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values | created_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | updated_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | hit_count int64 0 10.6M | has_video bool 2
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reverse-prefix-of-word | C++ BEST SOLUTION | c-best-solution-by-poxyprabal-9bps | \n\n# Code\n\nclass Solution {\npublic:\n string reversePrefix(string word, char ch) {\n int l = 0;\n for (int r = 0; r < word.size(); r++) {\n | poxyprabal | NORMAL | 2024-05-01T16:18:34.467264+00:00 | 2024-05-01T16:18:34.467285+00:00 | 385 | false | \n\n# Code\n```\nclass Solution {\npublic:\n string reversePrefix(string word, char ch) {\n int l = 0;\n for (int r = 0; r < word.size(); r++) {\n if (word[r] == ch) {\n while (l <= r) {\n swap(word[r], word[l]);\n l++;\n ... | 4 | 0 | ['C++'] | 0 |
reverse-prefix-of-word | Simple easy to understand solution in 0ms. | simple-easy-to-understand-solution-in-0m-gkeu | Intuition\nMy intition was to count the index of character and then reverse it to that index using a reverse function.\n\n# Approach\nfind the index of characte | ayushsachan7 | NORMAL | 2024-05-01T10:02:09.668166+00:00 | 2024-05-01T10:02:09.668194+00:00 | 309 | false | # Intuition\nMy intition was to count the index of character and then reverse it to that index using a reverse function.\n\n# Approach\nfind the index of character using for loop and reverse upto that index using swap function.\n\n# Complexity\n- Time complexity: O(N)\n\n- Space complexity: O(1)\n\n# Code\n```\nclass S... | 4 | 0 | ['Two Pointers', 'String', 'C++'] | 4 |
reverse-prefix-of-word | EXPLAINED SOLUTION | explained-solution-by-nurliaidin-ys5p | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n1. Prepare to Store Pre | Nurliaidin | NORMAL | 2024-05-01T09:54:25.042058+00:00 | 2024-05-01T09:54:25.042078+00:00 | 532 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Prepare to Store Prefix and Suffix:**\n - Initialize `pre` to store characters before `ch`.\n - Set up `suf` to store any characters after `ch`.\n\n2. **Searc... | 4 | 0 | ['C++', 'JavaScript'] | 3 |
reverse-prefix-of-word | String find/reverse solution | string-findreverse-solution-by-drgavriko-0q8p | Approach\n Describe your approach to solving the problem. \n\nFirst, we find the position of the first occurrence of the character \'ch\' in the string \'word\' | drgavrikov | NORMAL | 2024-05-01T07:10:31.463305+00:00 | 2024-05-01T07:11:00.476863+00:00 | 10 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n\nFirst, we find the position of the first occurrence of the character \'ch\' in the string \'word\' using the \'find\' function.\n\nIf the character is found (i.e., its index is not equal to std::string::npos), we use the \'std::reverse\' function fr... | 4 | 0 | ['C++'] | 0 |
reverse-prefix-of-word | Easy solution and iterative solution | easy-solution-and-iterative-solution-by-3iwig | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nIterative approach\nYou | Rishabhdwivedii | NORMAL | 2024-05-01T04:49:08.416858+00:00 | 2024-05-01T04:49:08.416879+00:00 | 541 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nIterative approach\nYou define a function reversePrefix that takes a string word and a character ch as input parameters and returns a string.\nYou initialize an empty ... | 4 | 0 | ['Python3'] | 4 |
reverse-prefix-of-word | easiest approach cpp | | easiest-approach-cpp-by-ankitlodhi-3y6i | 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 | ankitlodhi | NORMAL | 2024-05-01T04:14:16.693785+00:00 | 2024-05-01T04:14:16.693819+00:00 | 342 | 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. $$O(n)$$ --... | 4 | 0 | ['C++'] | 0 |
reverse-prefix-of-word | The solution really easy to got it | the-solution-really-easy-to-got-it-by-ja-sayw | Code\n\nfunc reversePrefix(word string, ch byte) string {\n var natija string\n for i := 0; i < len(word); i++ {\n if word[i] == ch { \n | Javohir_hasanov | NORMAL | 2024-05-01T03:59:40.744903+00:00 | 2024-05-01T03:59:40.744933+00:00 | 218 | false | # Code\n```\nfunc reversePrefix(word string, ch byte) string {\n var natija string\n for i := 0; i < len(word); i++ {\n if word[i] == ch { \n ch_index := i\n for i >= 0 {\n natija += string(word[i])\n i--\n } \n return natija + word[... | 4 | 0 | ['Go'] | 2 |
reverse-prefix-of-word | 💯Two Pointers🎯✅ | 98.63%🔥| 4 Easy Steps : 🎯| Simple (5 line) to understand💯Explained | two-pointers-9863-4-easy-steps-simple-5-4rrkd | Intuition\nsimply we are using 2\uFE0F\u20E3 pointers\uD83D\uDC96\uD83D\uDC96 left and right to traverse pointer in opposite direction and concurrently swaping | Prathamesh18X | NORMAL | 2024-05-01T03:34:04.908921+00:00 | 2024-05-01T03:34:04.908946+00:00 | 1,005 | false | # Intuition\nsimply we are using **2\uFE0F\u20E3 pointers**\uD83D\uDC96\uD83D\uDC96 `left` and `right` to traverse pointer in opposite direction and concurrently swaping each other to reverse the string till `left` is less than `right`...\uD83D\uDCAF\n\n# Dont forget to vote...\u2B06\uFE0F **me**\uD83E\uDD73\n### *and*... | 4 | 0 | ['Two Pointers', 'String', 'C', 'Python', 'C++', 'Java', 'Python3', 'Ruby', 'Kotlin', 'JavaScript'] | 3 |
reverse-prefix-of-word | ✅Beats 91%🔥 2 Solutions🔥🔥🔥Simple character matching solution. With no additional space | beats-91-2-solutionssimple-character-mat-cly4 | image.png\n\n\n# Intuition\nIf we can find the given charater ch in given string word, we have to reverse string till first ch including it. To do this, we can | Saketh3011 | NORMAL | 2024-05-01T01:31:10.330068+00:00 | 2024-05-01T07:01:33.198804+00:00 | 632 | false | image.png\n\n\n# Intuition\nIf we can find the given charater `ch` in given string `word`, we have to reverse string till first `ch` including it. To do this, we can find first `ch` index in `word` and then... | 4 | 0 | ['Two Pointers', 'String', 'C', 'Python', 'C++', 'Java', 'Python3'] | 1 |
reverse-prefix-of-word | Java || 100% beats | java-100-beats-by-saurabh_kumar1-h12l | 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 | saurabh_kumar1 | NORMAL | 2023-08-16T11:06:03.444839+00:00 | 2023-08-16T11:06:03.444864+00:00 | 80 | 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:0(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:0(1)\n<!-- Add your space complexity here, e.g. $$O... | 4 | 0 | ['Java'] | 1 |
reverse-prefix-of-word | [ Go Solution ]Great explanation and Full Description | go-solution-great-explanation-and-full-d-8zrr | Intuition\nThe problem asks to reverse the prefix of a string up to the first occurrence of a specified character. If the character does not exist in the string | sansaian | NORMAL | 2023-06-21T14:34:16.549695+00:00 | 2023-06-21T14:34:16.549721+00:00 | 454 | false | # Intuition\nThe problem asks to reverse the prefix of a string up to the first occurrence of a specified character. If the character does not exist in the string, we should return the original string. The problem can be solved by iterating through the string and once we find the specified character, we reverse the pre... | 4 | 0 | ['Two Pointers', 'Go'] | 0 |
reverse-prefix-of-word | reverse prefix python3 | reverse-prefix-python3-by-timsmyrnov-cy41 | Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(N)\n\n# Code\n\nclass Solution:\n def reversePrefix(self, word: str, ch: str) -> str:\n | timsmyrnov | NORMAL | 2023-05-01T07:40:58.023315+00:00 | 2023-05-01T07:40:58.023348+00:00 | 717 | false | # Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(N)\n\n# Code\n```\nclass Solution:\n def reversePrefix(self, word: str, ch: str) -> str:\n prefix = \'\'\n for i, c in enumerate(word):\n if c == ch:\n return (prefix + c)[::-1] + word[i+1:]\n prefix +... | 4 | 0 | ['Python3'] | 2 |
reverse-prefix-of-word | Sol: Reverse Prefix of Word [JAVA] | sol-reverse-prefix-of-word-java-by-subha-glbe | Code\n\nclass Solution {\n public String reversePrefix(String word, char ch) {\n String subString = word.substring(0, word.indexOf(ch)+1);\n\t\tString | subhajitbhattacharjee007 | NORMAL | 2022-12-28T12:18:56.394869+00:00 | 2022-12-28T12:19:13.747449+00:00 | 371 | false | # Code\n```\nclass Solution {\n public String reversePrefix(String word, char ch) {\n String subString = word.substring(0, word.indexOf(ch)+1);\n\t\tString remainingString = word.substring( word.indexOf(ch)+1, word.length());\n\t\tString s = "";\n\t\tfor( int i=subString.length()-1; i>=0; i--) {\n\t\t\ts=s+s... | 4 | 0 | ['Java'] | 2 |
reverse-prefix-of-word | C++ | Find and Reverse | Easy | c-find-and-reverse-easy-by-shreyanshxyz-kvb3 | \nclass Solution {\npublic:\n string reversePrefix(string word, char ch) {\n if(word.length() == 1) return word;\n \n for(int i = 0; i < | shreyanshxyz | NORMAL | 2022-08-10T13:50:38.788842+00:00 | 2022-08-10T13:50:38.788883+00:00 | 239 | false | ```\nclass Solution {\npublic:\n string reversePrefix(string word, char ch) {\n if(word.length() == 1) return word;\n \n for(int i = 0; i < word.size(); i++){\n if(word[i] == ch){\n reverse(word.begin(), word.begin() + i + 1);\n break;\n }\n ... | 4 | 0 | ['C'] | 0 |
reverse-prefix-of-word | C++ solution || 2 approaches || Solved using Stack(3ms) || using reverse function(0ms) | c-solution-2-approaches-solved-using-sta-ttv8 | \t\t\t\t\t\t\t\t\t----Approach 1(Stack)----\n\n\tclass Solution {\n\tpublic:\n\t\tstring reversePrefix(string word, char ch) {\n\n stack s;\n int | ap00rv_13 | NORMAL | 2022-07-08T04:00:56.179367+00:00 | 2022-07-08T04:00:56.179398+00:00 | 254 | false | \t\t\t\t\t\t\t\t\t----Approach 1(Stack)----\n\n\tclass Solution {\n\tpublic:\n\t\tstring reversePrefix(string word, char ch) {\n\n stack<char> s;\n int i=0, count = 0;\n int size = word.size();\n\n while(word[i] != ch){\n count++;\n if(count == size) return ... | 4 | 0 | ['Stack', 'C', 'C++'] | 1 |
reverse-prefix-of-word | c++ simple one line code | c-simple-one-line-code-by-sailakshmi1-0x70 | ```\nclass Solution {\npublic:\n string reversePrefix(string word, char ch) {\n int j=0;\n for(int i=0;i<word.size();i++){\n if(word[i]= | sailakshmi1 | NORMAL | 2022-06-18T08:41:16.997662+00:00 | 2022-06-18T08:41:16.997703+00:00 | 251 | false | ```\nclass Solution {\npublic:\n string reversePrefix(string word, char ch) {\n int j=0;\n for(int i=0;i<word.size();i++){\n if(word[i]==ch){\n reverse(word.begin(),word.begin()+i+1);\n break;\n }\n }\n return word;\n }\n}; | 4 | 0 | ['C'] | 1 |
reverse-prefix-of-word | Python 3 (25ms) | One Line Solution | Faster than 95% | Easy to Understand | python-3-25ms-one-line-solution-faster-t-6sla | \nclass Solution:\n def reversePrefix(self, word: str, ch: str) -> str:\n if ch not in word:\n return word\n return (\'\'.join(rever | MrShobhit | NORMAL | 2022-01-25T17:24:59.420287+00:00 | 2022-01-25T17:24:59.420331+00:00 | 545 | false | ```\nclass Solution:\n def reversePrefix(self, word: str, ch: str) -> str:\n if ch not in word:\n return word\n return (\'\'.join(reversed(word[:(word.index(ch)+1)]))+word[(word.index(ch))+1:])\n``` | 4 | 0 | ['String', 'Python', 'Python3'] | 4 |
reverse-prefix-of-word | [JavaScript] one line solution (86%,35%) | javascript-one-line-solution-8635-by-053-xz0d | Runtime: 76 ms, faster than 86.18% of JavaScript online submissions for Reverse Prefix of Word.\nMemory Usage: 38.9 MB, less than 35.53% of JavaScript online su | 0533806 | NORMAL | 2021-09-15T02:25:40.756831+00:00 | 2021-09-15T02:25:40.756869+00:00 | 641 | false | Runtime: 76 ms, faster than 86.18% of JavaScript online submissions for Reverse Prefix of Word.\nMemory Usage: 38.9 MB, less than 35.53% of JavaScript online submissions for Reverse Prefix of Word.\n```\nvar reversePrefix = function(word, ch) {\n return word.indexOf(ch) !== -1 ? word.split("").slice(0, word.indexOf(... | 4 | 0 | ['JavaScript'] | 2 |
reverse-prefix-of-word | C++ Simple and Easy 2-Line Solution, 0ms Faster than 100% | c-simple-and-easy-2-line-solution-0ms-fa-q8ym | We find the index of the first accurance of the char in word, then we use built-in function reverse which accepts a range to reverse.\n\nclass Solution {\npubli | yehudisk | NORMAL | 2021-09-13T14:34:55.396312+00:00 | 2021-09-13T14:34:55.396354+00:00 | 192 | false | We find the index of the first accurance of the char in word, then we use built-in function `reverse` which accepts a range to reverse.\n```\nclass Solution {\npublic:\n string reversePrefix(string word, char ch) {\n reverse(word.begin(), word.begin() + word.find(ch) + 1);\n return word;\n }\n};\n``... | 4 | 0 | ['C'] | 0 |
reverse-prefix-of-word | Python 3, straightforward | python-3-straightforward-by-silvia42-pwo9 | We are iterating over the string word until character ch is found.\nWe return reversed slice plus leftover from the string word.\n\nclass Solution:\n def rev | silvia42 | NORMAL | 2021-09-12T04:05:37.842968+00:00 | 2021-09-12T05:37:15.328076+00:00 | 357 | false | We are iterating over the string ```word``` until character ```ch``` is found.\nWe return reversed slice plus leftover from the string ```word```.\n```\nclass Solution:\n def reversePrefix(self, word: str, ch: str) -> str:\n for i in range(len(word)):\n if word[i]==ch:\n return word[... | 4 | 1 | [] | 2 |
reverse-prefix-of-word | beats 100% simple solution | beats-100-simple-solution-by-harshjat-x8uy | IntuitionThe problem requires reversing the prefix of the given string word up to the first occurrence of the character ch. A natural approach is to iterate thr | harshjat | NORMAL | 2025-03-03T09:46:42.869870+00:00 | 2025-03-03T09:46:42.869870+00:00 | 49 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem requires reversing the prefix of the given string word up to the first occurrence of the character ch. A natural approach is to iterate through the string, identify the first occurrence of ch, and reverse the substring from the ... | 3 | 0 | ['C++'] | 0 |
reverse-prefix-of-word | Real Solution using stack | real-solution-using-stack-by-piero24-r8h8 | Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(n)\n\n# Code\npython3 []\nclass Solution:\n def reversePrefix(self, word: str, ch: str) -> str:\ | Piero24 | NORMAL | 2024-09-27T13:48:54.903768+00:00 | 2024-09-27T13:48:54.903788+00:00 | 142 | false | # Complexity\n- Time complexity: $$O(n)$$\n\n- Space complexity: $$O(n)$$\n\n# Code\n```python3 []\nclass Solution:\n def reversePrefix(self, word: str, ch: str) -> str:\n head, stack = "", []\n for i, e in enumerate(word):\n if e == ch:\n head = e\n word = word... | 3 | 0 | ['Stack', 'C++', 'Java', 'Python3'] | 0 |
reverse-prefix-of-word | Beats 100.00% of users with C++ || Step By Step Explain || Easy To Understand || | beats-10000-of-users-with-c-step-by-step-8e05 | Abhiraj Pratap Singh\n\n---\n\n# if you like the solution please UPVOTE it....\n\n\n---\n\n# Intuition\n- The problem seems to involve reversing a substring of | abhirajpratapsingh | NORMAL | 2024-05-02T16:59:15.122512+00:00 | 2024-05-02T16:59:15.122546+00:00 | 13 | false | # Abhiraj Pratap Singh\n\n---\n\n# if you like the solution please UPVOTE it....\n\n\n---\n\n# Intuition\n- The problem seems to involve reversing a substring of a given string starting from the beginning up to the occurrence of a specified character.\n\n---\n\n](https://leetcode.com/problems/reverse-prefix-of-word/submissions/1246206292/)\n\n\n# Intuition\nThis problem involves reversing a substring of a given string up to a specified ch... | 3 | 0 | ['C++'] | 2 |
reverse-prefix-of-word | ✅Simple Solution || Multiple Approach || Easy Explanation || Beginner Friendly🔥🔥🔥 | simple-solution-multiple-approach-easy-e-h0gv | Intuition\nThe intuition behind this solution is to reverse the prefix of the given string word up to the first occurrence of the character ch. If the character | Sayan98 | NORMAL | 2024-05-01T06:52:06.467123+00:00 | 2024-05-02T05:44:56.251013+00:00 | 246 | false | # Intuition\nThe intuition behind this solution is to reverse the prefix of the given string word up to the first occurrence of the character ch. If the character ch is not present in the string, the entire string is returned unchanged.\n\n# Approach\n- Find the **index** of the first occurrence of the character ch in ... | 3 | 0 | ['Two Pointers', 'String', 'C#'] | 1 |
reverse-prefix-of-word | Beats 100% C++ solution | with explanantion | beats-100-c-solution-with-explanantion-b-okxg | Approach\n- Iterate over input string till we find required character.\n- Store and reverse string till current index\n- Return by adding it with remaining stri | anupsingh556 | NORMAL | 2024-05-01T06:51:12.135148+00:00 | 2024-05-01T06:51:39.546433+00:00 | 5 | false | # Approach\n- Iterate over input string till we find required character.\n- Store and reverse string till current index\n- Return by adding it with remaining string\n- If no occurence is found we can simply return original string\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n# Code\n```\nclass... | 3 | 0 | ['C++'] | 0 |
reverse-prefix-of-word | Python | Easy | python-easy-by-khosiyat-f9pt | see the Successfully Accepted Submission\n\n# Code\n\nclass Solution:\n def reversePrefix(self, word: str, ch: str) -> str:\n index = word.find(ch) # | Khosiyat | NORMAL | 2024-05-01T06:48:40.875554+00:00 | 2024-05-01T06:48:40.875591+00:00 | 362 | false | [see the Successfully Accepted Submission](https://leetcode.com/problems/reverse-prefix-of-word/submissions/1246310040/?source=submission-ac)\n\n# Code\n```\nclass Solution:\n def reversePrefix(self, word: str, ch: str) -> str:\n index = word.find(ch) # Find the index of the first occurrence of ch\n i... | 3 | 0 | ['Python3'] | 1 |
reverse-prefix-of-word | Best Solution || Beats 100% || Simplified Approach || O(n) || C++ || | best-solution-beats-100-simplified-appro-gbhd | # Intuition \n\n\n# Approach\n\n\n\nThe approach of the provided code is to reverse the prefix of a given string up to the first occurrence of a specific char | atharvviit | NORMAL | 2024-05-01T06:34:50.696515+00:00 | 2024-05-01T06:34:50.696546+00:00 | 116 | 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\nThe approach of the provided code is to reverse the prefix of a given string up to the first occurrence of a specific character. Here\'s a step-by-step br... | 3 | 0 | ['Two Pointers', 'String', 'C++', 'Java'] | 0 |
reverse-prefix-of-word | Easy to understand solution CPP, Java and Python | easy-to-understand-solution-cpp-java-and-sk7k | Intuition\n Describe your first thoughts on how to solve this problem. \nApply brute force i.e, do as you are asked to do \n\n# Approach\n Describe your approac | Harshit_PSIT | NORMAL | 2024-05-01T05:14:15.187369+00:00 | 2024-05-01T05:14:15.187401+00:00 | 687 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nApply brute force i.e, do as you are asked to do \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nAlgrithm to be performed-:\n- Find the index of the given character in the word string\n- Generate a substring from ... | 3 | 0 | ['Two Pointers', 'String', 'Python', 'C++', 'Java', 'Python3'] | 0 |
reverse-prefix-of-word | ✅ Simple 3 line solution beat 100% 🚀 | simple-3-line-solution-beat-100-by-cs_ba-335x | Code\n\nclass Solution {\npublic:\n string reversePrefix(string word, char ch) {\n int ind = word.find(ch);\n if (ind >= word.size())\n | cs_balotiya | NORMAL | 2024-05-01T04:30:51.316661+00:00 | 2024-05-01T04:31:15.930088+00:00 | 9 | false | # Code\n```\nclass Solution {\npublic:\n string reversePrefix(string word, char ch) {\n int ind = word.find(ch);\n if (ind >= word.size())\n return word; // IF CH IS NOT IN WORD;\n reverse(word.begin(), word.begin() + (ind + 1)); // REVERSE FROM STARTING TO POSITION OF CH +1;\n ... | 3 | 0 | ['Two Pointers', 'String', 'C++'] | 0 |
reverse-prefix-of-word | 🍉0ms||✅Beats 100%🔥|| Beginner friendly || Explained🙌 | 0msbeats-100-beginner-friendly-explained-i2f3 | Intuition\nNothing but we just traverse whole string until we find given ch in word\n\n# Approach\n1) Intialize index = 0; //for finding index\n2) So first we f | dharmikgohil | NORMAL | 2024-05-01T04:19:01.434942+00:00 | 2024-05-01T04:19:01.434964+00:00 | 207 | false | # Intuition\nNothing but we just traverse whole string until we find given ch in word\n\n# Approach\n1) Intialize index = 0; //for finding index\n2) So first we find the index of ch at which index it occuring in word\n3) If we find that index so we given it to index = i;\n4) Now let\'s create empty string and stores ch... | 3 | 0 | ['String', 'Python', 'C++', 'Java'] | 0 |
reverse-prefix-of-word | Full Detailed Explanation 🔥✅ | full-detailed-explanation-by-shivakumar1-knvo | Intuition\nSimple brute force, find the index of ch and just reverse from 0 to chth index.\n# Code Explanation\n\n\n - Two integer variables left and right are | shivakumar1 | NORMAL | 2024-05-01T03:09:29.498930+00:00 | 2024-05-01T03:09:29.498954+00:00 | 555 | false | # Intuition\nSimple brute force, find the index of `ch` and just reverse from 0 to `ch`th index.\n# Code Explanation\n\n\n - Two integer variables `left` and `right` are initialized to 0 and the index of the first occurrence of character `ch` in the string `word`, respectively.\n - A `StringBuilder` named `sb` is ini... | 3 | 0 | ['Two Pointers', 'String', 'Java'] | 1 |
reverse-prefix-of-word | Beginner friendly Easy JAVA solution🚀 | beginner-friendly-easy-java-solution-by-anjih | Intuition\nThe approach involves finding the index of the first occurrence of the given character ch in the string word. If the index is greater than 0, reverse | parthiv_77 | NORMAL | 2024-02-04T06:01:40.543075+00:00 | 2024-02-04T06:01:40.543101+00:00 | 123 | false | # Intuition\nThe approach involves finding the index of the first occurrence of the given character ch in the string word. If the index is greater than 0, reverse the substring from the beginning of the word up to that index, and concatenate it with the remaining portion of the word.\n\n# Approach\nFind the index of th... | 3 | 0 | ['Java'] | 0 |
reverse-prefix-of-word | Simple Solution | simple-solution-by-adwxith-wgtg | 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 | adwxith | NORMAL | 2023-11-15T15:03:36.875883+00:00 | 2023-11-15T15:03:36.875912+00:00 | 198 | 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 | ['JavaScript'] | 1 |
reverse-prefix-of-word | ✅Just in 3 lines | 0ms | Easy Approach | just-in-3-lines-0ms-easy-approach-by-ash-ek4e | Intuition\nWe just need the index of the character \'ch\', then just reverse the order from beginning to the end. That\'s it!!\n\n# Approach\nI\'m getting the i | ashitoshsable07 | NORMAL | 2023-06-08T21:29:33.640698+00:00 | 2023-06-08T21:29:33.640728+00:00 | 253 | false | # Intuition\nWe just need the **index** of the character \'ch\', then just reverse the order from beginning to the end. That\'s it!!\n\n# Approach\nI\'m getting the index of the character \'ch\' by using find function after that I\'m reversing the string from word.begin() to that index of ch.\n\n# Complexity\n- Time co... | 3 | 0 | ['C++'] | 1 |
reverse-prefix-of-word | Java Easy Solution | 0 ms - 100% beats | java-easy-solution-0-ms-100-beats-by-ako-l0vd | Approach\n Describe your approach to solving the problem. \n1. Find the index of the first occurrence of the character \'ch\' in the given string \'word\' using | akobirswe | NORMAL | 2023-05-30T13:53:34.094454+00:00 | 2023-05-30T13:53:34.094484+00:00 | 283 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n1. Find the index of the first occurrence of the character \'ch\' in the given string \'word\' using the `indexOf` method. If the character is not found (index is -1), return the original string \'word\' since there is nothing to reverse.\n2. Create a... | 3 | 0 | ['Array', 'Two Pointers', 'String', 'Java'] | 1 |
reverse-prefix-of-word | C# easy | c-easy-by-ghmarek-bi0m | Code\n\npublic class Solution {\n public string ReversePrefix(string word, char ch) {\n\n char[] tempCharArr = word.ToArray();\n\n Array.Revers | GHMarek | NORMAL | 2023-04-24T16:32:05.038599+00:00 | 2023-04-24T16:32:05.038632+00:00 | 39 | false | # Code\n```\npublic class Solution {\n public string ReversePrefix(string word, char ch) {\n\n char[] tempCharArr = word.ToArray();\n\n Array.Reverse(tempCharArr, 0, word.IndexOf(ch) + 1);\n\n return new string(tempCharArr);\n \n }\n}\n``` | 3 | 0 | ['C#'] | 0 |
reverse-prefix-of-word | Go easy solution | go-easy-solution-by-azizjon003-glh1 | \n# Complexity\n- Time complexity:59%\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:70%\n Add your space complexity here, e.g. O(n) \n\n# Co | Azizjon003 | NORMAL | 2023-04-12T08:48:16.737121+00:00 | 2023-04-12T08:48:16.737163+00:00 | 199 | false | \n# Complexity\n- Time complexity:59%\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:70%\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nfunc reversePrefix(word string, ch byte) string {\n a := []byte(word)\n\n for i := 0; i < len(word); i++ {\n if word... | 3 | 0 | ['Array', 'String', 'Go'] | 1 |
reverse-prefix-of-word | Easy Python3 Solution | easy-python3-solution-by-smisuol-s34b | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | smisuol | NORMAL | 2023-02-05T04:55:53.023500+00:00 | 2023-02-05T04:56:23.629684+00:00 | 621 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N)\n<!-- Add your space complexity here, e.g. $... | 3 | 0 | ['Python3'] | 2 |
reverse-prefix-of-word | Beats 99.38% | beats-9938-by-codequeror-xlcw | Upvote it :)\n\nclass Solution:\n def reversePrefix(self, word: str, ch: str) -> str:\n try:\n idx = word.index(ch)\n w = word[: | Codequeror | NORMAL | 2023-01-21T14:14:08.418562+00:00 | 2023-01-21T14:14:08.418603+00:00 | 624 | false | # Upvote it :)\n```\nclass Solution:\n def reversePrefix(self, word: str, ch: str) -> str:\n try:\n idx = word.index(ch)\n w = word[:idx + 1]\n return w[::-1] + word[idx + 1:]\n except: return word\n``` | 3 | 1 | ['Python', 'Python3'] | 0 |
reverse-prefix-of-word | JAVA || TWO POINTER APPROACH || BEATS 96% | java-two-pointer-approach-beats-96-by-sh-fdd0 | \n\n# Code\n\nclass Solution {\n public String reversePrefix(String word, char ch) {\n char[] array = word.toCharArray();\n int i = 0;\n | sharforaz_rahman | NORMAL | 2022-12-05T17:56:06.705542+00:00 | 2022-12-05T17:56:06.705578+00:00 | 808 | false | \n\n# Code\n```\nclass Solution {\n public String reversePrefix(String word, char ch) {\n char[] array = word.toCharArray();\n int i = 0;\n int j = word.length();\n int index = 0;\n while (i < j) {\n if (array[i] == ch) {\n index = i;\n brea... | 3 | 0 | ['Java'] | 1 |
reverse-prefix-of-word | Java Two pointers approach(1ms) | java-two-pointers-approach1ms-by-amrishr-3qg4 | public String reversePrefix(String word, char ch) {\n int start=0;\n char[] temp=word.toCharArray();\n for(int end=0;end<temp.length;end++) | AmrishRaaj | NORMAL | 2022-10-31T05:04:52.531531+00:00 | 2022-10-31T05:04:52.531571+00:00 | 653 | false | public String reversePrefix(String word, char ch) {\n int start=0;\n char[] temp=word.toCharArray();\n for(int end=0;end<temp.length;end++)\n {\n if(temp[end]==ch)\n {\n while(start<end)\n {\n char tempS=temp[start];\n ... | 3 | 0 | ['Java'] | 1 |
reverse-prefix-of-word | Javascript O(N) - One Iteration | javascript-on-one-iteration-by-videshgho-vb00 | \n/**\n * @param {string} word\n * @param {character} ch\n * @return {string}\n */\nvar reversePrefix = function(word, ch) {\n const index = word.indexOf(ch | videshghodarop | NORMAL | 2022-10-06T14:04:56.108354+00:00 | 2022-10-06T14:04:56.108394+00:00 | 358 | false | ```\n/**\n * @param {string} word\n * @param {character} ch\n * @return {string}\n */\nvar reversePrefix = function(word, ch) {\n const index = word.indexOf(ch);\n if (index === -1) return word;\n let result = \'\';\n\n for (let i = 0; i < word.length; i++) {\n if (i <= index) {\n result = word[i] + resu... | 3 | 0 | ['JavaScript'] | 0 |
reverse-prefix-of-word | Python Easy Solution in 5 Lines | python-easy-solution-in-5-lines-by-pulki-whm1 | \nclass Solution:\n def reversePrefix(self, word: str, ch: str) -> str:\n if ch not in word:\n return word\n i=word.index(ch)\n | pulkit_uppal | NORMAL | 2022-10-02T13:56:22.422498+00:00 | 2022-10-02T13:56:22.422539+00:00 | 503 | false | ```\nclass Solution:\n def reversePrefix(self, word: str, ch: str) -> str:\n if ch not in word:\n return word\n i=word.index(ch)\n word=word[i::-1]+word[i+1:]\n return word\n``` | 3 | 0 | ['Python'] | 2 |
reverse-prefix-of-word | c++ easy | c-easy-by-sailakshmi1-4sm5 | ```\nclass Solution {\npublic:\n string reversePrefix(string word, char ch) {\n int j=0;\n int n=word.length();\n for(int i=0;i<n;i++){\ | sailakshmi1 | NORMAL | 2022-06-15T14:15:02.983243+00:00 | 2022-06-15T14:15:43.256753+00:00 | 136 | false | ```\nclass Solution {\npublic:\n string reversePrefix(string word, char ch) {\n int j=0;\n int n=word.length();\n for(int i=0;i<n;i++){\n if(word[i]==ch){\n reverse(word.begin(),word.begin()+i+1);\n break;\n }\n }\n return word;\n }\... | 3 | 0 | ['C'] | 0 |
reverse-prefix-of-word | java 76% easy to understand! | java-76-easy-to-understand-by-m_israilov-evvi | \n\nclass Solution {\n public String reversePrefix(String word, char ch) {\n Integer a = word.indexOf(ch + "");\n StringBuilder b = new | m_israilovv | NORMAL | 2022-05-22T16:14:02.522223+00:00 | 2022-05-22T16:14:11.461870+00:00 | 88 | false | ```\n\n```class Solution {\n public String reversePrefix(String word, char ch) {\n Integer a = word.indexOf(ch + "");\n StringBuilder b = new StringBuilder(word.substring(0, a + 1));\n return b.reverse() + word.substring(a + 1, word.length());\n }\n} | 3 | 0 | [] | 2 |
reverse-prefix-of-word | Smart & fastest java solution using substring and without any loop 3 line code. 0 ms | smart-fastest-java-solution-using-substr-t1xi | \nclass Solution {\n public String reversePrefix(String word, char ch) \n {\n StringBuilder s = new StringBuilder();\n int index = word.inde | saurabh_173 | NORMAL | 2022-02-23T15:28:54.070531+00:00 | 2022-02-23T15:28:54.070574+00:00 | 196 | false | ```\nclass Solution {\n public String reversePrefix(String word, char ch) \n {\n StringBuilder s = new StringBuilder();\n int index = word.indexOf(ch);\n s.append(word.substring(0,index+1));\n s.reverse();\n s.append(word.substring(index+1));\n return s.toString();\n }... | 3 | 0 | ['String', 'Java'] | 0 |
reverse-prefix-of-word | [Python3/Golang] Easy to understand | python3golang-easy-to-understand-by-frol-ae78 | First of all we want to check if there is a ch in the word, then we have to return ch index plus 1, reverse the order and add the remainder of the word unchange | frolovdmn | NORMAL | 2021-09-15T08:43:59.331564+00:00 | 2025-01-19T14:37:37.689396+00:00 | 136 | false | First of all we want to check if there is a **ch** in the **word**, then we have to return **ch** index plus 1, reverse the order and add the remainder of the **word** unchanged
```python []
class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
return word[:word.index(ch) + 1][::-1] + word[wo... | 3 | 0 | ['Python', 'Go', 'Python3'] | 0 |
reverse-prefix-of-word | C++ very easy solution | c-very-easy-solution-by-pk_87-oapj | ```\nclass Solution {\npublic:\n string reversePrefix(string word, char ch) {\n \n int index=word.find(ch);\n if(index == -1)\n | pawan_mehta | NORMAL | 2021-09-12T04:09:27.889522+00:00 | 2021-09-12T04:09:27.889549+00:00 | 128 | false | ```\nclass Solution {\npublic:\n string reversePrefix(string word, char ch) {\n \n int index=word.find(ch);\n if(index == -1)\n return word;\n reverse(word.begin(),word.begin()+index+1);\n return word;\n }\n}; | 3 | 0 | [] | 2 |
reverse-prefix-of-word | 🔥🚀Two Easy Approach💯 || Time --O(n) | two-easy-approach-time-on-by-0xjwuvdyfc-h4eb | Reverse Prefix of a WordProblem StatementGiven a string word and a character ch, reverse the prefix of word that ends with ch. If ch does not exist in word, ret | 0XjwUvdYfc | NORMAL | 2025-03-24T07:52:28.629085+00:00 | 2025-03-24T07:52:28.629085+00:00 | 157 | false | ## Reverse Prefix of a Word
### Problem Statement
Given a string `word` and a character `ch`, reverse the prefix of `word` that ends with `ch`. If `ch` does not exist in `word`, return `word` unchanged.
### Approach
1. Use a stack to store characters until we find `ch`.
2. If `ch` is found, reverse the stored charact... | 2 | 0 | ['String', 'Stack', 'Java'] | 0 |
reverse-prefix-of-word | Beats 100% in runtime 🔥| Optimal O(1) solution ⏳| Very Easy Approach ✨ | beats-100-in-runtime-optimal-o1-solution-fa79 | IntuitionThe problem requires reversing a prefix of a given string up to the first occurrence of a specified character. The idea is to locate this character and | Dev_Sh | NORMAL | 2025-03-14T00:00:59.409499+00:00 | 2025-03-14T00:00:59.409499+00:00 | 70 | false | # Intuition
The problem requires reversing a prefix of a given string up to the first occurrence of a specified character. The idea is to locate this character and reverse the portion of the string from the beginning to that index.
# Approach
1. First, check if the character `ch` exists in the string using `word.find(... | 2 | 0 | ['C++'] | 0 |
reverse-prefix-of-word | beats 100%,super simple python3 code | beats-100super-simple-python3-code-by-pr-5zgk | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | PranaviGuddanti | NORMAL | 2025-02-03T13:53:03.957506+00:00 | 2025-02-03T13:53:03.957506+00:00 | 157 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 2 | 0 | ['Python3'] | 0 |
reverse-prefix-of-word | C++ 100% beating Brute Force | c-100-beating-brute-force-by-itsaditya7-bz3v | null | itsaditya7 | NORMAL | 2024-12-11T07:52:13.530642+00:00 | 2024-12-11T07:52:31.772900+00:00 | 45 | false | # Intuition\nJust find the ch.and reverse the string till the ith index.\n\n# Approach\nStart the code with onw for loop to find the character position from where we want to reverse the string.\n\n# Complexity\n- Time complexity:\nO(MN)\nwhere \nM:number at which the ch is found\n\n- Space complexity:\nO(1)\n\n# Code\n... | 2 | 0 | ['C++'] | 0 |
reverse-prefix-of-word | Effortless Character-Based Prefix Reversal in C++ | effortless-character-based-prefix-revers-4r6q | Intuition\n\nWhen thinking about how to reverse a part of a string based on the presence of a specific character, the first thought is to determine the position | MahmoudMohamedElsayed | NORMAL | 2024-10-10T19:04:10.813172+00:00 | 2024-10-10T19:04:10.813201+00:00 | 21 | false | # Intuition\n\nWhen thinking about how to reverse a part of a string based on the presence of a specific character, the first thought is to determine the position of that character. Once the character is found, we can reverse the portion of the string that precedes it. This will allow us to achieve the desired result e... | 2 | 0 | ['C++'] | 0 |
reverse-prefix-of-word | easy solution | 100.00% beats | easy-solution-10000-beats-by-q0art-3gfv | \n\ntypescript []\nconst reversePrefix = (string: string, char: string) => {\n const index = string.indexOf(char)\n\n return string.slice(0, index + 1).split( | q0art | NORMAL | 2024-09-24T22:03:57.498754+00:00 | 2024-09-24T22:03:57.498785+00:00 | 45 | false | \n\n```typescript []\nconst reversePrefix = (string: string, char: string) => {\n const index = string.indexOf(char)\n\n return string.slice(0, index + 1).split("").reverse... | 2 | 0 | ['TypeScript'] | 0 |
reverse-prefix-of-word | Q. 2000 | ✅Beginner-Friendly Python Code | 4️⃣ lined. | q-2000-beginner-friendly-python-code-4-l-lttx | Intuition\nThe problem asks us to reverse the prefix of a string up to and including the first occurrence of a given character ch. The most efficient approach i | charan1kh | NORMAL | 2024-08-08T16:29:17.798771+00:00 | 2024-09-21T08:01:37.318440+00:00 | 87 | false | # Intuition\nThe problem asks us to reverse the prefix of a string up to and including the first occurrence of a given character ch. The most efficient approach is to locate ch once, reverse the portion of the string up to its index, and then concatenate it with the remaining part of the string.\n\n\n# Approach\n1. Fin... | 2 | 0 | ['Python3'] | 2 |
reverse-prefix-of-word | Best and Easy to Understand ✅ || Beats 100% 🎯 | best-and-easy-to-understand-beats-100-by-ynlk | Intuition\n Describe your first thoughts on how to solve this problem. The task requires reversing a part of the string from the start up to and including the f | rckrockerz | NORMAL | 2024-05-23T09:37:10.623796+00:00 | 2024-05-23T09:37:10.623825+00:00 | 39 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->The task requires reversing a part of the string from the start up to and including the first occurrence of a given character. If the character does not exist in the string, the string remains unchanged. The idea is straightforward: find th... | 2 | 0 | ['Two Pointers', 'String', 'C++'] | 0 |
reverse-prefix-of-word | JAVA SOLUTION | java-solution-by-deleted_user-q2xe | Code\n\nclass Solution {\n public String reversePrefix(String word, char ch) {\n # initialise strings\n String ans = "";\n String rev = | deleted_user | NORMAL | 2024-05-03T16:04:08.987809+00:00 | 2024-05-03T16:04:08.987837+00:00 | 434 | false | # Code\n```\nclass Solution {\n public String reversePrefix(String word, char ch) {\n # initialise strings\n String ans = "";\n String rev = "";\n # iterate through word and find the character\n int n = word.length();\n for(int i = 0; i < n; i++) {\n if(word.charA... | 2 | 0 | ['Java'] | 0 |
reverse-prefix-of-word | Golang beats 100% | golang-beats-100-by-do2358-ju1d | Intuition\n Describe your first thoughts on how to solve this problem. \n- The reversePrefix function iterates through each character of the input word.\n- For | do2358 | NORMAL | 2024-05-02T15:20:03.901615+00:00 | 2024-05-02T15:20:03.901651+00:00 | 66 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- The reversePrefix function iterates through each character of the input word.\n- For each character at index i, it checks if it matches the target character ch.\n- If a match is found, the function proceeds to reverse the prefix.\n# App... | 2 | 0 | ['Go'] | 0 |
reverse-prefix-of-word | Simple Solution [Faster than 82% Runtime]✅ | simple-solution-faster-than-82-runtime-b-tu7j | Approach\n- Exit function block early, if word does not contain given ch\n- Get the index for ch in the word string\n- Split word into two subsequences, reverse | ProbablyLost | NORMAL | 2024-05-02T12:54:40.377943+00:00 | 2024-05-02T12:57:35.951565+00:00 | 52 | false | # Approach\n- Exit function block early, if `word` does not contain given `ch`\n- Get the `index` for ch in the word string\n- Split word into two subsequences, `reversedFirstPart` and `secondPart`\n- Concatenate and **return** the subsequences\n\n# Code\n```\nclass Solution {\n func reversePrefix(_ word: String, _ ... | 2 | 0 | ['String', 'Swift'] | 0 |
reverse-prefix-of-word | C++ easiest solution Beats 100% solutions. 2 pointer method | c-easiest-solution-beats-100-solutions-2-tudj | 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 | _chitransh_9 | NORMAL | 2024-05-01T18:02:50.425590+00:00 | 2024-05-01T18:02:50.425617+00:00 | 193 | 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)$$ --... | 2 | 0 | ['C++'] | 2 |
monotone-increasing-digits | Simple and very short C++ solution | simple-and-very-short-c-solution-by-zee-3eoc | \nclass Solution {\npublic:\n int monotoneIncreasingDigits(int N) {\n string n_str = to_string(N);\n \n int marker = n_str.size();\n | zee | NORMAL | 2017-12-03T04:22:26.099000+00:00 | 2018-08-16T19:01:56.570651+00:00 | 13,652 | false | ```\nclass Solution {\npublic:\n int monotoneIncreasingDigits(int N) {\n string n_str = to_string(N);\n \n int marker = n_str.size();\n for(int i = n_str.size()-1; i > 0; i --) {\n if(n_str[i] < n_str[i-1]) {\n marker = i;\n n_str[i-1] = n_str[i-1]... | 156 | 5 | [] | 18 |
monotone-increasing-digits | Clean Code | clean-code-by-gracemeng-zzyw | The result should be:\n\npreceding monotone increasing digits (ending digit should decrease by 1) \n\nfollowed by \n\nall remaining digits set to \'9\'\n\n.\n\n | gracemeng | NORMAL | 2018-07-17T07:00:20.750349+00:00 | 2020-08-21T15:33:16.652768+00:00 | 6,270 | false | The result should be:\n```\npreceding monotone increasing digits (ending digit should decrease by 1) \n```\nfollowed by \n```\nall remaining digits set to \'9\'\n```\n.\n\n`e.g. N = 12321, the result is 12299`.\n```\nmonotoneIncreasingEnd is finalized as : 2\ncurrent arrN : 12211\narrN is finalized as : 12299\n```\n***... | 96 | 1 | [] | 8 |
monotone-increasing-digits | Simple Python solution w/ Explanation | simple-python-solution-w-explanation-by-zadha | The idea is to go from the LSB to MSB and find the last digit, where an inversion happens.\nThere are 2 cases to consider:\n\ncase 1:\nIn 14267 , we see that in | Cubicon | NORMAL | 2017-12-03T04:04:22.478000+00:00 | 2018-09-12T07:23:11.031465+00:00 | 7,588 | false | The idea is to go from the LSB to MSB and find the last digit, where an inversion happens.\nThere are 2 cases to consider:\n\ncase 1:\nIn 14267 , we see that inversion happens at 4. In this case, then answer is obtained by reducing 4 to 3, and changing all the following digits to 9. \n=> 13999\n\ncase 2:\n1444267, here... | 64 | 3 | [] | 13 |
monotone-increasing-digits | Simple java solution with clear explanation. Very easy to understand. | simple-java-solution-with-clear-explanat-foaw | \nclass Solution {\n public int monotoneIncreasingDigits(int N) {\n //1. Convert the given integer to character array\n char[] ch = String.valu | sushant_chaudhari | NORMAL | 2018-05-04T00:05:28.074020+00:00 | 2018-05-04T00:05:28.074020+00:00 | 3,623 | false | ```\nclass Solution {\n public int monotoneIncreasingDigits(int N) {\n //1. Convert the given integer to character array\n char[] ch = String.valueOf(N).toCharArray();\n \n //2. Create a integer mark variable and initialize it to the length of the character array \n int mark = ch.l... | 42 | 2 | [] | 4 |
monotone-increasing-digits | Python, Straightforward Greedy Solution Explained | python-straightforward-greedy-solution-e-9zgs | Let\'s look at a few examples:\n\nInput Output\n55627 -> 55599\n55427 -> 49999\n99996 -> 89999 \n100 -> 99\n\nFrom the | ruxinzzz | NORMAL | 2022-02-12T01:34:18.333730+00:00 | 2022-02-18T20:37:25.571505+00:00 | 1,304 | false | Let\'s look at a few examples:\n```\nInput Output\n55627 -> 55599\n55427 -> 49999\n99996 -> 89999 \n100 -> 99\n```\nFrom the above examples we can see that if a number is **not** monotone increasing digits, we will need to convert as many digits as posible to `9` and decrease t... | 23 | 0 | ['Greedy', 'Python'] | 3 |
monotone-increasing-digits | cpp simple solution explanation with example beats 100% time and space | cpp-simple-solution-explanation-with-exa-3o0d | Runtime: 0 ms, faster than 100.00% of C++ online submissions for Monotone Increasing Digits.\nMemory Usage: 5.9 MB, less than 100.00% of C++ online submissions | _mrbing | NORMAL | 2020-05-25T12:59:05.879052+00:00 | 2020-05-25T12:59:05.879091+00:00 | 1,554 | false | Runtime: 0 ms, faster than 100.00% of C++ online submissions for Monotone Increasing Digits.\nMemory Usage: 5.9 MB, less than 100.00% of C++ online submissions for Monotone Increasing Digits.\nLogic : any number \'xxxxx..\' a number monotonically increasing largest but smaller than original number will be one less tha... | 22 | 0 | ['C++'] | 2 |
monotone-increasing-digits | C++ | BFS approach | c-bfs-approach-by-wh0ami-fg48 | \nclass Solution {\npublic:\n int monotoneIncreasingDigits(int N) {\n \n queue<long>q;\n for (int i = 1; i <= 9; i++)\n q.pus | wh0ami | NORMAL | 2020-06-18T05:34:06.752796+00:00 | 2020-06-18T05:34:06.752849+00:00 | 918 | false | ```\nclass Solution {\npublic:\n int monotoneIncreasingDigits(int N) {\n \n queue<long>q;\n for (int i = 1; i <= 9; i++)\n q.push(i);\n \n int res = INT_MIN;\n while (!q.empty()) {\n int t = q.front();\n q.pop();\n \n if... | 20 | 0 | [] | 3 |
monotone-increasing-digits | Simple and very short Java solution | simple-and-very-short-java-solution-by-b-f083 | //translated from the simple and very short c++ solution\n\n\n public int monotoneIncreasingDigits(int N) {\n\n if(N<=9)\n return N;\n | brucezu | NORMAL | 2017-12-03T09:00:47.355000+00:00 | 2017-12-03T09:00:47.355000+00:00 | 4,504 | false | //translated from the simple and very short c++ solution\n\n```\n public int monotoneIncreasingDigits(int N) {\n\n if(N<=9)\n return N;\n char[] x = String.valueOf(N).toCharArray();\n\n int mark = x.length;\n for(int i = x.length-1;i>0;i--){\n if(x[i]<x[i-1]){\n ... | 17 | 1 | [] | 1 |
monotone-increasing-digits | Python3 || 7 lines w/explanation || T/S: 82% / 87% | python3-7-lines-wexplanation-ts-82-87-by-pkys | \nclass Solution:\n def monotoneIncreasingDigits(self, n: int) -> int:\n\n n = str(n) # For example: n = 246621 | Spaulding_ | NORMAL | 2022-09-20T17:22:20.540912+00:00 | 2024-05-29T19:16:36.568363+00:00 | 786 | false | ```\nclass Solution:\n def monotoneIncreasingDigits(self, n: int) -> int:\n\n n = str(n) # For example: n = 246621 --> "246621"\n N = len(n)\n\n for i in range(N-1): # find the first break in monotonicity\n if n[i] > ... | 16 | 1 | ['Python', 'Python3'] | 1 |
monotone-increasing-digits | Simple Java, from back to front, no extra space and no conversion to char[] | simple-java-from-back-to-front-no-extra-fl047 | \nclass Solution {\n public int monotoneIncreasingDigits(int N) {\n int res = 0;\n int pre = Integer.MAX_VALUE;\n int offset = 1;\n | shibai86 | NORMAL | 2018-07-03T06:12:30.639364+00:00 | 2018-10-10T19:12:50.924558+00:00 | 1,052 | false | ```\nclass Solution {\n public int monotoneIncreasingDigits(int N) {\n int res = 0;\n int pre = Integer.MAX_VALUE;\n int offset = 1;\n while(N != 0) {\n int digit = N % 10;\n if(digit > pre) {\n res = digit * offset - 1;\n pre = digit - ... | 14 | 1 | [] | 3 |
monotone-increasing-digits | [Java/C++] 0ms || Awesome approach || 100% faster | javac-0ms-awesome-approach-100-faster-by-yak8 | Attention \u2757\nAvoid the following approach, it is too expensive because of n-- statements\u2757. \n\npublic int monotoneIncreasingDigits(int n) {\n w | im_obid | NORMAL | 2023-01-15T21:35:53.354207+00:00 | 2023-01-15T23:40:44.004404+00:00 | 1,731 | false | # Attention \u2757\nAvoid the following approach, it is too expensive because of ```n--``` statements\u2757. \n```\npublic int monotoneIncreasingDigits(int n) {\n while (!isSatisfies(n)) {\n n--;\n }\n return n;\n }\n\n public boolean isSatisfies(int n) {\n int k = 10;\n ... | 12 | 0 | ['C++', 'Java'] | 4 |
monotone-increasing-digits | Fast and simple 40ms Python solution using recursion | fast-and-simple-40ms-python-solution-usi-bxj8 | We scan through S from left to right:\n If it\'s monotonically increasing we just adding corresponding part to result\n If not then we simply decrease result by | rouroukerr | NORMAL | 2018-10-16T01:18:35.834038+00:00 | 2018-10-24T14:55:12.274831+00:00 | 802 | false | We scan through S from left to right:\n* If it\'s monotonically increasing we just adding corresponding part to result\n* If not then we simply decrease result by 1, which would result in some 9s in the tail\n* However, decrease by 1 might lead to our result being not monotonically increasing, so we run recursion base ... | 10 | 0 | [] | 3 |
monotone-increasing-digits | easy java | easy-java-by-0xdeadbeef-vszc | \n public int monotoneIncreasingDigits(int N) {\n char[] digit = (N + "").toCharArray();\n int flag = digit.length;\n for (int i = digit | 0xdeadbeef_ | NORMAL | 2017-12-14T14:54:30.249000+00:00 | 2017-12-14T14:54:30.249000+00:00 | 1,281 | false | ```\n public int monotoneIncreasingDigits(int N) {\n char[] digit = (N + "").toCharArray();\n int flag = digit.length;\n for (int i = digit.length - 1; i > 0; i--) {\n if (digit[i] < digit[i - 1]) {\n digit[i - 1]--;\n flag = i;\n }\n }\... | 8 | 0 | [] | 0 |
monotone-increasing-digits | 738: Solution with step by step explanation | 738-solution-with-step-by-step-explanati-so3v | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n\ndigits = list(str(n))\n\nWe first convert the integer n to its string r | Marlen09 | NORMAL | 2023-10-20T18:48:59.005595+00:00 | 2023-10-20T18:48:59.005621+00:00 | 540 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n```\ndigits = list(str(n))\n```\nWe first convert the integer n to its string representation and then create a list of its characters.\nThis step allows us to easily modify individual digits if needed.\n\n```\nn = len(digits... | 7 | 0 | ['Math', 'Greedy', 'Python', 'Python3'] | 0 |
monotone-increasing-digits | [C++] Simplest Greedy Solution Faster than 100% | c-simplest-greedy-solution-faster-than-1-u9hn | We need to return the largest number that is less than or equal to the given n, that has monotome increasing digits.\n\nAPPROACH :\n\n If the number n is monoto | Mythri_Kaulwar | NORMAL | 2022-02-18T04:06:12.029495+00:00 | 2022-02-18T04:06:53.233991+00:00 | 966 | false | We need to return the largest number that is less than or equal to the given ```n```, that has monotome increasing digits.\n\n**APPROACH :**\n\n* If the number ```n``` is monotone increasing then, we can return the number itself.\n* Otherwise, we start traversing from the last digit.\n* Wherever we find a digit that\'... | 6 | 0 | ['Greedy', 'C', 'C++'] | 1 |
monotone-increasing-digits | Solution in c++ | solution-in-c-by-ashish_madhup-dwxk | 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 | 2023-02-26T16:29:37.900514+00:00 | 2023-02-26T16:29:37.900618+00:00 | 673 | 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)$$ --... | 5 | 0 | ['C++'] | 0 |
monotone-increasing-digits | java simple | java-simple-by-jstm2022-lkrt | \nclass Solution {\n public int monotoneIncreasingDigits(int n) {\n char[] arr = String.valueOf(n).toCharArray();\n int start = arr.length;\n | JSTM2022 | NORMAL | 2022-04-26T02:00:47.759331+00:00 | 2022-04-26T02:00:47.759378+00:00 | 673 | false | ```\nclass Solution {\n public int monotoneIncreasingDigits(int n) {\n char[] arr = String.valueOf(n).toCharArray();\n int start = arr.length;\n for (int i = arr.length - 2; i >= 0; i --) {\n if (arr[i] > arr[i + 1]) {\n start = i + 1;\n arr[i] --;\n ... | 5 | 0 | ['Java'] | 0 |
monotone-increasing-digits | very intuitional Java solution using stack | very-intuitional-java-solution-using-sta-xo56 | ```\n public int monotoneIncreasingDigits(int N) {\n char[] num = String.valueOf(N).toCharArray();\n Stack stack = new Stack<>();\n boolea | peachfan | NORMAL | 2019-09-28T15:11:22.087463+00:00 | 2019-09-28T15:11:22.087504+00:00 | 273 | false | ```\n public int monotoneIncreasingDigits(int N) {\n char[] num = String.valueOf(N).toCharArray();\n Stack<Character> stack = new Stack<>();\n boolean find = false;\n for (int i = 0; i < num.length; i++) {\n char ch = num[i];\n while (!stack.isEmpty() && stack.peek() >... | 5 | 0 | [] | 0 |
monotone-increasing-digits | [Java] ✅ O(1) ✅ 1MS ✅ 100% ✅ FAST ✅ BEST ✅ CLEAN CODE | java-o1-1ms-100-fast-best-clean-code-by-d380f | Approach\n1. Looking at the numbers, the highest monotone increasing increasing is ending with 9 (unless number is already monotone: eg: 12345)\n2. To obtain nu | StefanelStan | NORMAL | 2024-10-14T08:47:49.801271+00:00 | 2024-10-14T08:47:49.801306+00:00 | 219 | false | # Approach\n1. Looking at the numbers, the highest monotone increasing increasing is ending with 9 (unless number is already monotone: eg: 12345)\n2. To obtain numbers ending with 9, subtract n % (j) + 1 from n, where j is a power of 10: 10,100, 1000.\n3. EG: \n - 332 - (332 % 10 + 1) = 332 - (2 + 1) = 332 - 3 = 329... | 4 | 0 | ['Java'] | 1 |
monotone-increasing-digits | O(n) Python solution - very simple logic | on-python-solution-very-simple-logic-by-axagf | Example: n = 4329762\nfor loop: i = 6 -> if condition: 2 < 6 is true: n_str = 4329762 - 63 = 4329699\nfor loop: i = 5 -> if condition: 9 < 6 is false\nfor loop | getsatnow | NORMAL | 2021-10-02T20:37:03.098544+00:00 | 2021-10-02T20:46:05.513469+00:00 | 287 | false | Example: n = 4329762\nfor loop: i = 6 -> if condition: 2 < 6 is true: n_str = 4329762 - 63 = 4329699\nfor loop: i = 5 -> if condition: 9 < 6 is false\nfor loop i = 4 -> if condition: 6 < 9 is true: n_str = 4329699 - 9700 = 4319999\nand so on...\n```\nclass Solution:\n def monotoneIncreasingDigits(self, n: int) -> in... | 4 | 0 | [] | 2 |
monotone-increasing-digits | Simple O(log n) Java solution with one pass | simple-olog-n-java-solution-with-one-pas-4nf5 | We scan from right to left, i.e., least significant to most significan, take 7634 fro example: \n1. if the digits are strictly increasing from left to right, we | linvava | NORMAL | 2020-06-16T04:56:46.888120+00:00 | 2020-06-16T04:58:07.530833+00:00 | 373 | false | We scan from right to left, i.e., least significant to most significan, take 7634 fro example: \n1. if the digits are strictly increasing from left to right, we do nothing, so 34 -> 34.\n2. if a one digit is smaller than any digit on the right, the result would be the digit minus 1, and fill remaining digits with 9. 63... | 4 | 0 | [] | 0 |
monotone-increasing-digits | Simple Python Solution | simple-python-solution-by-bigshen-romf | class Solution(object):\n def monotoneIncreasingDigits(self, N):\n """\n :type N: int\n :rtype: int\n """\n \n if N | bigshen | NORMAL | 2017-12-25T03:42:30.241000+00:00 | 2018-10-24T15:14:16.938836+00:00 | 795 | false | class Solution(object):\n def monotoneIncreasingDigits(self, N):\n """\n :type N: int\n :rtype: int\n """\n \n if N == 10:\n return 9\n if N < 20:\n return N\n num = [s for s in str(N)]\n i = len(num) - 1\n index = len(num) \... | 4 | 1 | [] | 0 |
monotone-increasing-digits | Easy Java Solution || Beats 95% | easy-java-solution-beats-95-by-ravikumar-2qc8 | 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 | ravikumar50 | NORMAL | 2023-09-07T14:36:09.951772+00:00 | 2023-09-07T14:36:09.951794+00:00 | 227 | 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 | ['Java'] | 1 |
monotone-increasing-digits | very easy to understand ! converting to string and solving in c++ | very-easy-to-understand-converting-to-st-65ij | Approach\n1.First convert the number into a string using to_string( ).\n2.Then check each digit in a for loop if the prev digit is smaller than or equal to the | Trivial_03 | NORMAL | 2023-03-21T19:39:41.690681+00:00 | 2023-03-21T19:39:41.690713+00:00 | 593 | false | # Approach\n1.First convert the number into a string using to_string( ).\n2.Then check each digit in a for loop if the prev digit is smaller than or equal to the next digit or not. If yes,then continue.\n3.If no then just make the rest leftout digits(to the right of index where condition is violated) to \'9\' and decre... | 3 | 0 | ['C++'] | 0 |
monotone-increasing-digits | 8 Line JS code | 8-line-js-code-by-yutingkung-tebm | \nvar monotoneIncreasingDigits = function(n) {\n let arr = String(n).split(\'\');\n for (let i=arr.length-2; i>=0; i--) {\n if (arr[i]>arr[i+1]) {\ | yutingkung | NORMAL | 2022-01-07T03:52:23.385616+00:00 | 2022-01-07T03:52:23.385660+00:00 | 151 | false | ```\nvar monotoneIncreasingDigits = function(n) {\n let arr = String(n).split(\'\');\n for (let i=arr.length-2; i>=0; i--) {\n if (arr[i]>arr[i+1]) {\n arr[i]--;\n for(let j=i+1; j<arr.length; j++) arr[j]=\'9\';\n }\n }\n return Number(arr.join(\'\'));\n};\n``` | 3 | 0 | ['JavaScript'] | 0 |
monotone-increasing-digits | 92 % ,Intuitive,Easy to Understand,GREEDY , Java | 92-intuitiveeasy-to-understandgreedy-jav-9045 | ```\nclass Solution {\n public int monotoneIncreasingDigits(int n) {\n char [] arr=String.valueOf(n).toCharArray();\n for(int i=arr.length-2;i> | rtvksingh | NORMAL | 2021-05-29T17:32:35.382147+00:00 | 2021-05-29T17:34:48.244907+00:00 | 141 | false | ```\nclass Solution {\n public int monotoneIncreasingDigits(int n) {\n char [] arr=String.valueOf(n).toCharArray();\n for(int i=arr.length-2;i>=0;i--){\n if(arr[i]<=arr[i+1])\n continue;\n else{\n arr[i]--;\n for(int j=i+1;j<arr.length;... | 3 | 1 | [] | 0 |
monotone-increasing-digits | Easy O(N) JavaScript with explanation | easy-on-javascript-with-explanation-by-f-log5 | \n\nconst monotoneIncreasingDigits = N => {\n \n // create any array of integers from number N\n const n = Array.from(\'\'+N, Number);\n \n let i = 0;\n \ | fl4sk | NORMAL | 2021-01-10T03:02:20.999461+00:00 | 2021-01-10T03:02:20.999498+00:00 | 276 | false | ```\n\nconst monotoneIncreasingDigits = N => {\n \n // create any array of integers from number N\n const n = Array.from(\'\'+N, Number);\n \n let i = 0;\n \n // find the cliff\n while(i < n.length-1 && n[i] <= n[i+1])\n i++;\n\n // decremnet the cliff and any values before the cliff which satisfy this condi... | 3 | 0 | ['JavaScript'] | 0 |
monotone-increasing-digits | StraightForward Greedy Python Solution | straightforward-greedy-python-solution-b-xpeh | \n def monotoneIncreasingDigits(self, N):\n """\n :type N: int\n :rtype: int\n """\n s = str(N)\n if len(s) < 2:\n | yuchenliu0513 | NORMAL | 2020-12-03T08:15:02.377923+00:00 | 2020-12-03T08:15:02.377965+00:00 | 562 | false | ```\n def monotoneIncreasingDigits(self, N):\n """\n :type N: int\n :rtype: int\n """\n s = str(N)\n if len(s) < 2:\n return N\n ans = ""\n pre = 0\n for i in range(len(s)):\n if int(s[i]) >= pre:\n ans += s[i]\n ... | 3 | 0 | ['Greedy', 'Python', 'Python3'] | 0 |
monotone-increasing-digits | c++,stack | cstack-by-sharma_amit-87ax | \nclass Solution {\n vector<int> convert_to_arr(int n){\n vector<int> v;\n while(n){\n int t = n%10;\n n = n/10;\n | sharma_amit | NORMAL | 2020-09-29T16:18:41.544555+00:00 | 2020-09-29T16:19:51.815516+00:00 | 381 | false | ```\nclass Solution {\n vector<int> convert_to_arr(int n){\n vector<int> v;\n while(n){\n int t = n%10;\n n = n/10;\n v.push_back(t);\n }\n reverse(v.begin(),v.end());\n return v;\n }\n \n int convert_int(vector<int> nums){\n int ans... | 3 | 0 | ['Stack', 'C'] | 0 |
monotone-increasing-digits | java 1ms solution with detailed explanation | java-1ms-solution-with-detailed-explanat-nwq4 | This problem can be solved in greedy strategy:\n1. At first, we use a list to represent the given number N where list.get(0) is the highest number.\n2. Then, we | pangeneral | NORMAL | 2019-05-09T00:38:42.625310+00:00 | 2019-05-09T00:38:42.625340+00:00 | 653 | false | This problem can be solved in greedy strategy:\n1. At first, we use a list to represent the given number ```N``` where list.get(0) is the highest number.\n2. Then, we traverse from 0 to list.size() - 1 to find the **first** i that list[i] > list[i + 1]\n3. If we don\'t find such i, then return the given number. Otherwi... | 3 | 0 | ['Math', 'Greedy', 'Java'] | 0 |
monotone-increasing-digits | Optimized Approach || 100% T.C || CPP | optimized-approach-100-tc-cpp-by-ganesh_-f7qu | 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 | Ganesh_ag10 | NORMAL | 2024-04-11T05:35:24.233137+00:00 | 2024-04-11T05:35:24.233173+00:00 | 555 | 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)$$ -->\n\n- Space complexity:O(n)\n<!-- Add your space complexity here, e.g. $$O... | 2 | 1 | ['C++'] | 0 |
monotone-increasing-digits | Easy C++ solution || Math Based | easy-c-solution-math-based-by-bharathgow-d43x | \n# Code\n\nclass Solution {\npublic:\n bool isMonotonic(int n){\n int k = 10;\n while(n > 0){\n if(k < n%10){\n retu | bharathgowda29 | NORMAL | 2023-12-27T15:43:01.273264+00:00 | 2023-12-27T15:43:01.273291+00:00 | 542 | false | \n# Code\n```\nclass Solution {\npublic:\n bool isMonotonic(int n){\n int k = 10;\n while(n > 0){\n if(k < n%10){\n return false;\n }\n else{\n k = n%10;\n n = n/10;\n }\n }\n\n return true;\n }\n\... | 2 | 0 | ['Math', 'Greedy', 'C++'] | 0 |
monotone-increasing-digits | Solution | solution-by-deleted_user-tga5 | C++ []\nclass Solution {\npublic:\n int monotoneIncreasingDigits(int n) {\n string s = to_string(n);\n while(1){\n bool flag = false | deleted_user | NORMAL | 2023-04-23T04:00:49.023701+00:00 | 2023-04-23T04:49:07.534319+00:00 | 873 | false | ```C++ []\nclass Solution {\npublic:\n int monotoneIncreasingDigits(int n) {\n string s = to_string(n);\n while(1){\n bool flag = false;\n for(int i = 1; i < s.length(); i++){\n if(s[i] >= s[i-1]) continue;\n else{\n s[i-1]--;\n ... | 2 | 0 | ['C++', 'Java', 'Python3'] | 0 |
monotone-increasing-digits | Simple C++ Sloution | simple-c-sloution-by-sunnyyad2002-fwzx | Intuition\n Describe your first thoughts on how to solve this problem. \nSuppose we have a non-negative integer N, we have to find the largest number that is le | sunnyyad2002 | NORMAL | 2023-03-17T15:00:05.898375+00:00 | 2023-03-17T15:00:05.898399+00:00 | 905 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSuppose we have a non-negative integer N, we have to find the largest number that is less than or equal to N with monotone increasing digits. We know that an integer has monotone increasing digits if and only if each pair of adjacent digi... | 2 | 0 | ['C++'] | 0 |
monotone-increasing-digits | Java faster than 100% | java-faster-than-100-by-ritik652018-jvnu | ```\nclass Solution {\n \n public int monotoneIncreasingDigits(int n) {\n \n for(int i=n;i>=0;i--){\n int k=i;\n int x=k%10; | ritik652018 | NORMAL | 2022-09-24T08:10:16.458388+00:00 | 2023-10-20T06:48:49.243064+00:00 | 424 | false | ```\nclass Solution {\n \n public int monotoneIncreasingDigits(int n) {\n \n for(int i=n;i>=0;i--){\n int k=i;\n int x=k%10;\n int flag=0;\n int count=1;\n \n while(k>1){\n k=k/10;\n int y= k%10;\n ... | 2 | 1 | ['Math', 'Java'] | 1 |
monotone-increasing-digits | Java Simple Solution without DP | java-simple-solution-without-dp-by-adity-uv1o | \nclass Solution {\n public int monotoneIncreasingDigits(int n) {\n int len = size(n);\n int[] dig = new int[len];\n len--;\n whi | Aditya001 | NORMAL | 2022-08-17T14:28:29.038537+00:00 | 2022-08-17T14:28:29.038605+00:00 | 995 | false | ```\nclass Solution {\n public int monotoneIncreasingDigits(int n) {\n int len = size(n);\n int[] dig = new int[len];\n len--;\n while(n>0){\n dig[len] = n%10;\n n /= 10;\n len--;\n }\n n = dig.length;\n int t = 10;\n\n while(t>... | 2 | 0 | ['Java'] | 0 |
monotone-increasing-digits | Easiest C++ solution | easiest-c-solution-by-kazuma0803-5h2r | \nclass Solution {\npublic:\n int monotoneIncreasingDigits(int n) {\\\n string s=to_string(n);\n int x=s.size();\n for (int i=s.size()-1 | Kazuma0803 | NORMAL | 2022-06-18T10:11:53.931180+00:00 | 2022-06-18T10:11:53.931210+00:00 | 336 | false | ```\nclass Solution {\npublic:\n int monotoneIncreasingDigits(int n) {\\\n string s=to_string(n);\n int x=s.size();\n for (int i=s.size()-1;i>0;i--) {\n if (s[i]<s[i-1]) {\n x=i;\n s[i-1]=s[i-1]-1;\n }\n }\n for (int i=x;i<s.size(... | 2 | 0 | ['C'] | 0 |
monotone-increasing-digits | C++ 100% Faster | c-100-faster-by-kunal_patil-1koh | ,,,\n\n int monotoneIncreasingDigits(int n) {\n \n vector v;\n \n while(n)\n {\n v.push_back(n%10);\n n | kunal_patil | NORMAL | 2022-02-09T06:50:08.896254+00:00 | 2022-02-09T06:50:08.896284+00:00 | 219 | false | ,,,\n\n int monotoneIncreasingDigits(int n) {\n \n vector<int> v;\n \n while(n)\n {\n v.push_back(n%10);\n n=n/10;\n }\n \n for(int i=0;i<v.size();i++)\n {\n if(i<v.size()-1 && v[i]<v[i+1])\n {\n v[... | 2 | 0 | ['Math', 'C'] | 0 |
monotone-increasing-digits | Simple C++ || Without changing into string || Stack Solution | simple-c-without-changing-into-string-st-4l5s | \nclass Solution {\npublic:\n int monotoneIncreasingDigits(int n) {\n int ans = 0;\n stack<int> st;\n \n while(n)\n {\n | goyalnaman198 | NORMAL | 2022-01-01T14:48:56.742441+00:00 | 2022-01-01T14:51:42.034462+00:00 | 236 | false | ```\nclass Solution {\npublic:\n int monotoneIncreasingDigits(int n) {\n int ans = 0;\n stack<int> st;\n \n while(n)\n {\n int last = n % 10; n /= 10;\n \n if(st.empty() or st.top() >= last)\n {\n st.push(last);\n ... | 2 | 0 | ['Stack', 'C'] | 0 |
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