob7 | csharp\npublic string GetSmallestString(int n, int rem)\n{\n\t char[] result = new char[n];\n\t Array.Fill(result, \'a\');\n\n\t // We have already filled array | christris | NORMAL | 2021-01-29T02:15:54.808001+00:00 | 2021-01-29T02:17:22.806458+00:00 | 95 | false | ```csharp\npublic string GetSmallestString(int n, int rem)\n{\n\t char[] result = new char[n];\n\t Array.Fill(result, \'a\');\n\n\t // We have already filled array with \'a\' with value 1 for each char\n\t rem -= n;\n\n\t for(int i = n - 1; i >= 0; i--)\n\t {\n\t\t int diff = Math.Min(rem, 25);\n\t\t result[i] = (char)... | 3 | 0 | [] | 0 |
smallest-string-with-a-given-numeric-value | [C++] Greedy Approach | c-greedy-approach-by-codedayday-p178 | Approach 1: Greedy Solution\nIdea:\nGenerate the initial string with all \'a\' characters. This will reduce k by n.\nThen, turn rightmost \'a\' into \'z\' (\'a\ | codedayday | NORMAL | 2021-01-28T15:12:32.183431+00:00 | 2021-01-28T15:12:32.183471+00:00 | 141 | false | Approach 1: Greedy Solution\nIdea:\nGenerate the initial string with all \'a\' characters. This will reduce k by n.\nThen, turn rightmost \'a\' into \'z\' (\'a\' + 25, or \'a\' + k) while k is positive.\n```\nclass Solution {\npublic:\n string getSmallestString(int n, int k) {\n string ans(n, \'a\');\n ... | 3 | 0 | [] | 1 |
smallest-string-with-a-given-numeric-value | [C++] Single-pass Solution Explained, ~100% Time, ~98% Space | c-single-pass-solution-explained-100-tim-iumc | Alright: while this problem might tempt you to actually compute all/some permutations, the size of the expected string is such that it should definitely make us | ajna | NORMAL | 2021-01-28T14:15:36.723067+00:00 | 2021-01-28T14:15:36.723093+00:00 | 201 | false | Alright: while this problem might tempt you to actually compute all/some permutations, the size of the expected string is such that it should definitely make us very wary about how to proceed in terms of performance.\n\nBut we can go through an actually smarter way, once we figure out that you average result will have ... | 3 | 0 | ['C', 'Combinatorics', 'Probability and Statistics', 'C++'] | 0 |
smallest-string-with-a-given-numeric-value | Rust 4ms solution | rust-4ms-solution-by-sugyan-xdfc | \nimpl Solution {\n pub fn get_smallest_string(n: i32, k: i32) -> String {\n let mut v = vec![0; n as usize];\n let mut k = k - n;\n for | sugyan | NORMAL | 2021-01-28T08:53:01.060448+00:00 | 2021-01-28T08:53:01.060498+00:00 | 78 | false | ```\nimpl Solution {\n pub fn get_smallest_string(n: i32, k: i32) -> String {\n let mut v = vec![0; n as usize];\n let mut k = k - n;\n for e in v.iter_mut().rev() {\n let m = std::cmp::min(25, k);\n *e = m as u8 + b\'a\';\n k -= m;\n }\n String::fr... | 3 | 0 | ['Rust'] | 1 |
smallest-string-with-a-given-numeric-value | [Java] O(logN) Solution, Divide And Conquer, 3ms Faster than 100% | java-ologn-solution-divide-and-conquer-3-3f95 | Use divide and conquer when adding \'a\' and \'z\'.\n\nclass Solution {\n public String getSmallestString(int n, int k) {\n int subK = k - n; \n | applechen | NORMAL | 2020-11-22T06:44:16.119975+00:00 | 2020-11-22T06:44:16.120009+00:00 | 326 | false | Use divide and conquer when adding \'a\' and \'z\'.\n```\nclass Solution {\n public String getSmallestString(int n, int k) {\n int subK = k - n; \n int numZ = subK / 25;\n char mid = (char)(\'a\' + subK % 25);\n int numA = n - numZ - 1;\n \n StringBuilder res = new StringBui... | 3 | 1 | ['Divide and Conquer', 'Java'] | 4 |
smallest-string-with-a-given-numeric-value | Python 3 | 5-line Greedy O(N) | Explanation | python-3-5-line-greedy-on-explanation-by-n6we | Understand the question\n- Question is asking for smallest string, that being said, \n\t- we want to have as most a on the left side as possible\n\t- If using a | idontknoooo | NORMAL | 2020-11-22T05:29:53.794647+00:00 | 2020-11-22T07:27:44.812235+00:00 | 227 | false | ### Understand the question\n- Question is asking for *smallest string*, that being said, \n\t- we want to have as most *a* on the left side as possible\n\t- If using *a* doesn\'t match given number, we can try some letter with larger number \n\t- *z* will be the last letter we want to use, so we want to have them all ... | 3 | 0 | ['Greedy', 'Python', 'Python3'] | 2 |
smallest-string-with-a-given-numeric-value | [C++] Simple O(N) Solution | Brute Force Solution | 2 Solutions | c-simple-on-solution-brute-force-solutio-fv09 | \nclass Solution\n{\npublic:\n // Best Solution\n string getSmallestString(int n, int k)\n {\n string res(n, \'a\');\n k -= n;\n i | ravireddy07 | NORMAL | 2020-11-22T04:51:07.569238+00:00 | 2020-11-22T04:52:08.277399+00:00 | 58 | false | ```\nclass Solution\n{\npublic:\n // Best Solution\n string getSmallestString(int n, int k)\n {\n string res(n, \'a\');\n k -= n;\n int midChar = 0;\n while (k > 0)\n {\n midChar = min(k, 25);\n k -= midChar;\n res[n - 1] += midChar;\n ... | 3 | 1 | [] | 0 |
smallest-string-with-a-given-numeric-value | Python simple O(N) math solution with comments | python-simple-on-math-solution-with-comm-wiwp | k = 73, n = 5\n526 -73 = 57\n\nz z z z z\n26 26 26 26 26\n| | | | |\n25 + 25 + 7 + 0 + 0 | SonicLLM | NORMAL | 2020-11-22T04:48:35.332032+00:00 | 2020-11-22T21:33:10.247673+00:00 | 286 | false | k = 73, n = 5\n5*26 -73 = 57\n```\nz z z z z\n26 26 26 26 26\n| | | | |\n25 + 25 + 7 + 0 + 0 = 57\n| | | | |\n1 1 19 0 0 \na a s z z\n```\n\n57 = 2 * 25 + 7 * 1\ndiv = 2, remain = 1\n=... | 3 | 0 | ['Math', 'Python'] | 3 |
smallest-string-with-a-given-numeric-value | [c++] easy with example | c-easy-with-example-by-sanjeev1709912-1f8s | My idea behind this code to take character gridily \nmeans first i try to pick 26 if possible \n\nNow the question is how to check if possible \nSo here is answ | sanjeev1709912 | NORMAL | 2020-11-22T04:32:17.625153+00:00 | 2020-11-22T05:08:24.565124+00:00 | 122 | false | My idea behind this code to take character gridily \nmeans first i try to pick 26 if possible \n\nNow the question is how to check if possible \nSo here is answer like if we pick 26 and the remaining value is geater then required number of element then it safe to pick 26 thats it\n\nok now see example \n```\nlet think ... | 3 | 0 | [] | 0 |
smallest-string-with-a-given-numeric-value | [Ruby] 3 lines. Intuitive + math solution with system of linear equations | ruby-3-lines-intuitive-math-solution-wit-pv0t | Algorithm\n\nSearch an answer in a form aa...aa[LETTER]zz...zz\nNumeric Value of that string is a number of a chars * 1 + number of z chars * 26 + value of LETT | shhavel | NORMAL | 2020-11-22T04:17:54.246832+00:00 | 2021-01-28T08:51:33.962698+00:00 | 203 | false | # Algorithm\n\nSearch an answer in a form `aa...aa[LETTER]zz...zz`\nNumeric Value of that string is a number of `a` chars * 1 + number of `z` chars * 26 + value of `LETTER`.\n\nMin value of a string of len `n` is `n` for string `aa...aaa`.\nAfter decreasing needed sum (value of string result) by n count of `z` is `s /... | 3 | 0 | ['Math', 'Ruby'] | 0 |
smallest-string-with-a-given-numeric-value | BEATS 99% || EASY with Eg|| In Depth Commented CODE. | beats-99-easy-with-eg-in-depth-commented-5oc4 | This provided code defines a function getSmallestString(n, k) that generates a lexicographically smallest string with the following properties:\n\n- Length: The | Abhishekkant135 | NORMAL | 2024-06-26T10:22:19.698413+00:00 | 2024-06-26T10:22:19.698440+00:00 | 218 | false | This provided code defines a function `getSmallestString(n, k)` that generates a lexicographically smallest string with the following properties:\n\n- **Length:** The string has a length of `n` characters.\n- **Maximum Zs:** The string can contain at most `k` characters \'z\'.\n\n**Functionality Breakdown:**\n\n1. **Ch... | 2 | 0 | ['String', 'Greedy', 'Java'] | 0 |
smallest-string-with-a-given-numeric-value | c++ | easy | fast | c-easy-fast-by-venomhighs7-stso | \n\n# Code\n\n//from hi-malik\nclass Solution {\npublic:\n string getSmallestString(int n, int k) {\n string res(n,\'a\');\n k -= n;\n \ | venomhighs7 | NORMAL | 2022-11-22T12:17:02.074263+00:00 | 2022-11-22T12:17:02.074298+00:00 | 224 | false | \n\n# Code\n```\n//from hi-malik\nclass Solution {\npublic:\n string getSmallestString(int n, int k) {\n string res(n,\'a\');\n k -= n;\n \n while(k > 0){\n res[--n] += min(25, k);\n k -= min(25, k);\n }\n return res;\n }\n};\n``` | 2 | 0 | ['C++'] | 1 |
smallest-string-with-a-given-numeric-value | Easy to understand with comments | easy-to-understand-with-comments-by-shas-76en | we start from the back , keep adding the highest letter possible to the result , highest possible letter is k-(remaining spaces)(remaining as we subtract the va | shashank_c10 | NORMAL | 2022-03-23T17:47:24.950988+00:00 | 2022-03-23T17:47:24.951048+00:00 | 200 | false | we start from the back , keep adding the highest letter possible to the result , highest possible letter is k-(remaining spaces)(remaining as we subtract the value of k by the value of each letter added, the remaining spaces are n - current length of result)\n```\ndef getSmallestString(self, n: int, k: int) -> str:\n ... | 2 | 0 | ['Greedy', 'Python', 'Python3'] | 0 |
smallest-string-with-a-given-numeric-value | ✅C++ || TC - O(N), SC- O(1) || 🗓️ Daily LeetCoding Challenge || March || Day 22 | c-tc-on-sc-o1-daily-leetcoding-challenge-0sja | Approach: \n1) Since we want to make a string lexicographically smaller so we will make a string of size n with \'a\'.\n2) Start from the right to left characte | shm_47 | NORMAL | 2022-03-22T18:08:49.871209+00:00 | 2022-03-22T18:19:37.371832+00:00 | 38 | false | **Approach:** \n1) Since we want to make a string lexicographically smaller so we will make a string of size n with \'a\'.\n2) Start from the right to left character, increase the char value by min(25, k) until k will reach to 0.\n\n```\nclass Solution {\npublic:\n string getSmallestString(int n, int k) {\n s... | 2 | 0 | [] | 0 |
smallest-string-with-a-given-numeric-value | ✅ C++ || Dry Run || Easy || 🗓️ Daily LeetCoding Challenge || March || Day 22 | c-dry-run-easy-daily-leetcoding-challeng-scvt | Please Upvote If It Helps\n\nAlgorithm\n Initialise an string of length n with all \'a\'s\n\n Start traversing from the end of the string while k is positive\n | mayanksamadhiya12345 | NORMAL | 2022-03-22T17:12:17.052153+00:00 | 2022-03-22T17:12:17.052200+00:00 | 34 | false | **Please Upvote If It Helps**\n\n**Algorithm**\n* **Initialise** an string of length **n** with all **\'a\'s**\n\n* **Start traversing** from the **end of the string** while **k is positive**\n* add the element of the string by **\'z\' if k >= 25**\n* else add the **ascii value character** \n* At the same time decrease... | 2 | 0 | [] | 1 |
smallest-string-with-a-given-numeric-value | Python | Reverse filling | Easy and clean solution | python-reverse-filling-easy-and-clean-so-j5oz | \nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n result, k, i = [\'a\']*n, k-n, n-1\n while k:\n | revanthnamburu | NORMAL | 2022-03-22T13:46:42.035521+00:00 | 2022-03-22T13:46:42.035565+00:00 | 143 | false | ```\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n result, k, i = [\'a\']*n, k-n, n-1\n while k:\n k = k + 1\n if k/26 >= 1:\n result[i], k, i = \'z\', k-26, i-1\n else:\n result[i], k =... | 2 | 0 | ['String', 'Python'] | 0 |
smallest-string-with-a-given-numeric-value | ✅ C++ Solution with explanation | O(n) | c-solution-with-explanation-on-by-karyga-45yw | Explanation\n\n1. Initialise a string ans all with \'a\'s (to ensure that the answer will be lexicographically small\n2. Start traversing the string from the en | karygauss03 | NORMAL | 2022-03-22T10:25:49.631977+00:00 | 2022-03-22T10:25:49.632001+00:00 | 32 | false | **Explanation**\n```\n1. Initialise a string ans all with \'a\'s (to ensure that the answer will be lexicographically small\n2. Start traversing the string from the end, and if k >= 25 we add \'z\' instead of \'a\' else we add \n\'b\' or \'c\' or .... \'y\' depending on k. (ans[i] += min(25, k)).\n3. We decrease the va... | 2 | 0 | [] | 0 |
smallest-string-with-a-given-numeric-value | C++|| easy to understand || with comments || greedy | c-easy-to-understand-with-comments-greed-kce5 | \nstring getSmallestString(int n, int k) {\n string ans="";\n for(int i=0;i<n;i++)\n {\n //for getting the lexicographically smalles | priyanshu12k5 | NORMAL | 2022-03-22T09:46:02.014049+00:00 | 2022-03-22T09:46:02.014082+00:00 | 71 | false | ```\nstring getSmallestString(int n, int k) {\n string ans="";\n for(int i=0;i<n;i++)\n {\n //for getting the lexicographically smallest string\n ans+=\'a\';\n }\n k=k-n;\n int j=n-1;\n while(k!=0 && j>=0)\n {\n// as we have added a no... | 2 | 0 | ['Greedy', 'C'] | 0 |
smallest-string-with-a-given-numeric-value | ✔️[C++] || 6 line Simple Code || Easy to understand || TC: O( n ) , SC: O( 1 ) | c-6-line-simple-code-easy-to-understand-8025t | ```\nstring getSmallestString(int n, int k) {\n string res(n,\'a\');\n for(int i=0;i=26) res[n-1-i]=\'z\' , k-=26;\n else res[n-1-i]=c | anant_0059 | NORMAL | 2022-03-22T09:34:45.675105+00:00 | 2022-03-22T09:34:45.675151+00:00 | 289 | false | ```\nstring getSmallestString(int n, int k) {\n string res(n,\'a\');\n for(int i=0;i<n;++i){\n int remain=k-(n-1-i);\n if(remain>=26) res[n-1-i]=\'z\' , k-=26;\n else res[n-1-i]=char(\'a\'+remain-1) , k=n-i-1;\n }\n return res;\n } | 2 | 0 | ['Greedy', 'C'] | 0 |
smallest-string-with-a-given-numeric-value | C++ Easy Explanation Optimal Approach | c-easy-explanation-optimal-approach-by-j-w827 | Approach\nstart with a string of all a\'s of length n as it is the lexicographically smallest string possible\nupdate k as k-n\n\nnow we traverse our string fro | jay_suk23 | NORMAL | 2022-03-22T09:27:50.231788+00:00 | 2022-03-22T09:27:50.231835+00:00 | 26 | false | **Approach**\nstart with a string of all a\'s of length n as it is the lexicographically smallest string possible\nupdate k as k-n\n\nnow we traverse our string from right to left and try to fill it with the maximum character possible (this will make sure that we get the lowest characters to the left)\n\n\n**fillling i... | 2 | 0 | ['String', 'Greedy', 'C'] | 0 |
smallest-string-with-a-given-numeric-value | c++ | Easy approach | Greedy | with explanation | with example | c-easy-approach-greedy-with-explanation-qz9m3 | Input: n = 5, k = 73\nOutput: "aaszz"\n\nstep 1: make string of length n=5 and fill it with \'a\'. after step 1 our ans string will look like ans="aaaaa"\nstep | akashjoshi99 | NORMAL | 2022-03-22T09:15:48.145528+00:00 | 2022-03-22T09:15:48.145562+00:00 | 140 | false | Input: n = 5, k = 73\nOutput: "aaszz"\n\nstep 1: make string of length n=5 and fill it with \'a\'. after step 1 our **ans** string will look like **ans="aaaaa"**\nstep 2: iterate from back and keep the track of **sumOfString** after replacing ith index with \'z\'. sumOfString(aaaaz) is 1+1+1+1+26=30.\nstep 3: if sumOfS... | 2 | 0 | ['Greedy', 'C'] | 0 |
smallest-string-with-a-given-numeric-value | Math with short proof | beats 98.62% / 99.54% | math-with-short-proof-beats-9862-9954-by-efnq | This solution is based on the fact that every output will always be of the form aaa...aaa?zzz...zzz regardless of the values of k and n.\n\nThis submission ran | sr_vrd | NORMAL | 2022-03-22T08:09:59.103240+00:00 | 2022-03-22T12:57:59.394816+00:00 | 55 | false | This solution is based on the fact that every output will always be of the form `aaa...aaa?zzz...zzz` *regardless* of the values of `k` and `n`.\n\nThis submission ran at **32 ms** and used **14.8 MB** of memory. At the time of submission it was **faster than 98.62%** Python3 submissions, and used **less memory than 99... | 2 | 0 | ['Math'] | 0 |
smallest-string-with-a-given-numeric-value | ✅ JAVA || 10 LINES || ONE TRAVERSAL || EASY || SIMPLE || | java-10-lines-one-traversal-easy-simple-q4dhh | JAVAEASY CLEAN & READABLE\n\n\n\n* Our main goal is to build a string lexiographically smallest.\n* So the smallest possible string for any n would be full of \ | bharathkalyans | NORMAL | 2022-03-22T07:50:01.168608+00:00 | 2022-03-22T07:50:01.168640+00:00 | 95 | false | ``` ```**JAVA**``` ```**EASY**``` ``` ``` ```**CLEAN & READABLE**``` ```\n\n\n```\n* Our main goal is to build a string lexiographically smallest.\n* So the smallest possible string for any n would be full of \'a\'.\n* Now we have to fuflill another condition that is we want the sum of characters to be some k.\n* By ... | 2 | 1 | ['Java'] | 1 |
smallest-string-with-a-given-numeric-value | C++ || Greedy Approch && Reverse filling || O(n) time | c-greedy-approch-reverse-filling-on-time-rtzl | \nclass Solution {\npublic:\n string getSmallestString(int n, int k) {\n \n string res = "";\n \n while(n>0){\n \n | pandeyji2023 | NORMAL | 2022-03-22T07:00:26.458645+00:00 | 2022-03-22T07:00:26.458670+00:00 | 26 | false | ```\nclass Solution {\npublic:\n string getSmallestString(int n, int k) {\n \n string res = "";\n \n while(n>0){\n \n for(int i=26;i>=1;i--){\n \n int rem = k-i;\n if(rem >= n-1){\n char c = i-1+\'a\';\n ... | 2 | 0 | [] | 0 |
smallest-string-with-a-given-numeric-value | You will love it || Java || Simple Solution || Well Commented | you-will-love-it-java-simple-solution-we-uzt8 | Very Easy Solution. I have commented all steps.\n\nclass Solution {\n public String getSmallestString(int n, int k) {\n char[] ch = new char[n]; // In | samyakdarshan97 | NORMAL | 2022-03-22T06:21:30.311106+00:00 | 2022-03-22T06:21:30.311147+00:00 | 175 | false | Very Easy Solution. I have commented all steps.\n```\nclass Solution {\n public String getSmallestString(int n, int k) {\n char[] ch = new char[n]; // Initially taking a character array and later we will convert it to String.\n int i=0;\n for(;i<n;i++){\n ch[i] = \'a\'; // Initially k... | 2 | 0 | ['String', 'Greedy', 'Java'] | 0 |
smallest-string-with-a-given-numeric-value | By Math | by-math-by-haiyunjin-ktx7 | This is simply a math problem.\n\n1. Fill all with "a".\n2. Then for each extra number, add from the right. The number of "z"\'s on the right can be calculated | haiyunjin | NORMAL | 2022-03-22T06:05:28.215800+00:00 | 2022-03-22T06:05:51.132263+00:00 | 31 | false | This is simply a math problem.\n\n1. Fill all with "a".\n2. Then for each extra number, add from the right. The number of "z"\'s on the right can be calculated as `extra/25`.\n3. The letter right after "a"\'s can also be calculated by `\'a\' + extra%25`\n4. Then just add "a"\'s, first non-\'a\' and fill rest with "z".\... | 2 | 0 | ['Math', 'Java'] | 0 |
smallest-string-with-a-given-numeric-value | ✅ Easy-Peasy Solution | Greedy | easy-peasy-solution-greedy-by-igi17-pbaq | Intuition- At each index of string take smallest character possible such that remaining character \n are possible to fill with remaining k value. | igi17 | NORMAL | 2022-03-22T05:56:39.141779+00:00 | 2022-03-22T05:59:54.515901+00:00 | 78 | false | ***Intuition***- At each index of string take smallest character possible such that remaining character \n are possible to fill with remaining k value.\n\n\tclass Solution {\n\tpublic:\n\t\tstring getSmallestString(int n, int k) {\n\t\t\tstring ans;\n\t\t\tfor(int i=0;i<n;i++){\n\t\t\t\tfor(int j=0;j<26;... | 2 | 0 | ['Greedy', 'C'] | 0 |
smallest-string-with-a-given-numeric-value | C++ | one line solution, fast solution and explanation |O(n) or can O(1)? | c-one-line-solution-fast-solution-and-ex-tfp2 | 1. One line solution\nThe solution can be combined by three kind of string, so we have the one line solution. \nC++\nclass Solution {\npublic:\n string getSm | milochen | NORMAL | 2022-03-22T05:39:13.563841+00:00 | 2022-03-22T05:39:13.563867+00:00 | 141 | false | # 1. One line solution\nThe solution can be combined by three kind of string, so we have the one line solution. \n```C++\nclass Solution {\npublic:\n string getSmallestString(int n, int k) {\n\t\treturn string(n-(k-n)/25-((k-n)%25!=0), \'a\') + string(((k-n)%25!=0),\'a\'+(k-n)%25) + string((k-n)/25,\'z\');\n\t}\n};\... | 2 | 0 | ['C'] | 0 |
smallest-string-with-a-given-numeric-value | Greedy Python Solution with Explanation | greedy-python-solution-with-explanation-c78xw | Observation:\nFor any test case we will have some number of \'a\' in the start one or zero character from \'b\' to \'y\' in the middle and some number of \'z\' | ancoderr | NORMAL | 2022-03-22T04:47:28.337047+00:00 | 2022-03-22T04:52:46.093766+00:00 | 195 | false | **Observation:**\nFor any test case we will have some number of \'a\' in the start one or zero character from \'b\' to \'y\' in the middle and some number of \'z\' in the end.\n\n**Intuition:**\nWe want the lexicographically smallest string. If we have a choice to either chose b/w **`az`** and **`by`**, we will chose *... | 2 | 0 | ['Greedy', 'Python', 'Python3'] | 0 |
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