question_slug stringlengths 3 77 | title stringlengths 1 183 | slug stringlengths 12 45 | summary stringlengths 1 160 ⌀ | author stringlengths 2 30 | certification stringclasses 2
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
classes | content stringlengths 4 576k | upvotes int64 0 11.5k | downvotes int64 0 358 | tags stringlengths 2 193 | comments int64 0 2.56k |
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score-of-a-string | Easy 3 line code | easy-3-line-code-by-nithilab2106-lzat | IntuitionThe question asked us to find the absolute difference of each letter with the next and sum up all these diffrencesApproachWhen we try storing a charcte | nithilab2106 | NORMAL | 2025-01-14T14:06:09.089564+00:00 | 2025-01-14T14:06:09.089564+00:00 | 98 | false | # Intuition
The question asked us to find the absolute difference of each letter with the next and sum up all these diffrences
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
When we try storing a charcter in an integer varaible or perform mathematical functions java automatically conve... | 1 | 0 | ['String', 'Java'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | C++/Python O(n) | cpython-on-by-votrubac-4o9o | We do not need more than 3 letters to build a non-repeating character sequence.\n\nFor Python, we can use set difference to determine which one to use.\n\nPytho | votrubac | NORMAL | 2020-09-06T04:01:02.465128+00:00 | 2020-09-06T04:59:25.577416+00:00 | 10,276 | false | We do not need more than 3 letters to build a non-repeating character sequence.\n\nFor Python, we can use set difference to determine which one to use.\n\n**Python**\n```python\ndef modifyString(self, s: str) -> str:\n\tres, prev = "", \'?\'\n\tfor i, c in enumerate(s):\n\t\tnext = s[i + 1] if i + 1 < len(s) else \'?\'... | 117 | 1 | [] | 11 |
replace-all-s-to-avoid-consecutive-repeating-characters | Java Simple O(n) loop | java-simple-on-loop-by-hobiter-2uqh | for each char, just try \u2018a\u2019, \u2018b\u2019, \u2018c\u2019, and select the one not the same as neighbors.\n\n\n public String modifyString(String s) | hobiter | NORMAL | 2020-09-06T04:01:14.868400+00:00 | 2020-09-06T04:01:14.868459+00:00 | 8,211 | false | for each char, just try \u2018a\u2019, \u2018b\u2019, \u2018c\u2019, and select the one not the same as neighbors.\n\n```\n public String modifyString(String s) {\n char[] arr = s.toCharArray();\n for (int i = 0; i < arr.length; i++) {\n if (arr[i] == \'?\') {\n for (int j = 0... | 81 | 3 | [] | 7 |
replace-all-s-to-avoid-consecutive-repeating-characters | [Python3] one of three letters | python3-one-of-three-letters-by-ye15-qqaj | \n\nclass Solution:\n def modifyString(self, s: str) -> str:\n s = list(s)\n for i in range(len(s)):\n if s[i] == "?": \n | ye15 | NORMAL | 2020-09-06T04:02:34.600206+00:00 | 2020-09-06T04:02:34.600246+00:00 | 4,825 | false | \n```\nclass Solution:\n def modifyString(self, s: str) -> str:\n s = list(s)\n for i in range(len(s)):\n if s[i] == "?": \n for c in "abc": \n if (i == 0 or s[i-1] != c) and (i+1 == len(s) or s[i+1] != c): \n s[i] = c\n ... | 79 | 1 | ['Python3'] | 8 |
replace-all-s-to-avoid-consecutive-repeating-characters | c++ | 0ms Code | easy to understand | with explanation | c-0ms-code-easy-to-understand-with-expla-qb7x | \n\'\'\'\n\n\t\tclass Solution {\n\t\tpublic:\n\t\tstring modifyString(string s) {\n int n = s.length();\n for(int i=0;i<n;i++){\n \n | pranjalmittal21 | NORMAL | 2020-09-07T09:10:08.596855+00:00 | 2020-09-24T09:25:30.018041+00:00 | 2,094 | false | \n\'\'\'\n\n\t\tclass Solution {\n\t\tpublic:\n\t\tstring modifyString(string s) {\n int n = s.length();\n for(int i=0;i<n;i++){\n \n //checking every character, if character is \'?\'\n if(s[i] == \'?\'){\n \n //loop over every alphabet\n ... | 29 | 0 | ['C', 'C++'] | 5 |
replace-all-s-to-avoid-consecutive-repeating-characters | Super easy solution, O(n), beats 100% | super-easy-solution-on-beats-100-by-eaim-44wt | \npublic String modifyString(String s) {\n if (s == null || s.isEmpty()) return "";\n \n char[] chars = s.toCharArray();\n for (int i=0; i<chars.len | eaiman | NORMAL | 2020-09-06T07:11:09.432675+00:00 | 2020-09-06T07:11:38.793083+00:00 | 3,524 | false | ```\npublic String modifyString(String s) {\n if (s == null || s.isEmpty()) return "";\n \n char[] chars = s.toCharArray();\n for (int i=0; i<chars.length; i++) {\n if (chars[i] == \'?\') {\n for (char j=\'a\'; j<=\'z\'; j++) {\n chars[i] = j;\n if (i>0 && cha... | 21 | 0 | ['Java'] | 6 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python simple solution | python-simple-solution-by-tovam-kwxf | Python :\n\n\ndef modifyString(self, s: str) -> str:\n\ts = list(s)\n\n\tfor i in range(len(s)):\n\t\tif s[i] == \'?\':\n\t\t\tfor c in "abc":\n\t\t\t\tif (i == | TovAm | NORMAL | 2021-11-03T22:52:46.999158+00:00 | 2021-11-03T22:52:46.999201+00:00 | 1,333 | false | **Python :**\n\n```\ndef modifyString(self, s: str) -> str:\n\ts = list(s)\n\n\tfor i in range(len(s)):\n\t\tif s[i] == \'?\':\n\t\t\tfor c in "abc":\n\t\t\t\tif (i == 0 or s[i - 1] != c) and (i + 1 == len(s) or s[i + 1] != c):\n\t\t\t\t\ts[i] = c\n\t\t\t\t\tbreak\n\n\treturn "".join(s)\n```\n\n**Like it ? please upvot... | 12 | 0 | ['Python', 'Python3'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | C# | c-by-mhorskaya-7x0o | \npublic string ModifyString(string s) {\n\tvar chars = s.ToArray();\n\n\tfor (var i = 0; i < s.Length; i++) {\n\t\tif (chars[i] != \'?\') continue;\n\n\t\tvar | mhorskaya | NORMAL | 2020-09-09T10:29:27.969362+00:00 | 2020-09-09T10:29:27.969392+00:00 | 429 | false | ```\npublic string ModifyString(string s) {\n\tvar chars = s.ToArray();\n\n\tfor (var i = 0; i < s.Length; i++) {\n\t\tif (chars[i] != \'?\') continue;\n\n\t\tvar left = i > 0 ? chars[i - 1] : (char?)null;\n\t\tvar right = i < s.Length - 1 ? chars[i + 1] : (char?)null;\n\n\t\tif (left != \'a\' && right != \'a\') chars[... | 10 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | EASY C++ solution with comments || O(n) | easy-c-solution-with-comments-on-by-sky9-sjgv | \nclass Solution {\npublic:\n string modifyString(string s) {\n for(int i = 0;i < s.size();i++){\n if(s[i] == \'?\'){\n \n | sky97 | NORMAL | 2020-09-06T04:04:21.383703+00:00 | 2020-09-06T04:04:21.383768+00:00 | 1,141 | false | ```\nclass Solution {\npublic:\n string modifyString(string s) {\n for(int i = 0;i < s.size();i++){\n if(s[i] == \'?\'){\n \n if(i == 0){ // if we are starting then we have to check only next character\n if(i + 1 < s.size()){\n ... | 10 | 0 | [] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | 1576. Replace All ?'s to Avoid Consecutive Re..., Time Complexity: O(N), Space Complexity: O(1) | 1576-replace-all-s-to-avoid-consecutive-5lyai | IntuitionApproachComplexity
Time complexity: O(N)
Space complexity: O(1)
Code | richardmantikwang | NORMAL | 2024-12-19T10:44:55.036685+00:00 | 2024-12-19T10:47:29.251629+00:00 | 141 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->... | 9 | 0 | ['Python3'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | [Java] a, b, c solution beats 100% - comment explained | java-a-b-c-solution-beats-100-comment-ex-1xpc | \nclass Solution {\n public String modifyString(String s) {\n char[] ch = s.toCharArray();\n for (int i = 0;i<ch.length;i++){\n if ( | vinsinin | NORMAL | 2021-02-23T07:40:34.426814+00:00 | 2021-02-23T07:40:34.426850+00:00 | 1,070 | false | ```\nclass Solution {\n public String modifyString(String s) {\n char[] ch = s.toCharArray();\n for (int i = 0;i<ch.length;i++){\n if (ch[i] == \'?\'){\n for (char j = \'a\'; j <= \'c\';j++){\n if (i > 0 && ch[i-1] == j) continue; //skip if previous characte... | 9 | 1 | ['Java'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | python3 beginner friendly | python3-beginner-friendly-by-m0u1ea5-rrwr | python\nclass Solution:\n def modifyString(self, s: str) -> str:\n if len(s)==1:\n return \'a\' \n s=list(s)\n for i in range | M0u1ea5 | NORMAL | 2020-12-07T05:55:18.202491+00:00 | 2021-03-20T12:30:07.042145+00:00 | 937 | false | ```python\nclass Solution:\n def modifyString(self, s: str) -> str:\n if len(s)==1:\n return \'a\' \n s=list(s)\n for i in range(len(s)):\n if s[i] == \'?\':\n for x in \'abc\': \n if i == 0 and s[i+1] != x: \n ... | 9 | 0 | [] | 2 |
replace-all-s-to-avoid-consecutive-repeating-characters | [C++] Simple Brute Force Solution | Self-Explanatory | c-simple-brute-force-solution-self-expla-c7gd | \nclass Solution\n{\npublic:\n string modifyString(string s)\n {\n char temp1, temp2;\n int index;\n for (int k = 0; k < s.length(); | ravireddy07 | NORMAL | 2020-09-06T04:41:03.755820+00:00 | 2021-03-12T17:05:52.045818+00:00 | 836 | false | ```\nclass Solution\n{\npublic:\n string modifyString(string s)\n {\n char temp1, temp2;\n int index;\n for (int k = 0; k < s.length(); ++k)\n {\n if (s[k] == \'?\')\n {\n index = k;\n for (int i = 0; i < 26; ++i)\n {\n... | 9 | 1 | ['C'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | 💻✅BEATS 100%⌛⚡[C++/Java/Py3/JS]🍨| EASY n CLEAN EXPLANATION⭕💌 | beats-100cjavapy3js-easy-n-clean-explana-blry | IntuitionThe problem requires replacing each '?' in the given string with a lowercase English letter such that no two adjacent characters are the same. Since th | Fawz-Haaroon | NORMAL | 2025-03-20T05:13:34.799380+00:00 | 2025-03-20T05:13:34.799380+00:00 | 60 | false | # Intuition
The problem requires replacing each `'?'` in the given string with a lowercase English letter such that no two adjacent characters are the same. Since there are only 26 possible letters, we can greedily replace `'?'` with the smallest possible letter that does not match its adjacent characters.
# Approach
... | 8 | 0 | ['String', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python 3 Solution for Beginners ( 3 letter soln) | python-3-solution-for-beginners-3-letter-geau | \nclass Solution:\n def modifyString(self, s: str) -> str:\n if len(s)==1: #if input contains only a \'?\'\n if s[0]==\'?\':\n | nikhilsmanu | NORMAL | 2020-09-06T04:37:07.589668+00:00 | 2020-09-06T04:41:53.043467+00:00 | 1,448 | false | ```\nclass Solution:\n def modifyString(self, s: str) -> str:\n if len(s)==1: #if input contains only a \'?\'\n if s[0]==\'?\':\n return \'a\'\n s=list(s)\n for i in range(len(s)):\n if s[i]==\'?\':\n for c in \'abc\':\n if ... | 8 | 1 | ['Python', 'Python3'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | C++ O(n) 0ms solution | c-on-0ms-solution-by-ashshetty333-w2tq | \nclass Solution {\npublic:\n string modifyString(string s) {\n for(int i=0; i<s.size(); ++i)\n {\n if(s[i]==\'?\')\n {\n | ashshetty333 | NORMAL | 2020-09-09T09:54:45.986671+00:00 | 2020-09-09T09:54:45.986700+00:00 | 411 | false | ```\nclass Solution {\npublic:\n string modifyString(string s) {\n for(int i=0; i<s.size(); ++i)\n {\n if(s[i]==\'?\')\n {\n s[i]=\'a\';\n while((i>0 && s[i]==s[i-1]) || ((i+1)<s.size() && s[i]==s[i+1]))\n {\n s[i]++... | 7 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python | Easy Solution✅ | python-easy-solution-by-gmanayath-ettc | Code\u2705\n\nclass Solution:\n def modifyString(self, s: str) -> str:\n if s =="?":\n return "a"\n if len(s) == 1:\n ret | gmanayath | NORMAL | 2023-02-03T07:50:55.772381+00:00 | 2023-02-03T07:50:55.772434+00:00 | 971 | false | # Code\u2705\n```\nclass Solution:\n def modifyString(self, s: str) -> str:\n if s =="?":\n return "a"\n if len(s) == 1:\n return s\n \n s_list = [x for x in s]\n for index, letter in enumerate(s_list):\n replace_char = {"a","b","c"} \n ... | 6 | 0 | ['String', 'Python', 'Python3'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Easy to understand Java solution beats 100% | easy-to-understand-java-solution-beats-1-g9cw | \nclass Solution {\n public String modifyString(String s) {\n char[] S = s.toCharArray();\n for (int i = 0; i < S.length; i++) {\n i | endianless | NORMAL | 2021-03-03T02:00:47.951507+00:00 | 2021-03-03T02:00:47.951564+00:00 | 427 | false | ```\nclass Solution {\n public String modifyString(String s) {\n char[] S = s.toCharArray();\n for (int i = 0; i < S.length; i++) {\n if (S[i] != \'?\') continue;\n char prev = i > 0 ? S[i - 1] : (char) (\'a\' - 1);\n char next = i < S.length - 1 ? S[i + 1] : (char) (\'... | 6 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | python solution | python-solution-by-akaghosting-ia2t | \tclass Solution:\n\t\tdef modifyString(self, s: str) -> str:\n\t\t\tif s == "?":\n\t\t\t\treturn "a"\n\t\t\tletters = "abcdefghijklmnopqrstuvwxyz"\n\t\t\tres = | akaghosting | NORMAL | 2020-09-06T04:23:17.749337+00:00 | 2020-09-06T04:23:17.749392+00:00 | 555 | false | \tclass Solution:\n\t\tdef modifyString(self, s: str) -> str:\n\t\t\tif s == "?":\n\t\t\t\treturn "a"\n\t\t\tletters = "abcdefghijklmnopqrstuvwxyz"\n\t\t\tres = ""\n\t\t\tfor i in range(len(s)):\n\t\t\t\tif s[i] != "?":\n\t\t\t\t\tres += s[i]\n\t\t\t\telse:\n\t\t\t\t\tif i == 0:\n\t\t\t\t\t\tfor j in letters:\n\t\t\t\t... | 6 | 1 | [] | 2 |
replace-all-s-to-avoid-consecutive-repeating-characters | JavaScript | javascript-by-adrianlee0118-i2px | \nconst modifyString = s => {\n for (let i = 0; i < s.length; i++){ //Iterating through the string\n if (s[i] === \'?\'){ | adrianlee0118 | NORMAL | 2020-09-06T04:00:26.517857+00:00 | 2020-09-06T04:00:26.517903+00:00 | 441 | false | ```\nconst modifyString = s => {\n for (let i = 0; i < s.length; i++){ //Iterating through the string\n if (s[i] === \'?\'){ //If a \'?\' is encountered\n let rep = \'\';\n if (s[i-1] !== \'a\' && s[i+1] !== \'a\') rep = \'a\'; //determine the... | 6 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | JAVA | Easy solution ✅ | java-easy-solution-by-sourin_bruh-6pj7 | Please Upvote :D\njava []\nclass Solution {\n public String modifyString(String s) {\n if (s.equals("?")) { // edge case\n return "a";\n | sourin_bruh | NORMAL | 2023-01-22T18:06:21.683340+00:00 | 2023-01-22T18:11:08.080962+00:00 | 937 | false | # Please Upvote :D\n``` java []\nclass Solution {\n public String modifyString(String s) {\n if (s.equals("?")) { // edge case\n return "a";\n }\n\n char[] a = s.toCharArray();\n for (int i = 0; i < a.length; i++) {\n if (a[i] == \'?\') {\n // corne... | 4 | 0 | ['String', 'Java'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | Simple Python solution | simple-python-solution-by-c_n-ci88 | Idea:\nFor every ? in the string, recplace it with either a, b, or c, as there are only two sides for \neach letter therefor there is no need for other letters | C_N_ | NORMAL | 2021-11-05T11:30:01.287339+00:00 | 2021-11-29T21:19:39.308719+00:00 | 458 | false | **Idea:**\nFor every `?` in the string, recplace it with either `a, b, or c`, as there are only two sides for \neach letter therefor there is no need for other letters in the alphabet to recplace it in order to receive a valid string.\n(The spaces added to the beginning and end of the string are in order to skip checki... | 4 | 0 | ['Python3'] | 2 |
replace-all-s-to-avoid-consecutive-repeating-characters | EASY JAVA CODE | easy-java-code-by-190030664-peiy | \nclass Solution {\n public String modifyString(String s) {\n char c;\n char[] chars = s.toCharArray();\n for (int i=0; i<s.length(); i+ | 190030664 | NORMAL | 2021-03-13T05:10:07.012015+00:00 | 2021-03-13T05:10:07.012052+00:00 | 675 | false | ```\nclass Solution {\n public String modifyString(String s) {\n char c;\n char[] chars = s.toCharArray();\n for (int i=0; i<s.length(); i++){\n if(chars[i] == \'?\'){\n for(c=\'a\'; c<=\'z\'; c++){\n if((i == 0 || c != chars[i-1]) && (i == s.length()... | 4 | 0 | ['String', 'Java'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | simple and easy C++ solution 😍❤️🔥 | simple-and-easy-c-solution-by-shishirrsi-uxt1 | if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n\nclass Solution {\npublic:\n bool isValidIndex(int i, int n)\n {\n return (i >= 0 | shishirRsiam | NORMAL | 2024-06-14T15:20:51.096025+00:00 | 2024-06-14T15:20:51.096058+00:00 | 273 | false | # if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n```\nclass Solution {\npublic:\n bool isValidIndex(int i, int n)\n {\n return (i >= 0 and i < n);\n }\n string modifyString(string s) \n {\n int n = s.size();\n if(n == 1 and s[0] == \'?\') return "a";\n for(int... | 3 | 0 | ['String', 'Simulation', 'C++'] | 3 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python random choice solution | python-random-choice-solution-by-inversi-z2xh | Intuition\n Describe your first thoughts on how to solve this problem. \nI came up with this random choice solution. \nHowever, I realized that we only need thr | inversion39 | NORMAL | 2022-12-20T03:55:53.017538+00:00 | 2022-12-20T03:55:53.017578+00:00 | 665 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nI came up with this random choice solution. \nHowever, I realized that we only need three letters when I checked out the other solutions. LOL\nAnyway, the upper lengh of the string is 100. \nSo I guess there is no big difference in time c... | 3 | 0 | ['Python3'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | javaScript, Simple Solution | javascript-simple-solution-by-petr12332-4q2p | var modifyString = function (s) {\n let arr = s.split("");\n for (let i = 0; i <= arr.length; i++) {\n let res = ["a", "b", "c"];\n if (arr[i] === "?") | Petr12332 | NORMAL | 2022-12-05T12:01:11.085837+00:00 | 2022-12-05T13:21:47.986569+00:00 | 680 | false | var modifyString = function (s) {\n let arr = s.split("");\n for (let i = 0; i <= arr.length; i++) {\n let res = ["a", "b", "c"];\n if (arr[i] === "?") {\n if (arr[i - 1] === "a" || arr[i + 1] === "a") {\n res.splice(res.indexOf("a"), 1);\n }\n if (arr[i - 1] === "b" || arr[i + 1] === "b")... | 3 | 0 | ['JavaScript'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | JavaScript Simple Solution (Faster than 99.4%) | javascript-simple-solution-faster-than-9-3vx0 | javascript\nconst modifyString = function(s) {\n let res = [...s];\n \n for (let i = 0; i < res.length; i++) {\n if (res[i] !== "?") continue;\n if (re | Cookie_Ryu | NORMAL | 2021-07-21T14:21:17.187636+00:00 | 2021-07-21T14:21:17.187673+00:00 | 304 | false | ```javascript\nconst modifyString = function(s) {\n let res = [...s];\n \n for (let i = 0; i < res.length; i++) {\n if (res[i] !== "?") continue;\n if (res[i-1] !== "a" && res[i+1] !== "a") { res[i] = "a"; continue; }\n if (res[i-1] !== "b" && res[i+1] !== "b") { res[i] = "b"; continue; }\n res[i] = "c";... | 3 | 0 | ['JavaScript'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | As easy as "abc" | as-easy-as-abc-by-ecgan-x5w4 | js\nconst convertChar = (arr, i) => {\n if (arr[i] !== \'?\') {\n return arr[i]\n }\n\n if (arr[i - 1] !== \'a\' && arr[i + 1] !== \'a\') {\n return \' | ecgan | NORMAL | 2020-09-06T09:05:29.143489+00:00 | 2020-09-06T09:07:44.805888+00:00 | 449 | false | ```js\nconst convertChar = (arr, i) => {\n if (arr[i] !== \'?\') {\n return arr[i]\n }\n\n if (arr[i - 1] !== \'a\' && arr[i + 1] !== \'a\') {\n return \'a\'\n }\n\n if (arr[i - 1] !== \'b\' && arr[i + 1] !== \'b\') {\n return \'b\'\n }\n\n return \'c\'\n}\n\n/**\n * @param {string} s\n * @return {strin... | 3 | 0 | ['JavaScript'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Simple Python, We just need 3 characters('a', 'b', 'c') | simple-python-we-just-need-3-charactersa-029y | Okay so the idea is basically, we dont want any 2 characters to be consecutive, so all we have to make sure that: if both the adjacent characters are not \'a\', | shubhitt | NORMAL | 2020-09-06T04:06:14.668803+00:00 | 2020-09-06T04:06:14.668905+00:00 | 247 | false | Okay so the idea is basically, we dont want any 2 characters to be consecutive, so all we have to make sure that: if both the adjacent characters are not \'a\', we can replace \'?\' by \'a\'\nsimilarly with \'b\' and \'c\'\n\n```\nclass Solution:\n def modifyString(self, s: str) -> str:\n s = \'1\' + s + \'1\... | 3 | 1 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Very simple & easy to understand. | very-simple-easy-to-understand-by-srh_ab-cf72 | \n# Code\npython3 []\nclass Solution:\n def modifyString(self, s: str) -> str:\n # Convert the string to a list of characters since strings are immuta | srh_abhay | NORMAL | 2024-11-18T02:52:18.266817+00:00 | 2024-11-18T02:52:47.874760+00:00 | 98 | false | \n# Code\n```python3 []\nclass Solution:\n def modifyString(self, s: str) -> str:\n # Convert the string to a list of characters since strings are immutable\n s = list(s)\n \n # Helper function to get a valid character for replacement\n def get_valid_char(i):\n # We need... | 2 | 0 | ['Python3'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python Easy Solution | python-easy-solution-by-sumedh0706-b1nu | Code\n\nclass Solution:\n def modifyString(self, s: str) -> str:\n new=s\n for i in range(len(s)):\n if new[i]=="?":\n | Sumedh0706 | NORMAL | 2023-06-30T10:41:47.010466+00:00 | 2023-06-30T10:41:47.010494+00:00 | 189 | false | # Code\n```\nclass Solution:\n def modifyString(self, s: str) -> str:\n new=s\n for i in range(len(s)):\n if new[i]=="?":\n if len(new)==1:\n return "p"\n if i==0:\n lis=[new[i+1]]\n if i==len(s)-1:\n ... | 2 | 0 | ['Python3'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | CPP || EASY || 3ms | cpp-easy-3ms-by-peeronappper-2iuo | \nclass Solution {\npublic:\n string modifyString(string s) {\n for(int i=0;i<s.length();i++){\n if(i==0 && s[i]==\'?\') s[i]=(s[i+1]+1)%26 | PeeroNappper | NORMAL | 2022-06-13T11:08:09.207677+00:00 | 2022-06-13T11:08:09.207717+00:00 | 235 | false | ```\nclass Solution {\npublic:\n string modifyString(string s) {\n for(int i=0;i<s.length();i++){\n if(i==0 && s[i]==\'?\') s[i]=(s[i+1]+1)%26+\'a\';\n else if(i==s.length()-1 && s[i]==\'?\') s[i]=(s[i-1]-\'a\'+1)%26+\'a\';\n else if(s[i]==\'?\'){\n int p=1;\n ... | 2 | 0 | ['C++'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Golang Simple O(n) | golang-simple-on-by-alfinm01-aerq | \nfunc modifyString(s string) string {\n str := strings.Split(s, "")\n chars := []string{"a", "b", "c"}\n\t\n for i := range str {\n if str[i] ! | alfinm01 | NORMAL | 2021-11-08T02:39:45.218523+00:00 | 2022-07-25T09:50:06.891231+00:00 | 147 | false | ```\nfunc modifyString(s string) string {\n str := strings.Split(s, "")\n chars := []string{"a", "b", "c"}\n\t\n for i := range str {\n if str[i] != "?" {\n\t\t\tcontinue\n\t\t}\n\t\t\n\t\tfor _, c := range chars {\n\t\t\tif (i == 0 || str[i-1] != c) && (i == len(s)-1 || str[i+1] != c) {\n\t\t\t\tstr[i]... | 2 | 0 | ['Go'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | [Python3] Faster than 90% submissions, simple code | python3-faster-than-90-submissions-simpl-wxg5 | Execution Result\n\n\nRuntime: 28 ms, faster than 90.93% of Python3 online submissions for Replace All ?\'s to Avoid Consecutive Repeating Characters.\nMemory U | GauravKK08 | NORMAL | 2021-07-05T14:49:52.039789+00:00 | 2021-07-05T14:50:51.529125+00:00 | 155 | false | `Execution Result`\n\n```\nRuntime: 28 ms, faster than 90.93% of Python3 online submissions for Replace All ?\'s to Avoid Consecutive Repeating Characters.\nMemory Usage: 14.1 MB, less than 92.19% of Python3 online submissions for Replace All ?\'s to Avoid Consecutive Repeating Characters.\n```\n\n`Solution`\n```\nclas... | 2 | 0 | ['Python3'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | C++ 100% Faster | c-100-faster-by-ayush_gupta4-k3gy | \nclass Solution {\npublic:\n string modifyString(string s) {\n int i,n=s.size();\n for(i=0;i<n;i++){\n if(s[i]==\'?\'){\n | ayush_gupta4 | NORMAL | 2021-06-05T08:49:05.138769+00:00 | 2021-06-05T08:49:05.138801+00:00 | 97 | false | ```\nclass Solution {\npublic:\n string modifyString(string s) {\n int i,n=s.size();\n for(i=0;i<n;i++){\n if(s[i]==\'?\'){\n s[i]=\'a\';\n while((i>0 && s[i]==s[i-1]) || (i+1<n && s[i]==s[i+1])) s[i]++;\n }\n }\n return s;\n }\n};\n`... | 2 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Javascript easy to read | javascript-easy-to-read-by-fbecker11-2low | \nvar modifyString = function(s) {\n const arr = s.split(\'\')\n let res = \'\'\n for(let i=0;i<s.length;i++){\n const curr = arr[i]\n if(curr !== \'?\ | fbecker11 | NORMAL | 2021-05-01T14:40:17.607185+00:00 | 2021-05-01T14:41:22.811547+00:00 | 231 | false | ```\nvar modifyString = function(s) {\n const arr = s.split(\'\')\n let res = \'\'\n for(let i=0;i<s.length;i++){\n const curr = arr[i]\n if(curr !== \'?\') continue\n const prevCharCode = (arr[i-1] || \'\').charCodeAt(0)\n const nextCharCode = (arr[i+1] || \'\').charCodeAt(0)\n let random = 0\n wh... | 2 | 0 | ['JavaScript'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | C# solution | c-solution-by-daraa-3h4j | \nStringBuilder sb = new StringBuilder();\n sb.Append(s);\n for(int i=0;i<sb.Length;i++){\n if(s[i]==\'?\'){\n for(char | daraa | NORMAL | 2021-04-24T10:29:58.930272+00:00 | 2021-04-24T10:29:58.930303+00:00 | 89 | false | ```\nStringBuilder sb = new StringBuilder();\n sb.Append(s);\n for(int i=0;i<sb.Length;i++){\n if(s[i]==\'?\'){\n for(char j=\'a\'; j<=\'z\';j++){\n if(i-1>=0 && j == sb[i-1]) continue;\n if(i+1<sb.Length && j == sb[i+1]) continue;\n ... | 2 | 0 | ['String'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | C++ 0ms | c-0ms-by-leetweeb-mh3o | ```\nclass Solution {\npublic:\n string modifyString(string s) {\n int n = s.size();\n for(int i=0;i=0)\n prev = s[i-1];\n | LeetWeeb | NORMAL | 2021-04-13T08:37:25.220937+00:00 | 2021-04-13T08:37:25.220977+00:00 | 69 | false | ```\nclass Solution {\npublic:\n string modifyString(string s) {\n int n = s.size();\n for(int i=0;i<n;i++)\n {\n if(s[i] == \'?\')\n {\n char c = \'a\';\n char prev = \'a\',next = \'a\';\n if(i-1 >=0)\n prev =... | 2 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | easy to understand java | easy-to-understand-java-by-d_eat_h-teko | ```\nclass Solution {\n public String modifyString(String s) {\n char[] arr = s.toCharArray();\n for(int i = 0; i < arr.length; i++){\n | D_eat_H | NORMAL | 2021-04-11T01:34:36.816454+00:00 | 2021-04-11T01:34:36.816493+00:00 | 79 | false | ```\nclass Solution {\n public String modifyString(String s) {\n char[] arr = s.toCharArray();\n for(int i = 0; i < arr.length; i++){\n if(s.charAt(i) != \'?\')\n continue;\n for(char ch = \'a\'; ch <= \'z\'; ch++){\n if(satisfies(arr, ch, i)){\n ... | 2 | 1 | [] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | C# straightforward solution 76ms | c-straightforward-solution-76ms-by-iiii1-uhgl | \npublic class Solution {\n public string ModifyString(string s) {\n char[] letters = new char[]{\'a\',\'b\',\'c\',\'d\',\'e\',\'f\',\'g\',\'h\',\'i\' | iiii110270 | NORMAL | 2021-03-25T06:07:53.751056+00:00 | 2021-03-25T06:07:53.751089+00:00 | 69 | false | ```\npublic class Solution {\n public string ModifyString(string s) {\n char[] letters = new char[]{\'a\',\'b\',\'c\',\'d\',\'e\',\'f\',\'g\',\'h\',\'i\',\n \'j\',\'k\',\'l\',\'m\',\'n\',\'o\',\'p\',\'q\',\'r\',\n \'s\',\'t\',\'u\',\'v\',\'... | 2 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | c++ simple solution (0ms) | c-simple-solution-0ms-by-dilipsuthar60-umh1 | \nclass Solution {\npublic:\n string modifyString(string s) \n {\n int n=s.size();\n for(int i=0;i<n;i++)\n {\n if(s[i]==\ | dilipsuthar17 | NORMAL | 2021-02-04T15:48:48.810259+00:00 | 2021-02-04T15:48:48.810310+00:00 | 112 | false | ```\nclass Solution {\npublic:\n string modifyString(string s) \n {\n int n=s.size();\n for(int i=0;i<n;i++)\n {\n if(s[i]==\'?\')\n {\n s[i]=\'a\';\n while(((i>0)&&(s[i-1]==s[i]))||(i+1<n&&(s[i+1]==s[i])))\n {\n ... | 2 | 0 | [] | 2 |
replace-all-s-to-avoid-consecutive-repeating-characters | Java Straightforward Solution | java-straightforward-solution-by-ensifer-36u8 | \nclass Solution {\n public String modifyString(String s) {\n char[] res = s.toCharArray();\n for(int i = 0; i < res.length; i++) {\n | ensiferum | NORMAL | 2021-01-30T19:53:00.086126+00:00 | 2021-01-31T00:39:32.631544+00:00 | 177 | false | ```\nclass Solution {\n public String modifyString(String s) {\n char[] res = s.toCharArray();\n for(int i = 0; i < res.length; i++) {\n char c = res[i];\n if(c == \'?\') {\n char repl = \'a\';\n while((i > 0 && res[i - 1] == repl)\n ... | 2 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | simple solution - python | simple-solution-python-by-waveletus-jvpy | \nclass Solution:\n def modifyString(self, s: str) -> str:\n res = list(s)\n \n for i in range(len(res)):\n if res[i] == \'?\ | waveletus | NORMAL | 2020-11-23T04:01:42.448679+00:00 | 2020-11-23T04:01:42.448705+00:00 | 155 | false | ```\nclass Solution:\n def modifyString(self, s: str) -> str:\n res = list(s)\n \n for i in range(len(res)):\n if res[i] == \'?\':\n left = res[i-1] if i > 0 else \'\'\n right = res[i+1] if i < len(s) - 1 else \'\'\n \n for c... | 2 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python solution faster than 100 percent! | python-solution-faster-than-100-percent-2clfn | \nclass Solution:\n def modifyString(self, s: str) -> str:\n ans=\'\'\n if len(s)==1:\n if s[0]==\'?\':\n return \'a\ | man_it | NORMAL | 2020-09-07T10:28:57.130515+00:00 | 2020-09-07T10:28:57.130569+00:00 | 628 | false | ```\nclass Solution:\n def modifyString(self, s: str) -> str:\n ans=\'\'\n if len(s)==1:\n if s[0]==\'?\':\n return \'a\'\n else:\n return s\n \n if s[0]=="?":\n for j in range(97,124):\n if ord(s[1])!=j:\n ... | 2 | 0 | ['Python', 'Python3'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | C++ O(N) One Pass with explanation | c-on-one-pass-with-explanation-by-lzl124-w73e | See my latest update in repo LeetCode\n\n## Solution 1.\n\nWhen we see a s[i] == \'?\', we set it as s[i - 1] + 1 and round it to \'a\' if necessary.\n\nWhen s[ | lzl124631x | NORMAL | 2020-09-06T04:08:40.041625+00:00 | 2020-09-06T04:11:55.882085+00:00 | 187 | false | See my latest update in repo [LeetCode](https://github.com/lzl124631x/LeetCode)\n\n## Solution 1.\n\nWhen we see a `s[i] == \'?\'`, we set it as `s[i - 1] + 1` and round it to `\'a\'` if necessary.\n\nWhen `s[i] == \'?\' && s[i + 1] != \'?\'`, there is a chance of conflict with `s[i + 1]`. If there is a conflict, we si... | 2 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | 100% beats in c++ solution. | 100-beats-in-c-solution-by-jahidulcse15-80im | IntuitionApproachComplexity
Time complexity:O(N)
Space complexity:O(N)
Code | jahidulcse15 | NORMAL | 2025-04-12T05:50:24.091296+00:00 | 2025-04-12T05:50:24.091296+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:O(N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:O(N)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
... | 1 | 0 | ['C++'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | <<BEAT 100% SOLUTIONS>> | beat-100-solutions-by-dakshesh_vyas123-09mu | PLEASE UPVOTE MECode | Dakshesh_vyas123 | NORMAL | 2025-03-18T16:41:35.442207+00:00 | 2025-03-18T16:41:35.442207+00:00 | 21 | false | # PLEASE UPVOTE ME
# Code
```cpp []
class Solution {
public:
string modifyString(string s) {
for (auto i = 0; i < s.size(); ++i)
if (s[i] == '?')
for (s[i] = 'a'; s[i] <= 'c'; ++s[i])
if ((i == 0 || s[i - 1] != s[i]) && (i == s.size() - 1 || s[i + 1] != s[i]))
... | 1 | 0 | ['C++'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Easy Solution for beginners || Beats 100%+✨ || ✌️🥱 | easy-solution-for-beginners-beats-100-by-0at5 | \n# Approach:\n Describe your approach to solving the problem. \nTo solve this, we turn the string into a list to easily change its characters. As we go through | 07_deepak | NORMAL | 2024-10-22T12:44:42.643009+00:00 | 2024-10-22T12:45:29.805009+00:00 | 16 | false | \n# Approach:\n<!-- Describe your approach to solving the problem. -->\n*To solve this, we turn the string into a list to easily change its characters. As we go through the string, we replace each `\'?\'` with a letter, making sure it\u2019s different from its neighbors. For the first and last characters, we only check... | 1 | 1 | ['Python', 'Python3'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Easy C++ Solution | easy-c-solution-by-bharijamegha-tu5g | 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 | bharijamegha | NORMAL | 2024-10-05T11:05:06.013768+00:00 | 2024-10-05T11:05:06.013796+00:00 | 56 | 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. $$O(n)$$ --... | 1 | 0 | ['C++'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | abba accepted !!! Ganapati Bappa Morya ,Mangal Murthi Morya !!!!! :D | abba-accepted-ganapati-bappa-morya-manga-qatf | 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 | yesyesem | NORMAL | 2024-09-06T18:59:06.257258+00:00 | 2024-09-06T18:59:06.257284+00:00 | 33 | 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. $$O(n)$$ --... | 1 | 0 | ['C++'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Dart Solution | dart-solution-by-abbos2101-c0i8 | \nclass Solution {\n String modifyString(String s) {\n final letters = List.generate(26, (i) => String.fromCharCode(i + 97));\n int index = s.indexOf(\'? | abbos2101 | NORMAL | 2024-07-10T22:10:47.249780+00:00 | 2024-07-10T22:10:47.249808+00:00 | 4 | false | ```\nclass Solution {\n String modifyString(String s) {\n final letters = List.generate(26, (i) => String.fromCharCode(i + 97));\n int index = s.indexOf(\'?\');\n while (index != -1) {\n String first = \'\';\n String second = \'\';\n if (index > 0) first = s[index - 1];\n if (index < s.len... | 1 | 0 | ['Dart'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | [Python] ugly code | python-ugly-code-by-pbelskiy-09t9 | \nclass Solution:\n def modifyString(self, s: str) -> str:\n a = list(s)\n\n for i in range(len(a)):\n if a[i] != \'?\':\n | pbelskiy | NORMAL | 2024-05-16T07:27:53.173020+00:00 | 2024-05-16T07:27:53.173052+00:00 | 14 | false | ```\nclass Solution:\n def modifyString(self, s: str) -> str:\n a = list(s)\n\n for i in range(len(a)):\n if a[i] != \'?\':\n continue\n\n used = set()\n if i > 0:\n used.add(a[i - 1])\n if i + 1 < len(s):\n used.a... | 1 | 0 | ['Python'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Simple java code 1 ms beats 100 % && 42 mb beats 88.78 % | simple-java-code-1-ms-beats-100-42-mb-be-rhyc | \n# Complexity\n\n- image.png\n# Code\n\nclass Solution {\n public char help(char p,char n){\n for(char c=\'b\';c<=\'y\';c++){\n if(c!=p&&c | Arobh | NORMAL | 2024-03-25T06:23:22.040499+00:00 | 2024-03-25T06:23:22.040518+00:00 | 268 | false | \n# Complexity\n\n- image.png\n# Code\n```\nclass Solution {\n public char help(char p,char n){\n for(char c=\'b\';c<=\'y\';c++){\n if(c!=p&&c!=n) return c;\n }\n return ... | 1 | 0 | ['String', 'Java'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Easy to understand solution | easy-to-understand-solution-by-pratik_pa-eghl | 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 | Pratik_Pathak_05 | NORMAL | 2024-02-10T15:57:38.438396+00:00 | 2024-02-10T15:57:38.438424+00:00 | 14 | 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. $$O(n)$$ --... | 1 | 0 | ['Java'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Easy / Simple [ C | C++ | Java ] -- Dry Run Beats 100% in runtime & memory.. | easy-simple-c-c-java-dry-run-beats-100-i-kovd | Intuition\n\nC++ []\nclass Solution {\npublic:\n string modifyString(string s) {\n for (auto i = 0; i < s.size(); ++i)\n if (s[i] == \'?\') | Edwards310 | NORMAL | 2024-02-04T08:04:17.850195+00:00 | 2024-02-04T08:04:17.850219+00:00 | 26 | false | # Intuition\n\n```C++ []\nclass Solution {\npublic:\n string modifyString(string s) {\n for (auto i = 0; i < s.size(); ++i)\n if (s[i] == \'?\')\n ... | 1 | 0 | ['String', 'C', 'C++', 'Java'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | JS || Soution by Bharadwaj | js-soution-by-bharadwaj-by-manu-bharadwa-r401 | Intuition\nfunction modifyString(s) {\n // Array of lowercase letters for replacement\n const letters = [\'a\', \'b\', ..., \'z\'];\n\n // Iterate through ea | Manu-Bharadwaj-BN | NORMAL | 2024-01-11T06:39:17.647960+00:00 | 2024-01-11T06:39:17.647982+00:00 | 51 | false | # Intuition\nfunction modifyString(s) {\n // Array of lowercase letters for replacement\n const letters = [\'a\', \'b\', ..., \'z\'];\n\n // Iterate through each character in the string\n for (let i = 0; i < s.length; i++) {\n // If the character is \'?\', replace it with a distinct letter\n if (s[i] === \'?\... | 1 | 0 | ['JavaScript'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | beats 100% cpp soln | beats-100-cpp-soln-by-vermasachin6101-5ees | Intuition\nIf the input string has length 1 and the only character is \'?\', it replaces it with \'a\'. Otherwise, it returns the original string.\n\nIf the fir | gyuyul | NORMAL | 2023-12-20T15:08:55.262963+00:00 | 2023-12-20T15:08:55.262986+00:00 | 12 | false | # Intuition\nIf the input string has length 1 and the only character is \'?\', it replaces it with \'a\'. Otherwise, it returns the original string.\n\nIf the first character of the string is \'?\', it replaces it with \'a\' if the next character is \'a\', and with \'b\' otherwise.\n\nFor each \'?\' in the middle of th... | 1 | 0 | ['C++'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Easy Solution for beginners || Beats 95%+✨💫 || 🧛✌️ | easy-solution-for-beginners-beats-95-by-ech8m | Intuition\nIt seems that you want to replace the question mark (\'?\') in the input string \'s\' with lowercase English alphabets (\'a\' to \'z\') in such a way | EraOfKaushik003 | NORMAL | 2023-07-23T09:00:07.826303+00:00 | 2023-07-23T09:02:12.042393+00:00 | 453 | false | # Intuition\nIt seems that you want to replace the question mark (\'?\') in the input string \'s\' with lowercase English alphabets (\'a\' to \'z\') in such a way that adjacent characters are not the same. The code uses a dictionary to keep track of the positions of the question marks and then iterates through the stri... | 1 | 0 | ['String', 'Python3'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Easy, Simple and Fastest way in C++ | easy-simple-and-fastest-way-in-c-by-ashu-3ncx | Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n\nclass Solution {\npublic:\n string modifyString(string s) {\n int len = | ashujain | NORMAL | 2023-06-25T04:03:21.986686+00:00 | 2023-06-25T04:03:21.986705+00:00 | 113 | false | # Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\npublic:\n string modifyString(string s) {\n int len = s.size();\n for(int i=0; i<s.size(); i++) {\n char c1, c2;\n if(s[i] == \'?\') {\n if(i-1>=0 && i+1<len) {\n ... | 1 | 0 | ['C++'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | Java O(n) | 100% | java-on-100-by-im_obid-3k2e | We do not need more than 3 letters to build a non-repeating character sequence.\n\n# Code\n\nclass Solution {\n public String modifyString(String s) {\n\n | im_obid | NORMAL | 2023-05-16T04:07:53.139485+00:00 | 2023-05-16T04:08:24.736906+00:00 | 64 | false | **We do not need more than 3 letters to build a non-repeating character sequence.**\n\n# Code\n```\nclass Solution {\n public String modifyString(String s) {\n\n if(s.length()==1) return s.equals("?")?"a":s;\n\n char[] ch = s.toCharArray();\n\n for(int i=1;i<ch.length-1;i++){\n if(ch[... | 1 | 0 | ['Java'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | c++ simple solution 100% beats in 0ms. | c-simple-solution-100-beats-in-0ms-by-co-5e9e | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n1- iterate over the str | code_breaker_42 | NORMAL | 2023-05-05T14:04:37.901933+00:00 | 2023-05-05T14:04:37.901970+00:00 | 598 | 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- iterate over the string.\n2-than make a function of default arguments to handle multiple cases in one function.\n3-function contain a for loop fron a to z.\n4-if(i ... | 1 | 0 | ['C++'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | Replace All ?'s to Avoid Consecutive Repeating Characters Solution Java | replace-all-s-to-avoid-consecutive-repea-iv7t | class Solution {\n public String modifyString(String s) {\n char[] array = s.toCharArray();\n int length = array.length;\n if (array[0] | bhupendra786 | NORMAL | 2022-09-15T08:13:33.973164+00:00 | 2022-09-15T08:13:33.973204+00:00 | 310 | false | class Solution {\n public String modifyString(String s) {\n char[] array = s.toCharArray();\n int length = array.length;\n if (array[0] == \'?\') {\n if (length == 1)\n array[0] = \'a\';\n else {\n if (array[1] == \'a\')\n ar... | 1 | 0 | ['String'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python, less condition approach | python-less-condition-approach-by-maleki-r72o | \nclass Solution:\n def modifyString(self, s: str) -> str:\n s = list("0"+s+"0")\n for i, l in enumerate(s[:-1]):\n if l == "?":\n | maleki_shoja | NORMAL | 2022-08-30T15:38:20.328515+00:00 | 2022-08-30T15:38:20.328556+00:00 | 396 | false | ```\nclass Solution:\n def modifyString(self, s: str) -> str:\n s = list("0"+s+"0")\n for i, l in enumerate(s[:-1]):\n if l == "?":\n for c in "abc":\n if c != s[i-1] and c != s[i+1]:\n s[i] = c\n break\n ... | 1 | 0 | ['Python'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Java Solution | Runtime: 2ms | java-solution-runtime-2ms-by-harsh_kode-quiu | \nclass Solution {\n public String modifyString(String s) {\n char[] ch = s.toCharArray();\n for(int i=0;i<s.length();i++){\n if(ch[ | Harsh_Kode | NORMAL | 2022-08-26T20:18:29.304291+00:00 | 2022-08-26T20:18:29.304361+00:00 | 443 | false | ```\nclass Solution {\n public String modifyString(String s) {\n char[] ch = s.toCharArray();\n for(int i=0;i<s.length();i++){\n if(ch[i]==\'?\')\n for(char j=\'a\';j<=\'c\';j++){ // j = (a-z) is not required.\n if(i>0 && ch[i-1]==j) continue; // skip if pre... | 1 | 0 | ['String', 'Java'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | [C++] || Very easy to understand | c-very-easy-to-understand-by-riteshkhan-nfwh | \nclass Solution {\n char update_char(char x, char y){\n char c = \'a\';\n while(c==x || c==y){\n c++;\n }\n return c; | RiteshKhan | NORMAL | 2022-07-25T08:43:33.594589+00:00 | 2022-07-25T08:53:38.819054+00:00 | 215 | false | ```\nclass Solution {\n char update_char(char x, char y){\n char c = \'a\';\n while(c==x || c==y){\n c++;\n }\n return c;\n }\npublic:\n string modifyString(string s) {\n int l = s.length();\n char c;\n if(s[0] == \'?\'){\n c = update_char(... | 1 | 0 | ['C'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | 🐌 C# - straightforward solution | c-straightforward-solution-by-user7166uf-3gr3 | \npublic class Solution\n{\n public string ModifyString(string s)\n {\n char[] characters = s.ToCharArray();\n\n for (int index = 0; index < | user7166Uf | NORMAL | 2022-06-12T12:39:50.011930+00:00 | 2022-06-12T12:39:50.011970+00:00 | 62 | false | ```\npublic class Solution\n{\n public string ModifyString(string s)\n {\n char[] characters = s.ToCharArray();\n\n for (int index = 0; index < characters.Length; index++)\n {\n if (characters[index] is not \'?\') continue;\n\n char? previousCharacter = index > 0 ? chara... | 1 | 0 | ['C#'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Tried a lot but only 749/776 passed :( | tried-a-lot-but-only-749776-passed-by-ch-jcpg | \nclass Solution {\npublic:\n string modifyString(string s) {\n string str="";\n for(int i=0;i<s.size();i++){\n if(s[i]==\'?\')\n | charu794 | NORMAL | 2022-05-07T11:01:59.222831+00:00 | 2022-05-07T11:01:59.222868+00:00 | 47 | false | ```\nclass Solution {\npublic:\n string modifyString(string s) {\n string str="";\n for(int i=0;i<s.size();i++){\n if(s[i]==\'?\')\n {\n int x,y;\n if(i==0){\n x= s[i+1];\n if(x<97 || x>122)\n ... | 1 | 0 | [] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | C++ solution | c-solution-by-charu794-ia25 | \nclass Solution {\npublic:\n string modifyString(string s) {\n \n for(int i=0;i<s.size();i++)\n {\n if(s[i]==\'?\')\n | charu794 | NORMAL | 2022-05-07T11:00:31.805600+00:00 | 2022-05-07T11:00:31.805626+00:00 | 87 | false | ```\nclass Solution {\npublic:\n string modifyString(string s) {\n \n for(int i=0;i<s.size();i++)\n {\n if(s[i]==\'?\')\n {\n for(char ch=\'a\';ch<=\'z\';ch++)\n {\n if(i==0 && s[i+1]!=ch)\n {\n ... | 1 | 0 | [] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | java || easy to understand | java-easy-to-understand-by-pratham_18-zos6 | \nclass Solution {\n public String modifyString(String s) {\n char arr[]=new char[s.length()];\n for(int i=0;i<s.length();i++){\n arr | Pratham_18 | NORMAL | 2022-04-19T07:38:21.417311+00:00 | 2022-04-19T07:38:21.417354+00:00 | 171 | false | ```\nclass Solution {\n public String modifyString(String s) {\n char arr[]=new char[s.length()];\n for(int i=0;i<s.length();i++){\n arr[i]=s.charAt(i);\n }\n for(int i=0;i<arr.length;i++){\n if(arr[i]==\'?\'){\n for(char j=\'a\';j<=\'c\';j++){\n ... | 1 | 0 | ['Java'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | python clear and fast solution | python-clear-and-fast-solution-by-tirtha-xcdx | \n````\nclass Solution:\n def modifyString(self, s: str) -> str:\n a=list(s)+["."]\n for i in range(len(a)-1):\n if a[i]=="?":\n | tirtha_20 | NORMAL | 2022-03-11T12:10:00.096226+00:00 | 2022-03-11T12:10:00.096273+00:00 | 77 | false | \n````\nclass Solution:\n def modifyString(self, s: str) -> str:\n a=list(s)+["."]\n for i in range(len(a)-1):\n if a[i]=="?":\n if a[i+1]!="a" and a[i-1]!="a":\n ... | 1 | 0 | [] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | 2 C++ Solutions || Different Methods each time for replacement of '?' || Most Optimized Approach | 2-c-solutions-different-methods-each-tim-fvbt | Approach 1\nIn this approach when we encounter a \'?\' we try from the first character \'a\', if the character satisfies the condition of the question i.e. not | dee_stroyer | NORMAL | 2021-12-18T12:50:22.873634+00:00 | 2021-12-18T12:50:22.873675+00:00 | 172 | false | **Approach 1**\nIn this approach when we encounter a \'?\' we try from the first character \'a\', if the character satisfies the condition of the question i.e. not similar to the consecutive one, then we replace the ? with the character, otherwise we increment the current char try doing the same as we did with a.\n```\... | 1 | 0 | ['C', 'C++'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | C++ Simple and Short Straight-Forward Solution | c-simple-and-short-straight-forward-solu-0ier | Idea:\nWe only need three characters for replacement, because there can\'t be more than one character at each side of the ?.\nSo we try each letter and replace. | yehudisk | NORMAL | 2021-11-03T22:30:20.622478+00:00 | 2021-11-03T22:30:20.622509+00:00 | 182 | false | **Idea:**\nWe only need three characters for replacement, because there can\'t be more than one character at each side of the `?`.\nSo we try each letter and replace.\n\n**Time Complexity:** O(n)\n**Space Complexity:** None\n```\nclass Solution {\npublic:\n string modifyString(string s) {\n for (int i = 0; i ... | 1 | 0 | ['C'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | Java- Easy Solution | java-easy-solution-by-srajen488-k4s0 | \nclass Solution {\n public String modifyString(String s) {\n char[] str = s.toCharArray();\n for(int i =0; i < str.length; i ++)\n {\n | srajen488 | NORMAL | 2021-09-07T02:32:40.935085+00:00 | 2021-09-07T02:32:40.935122+00:00 | 170 | false | ```\nclass Solution {\n public String modifyString(String s) {\n char[] str = s.toCharArray();\n for(int i =0; i < str.length; i ++)\n {\n if(str[i] == \'?\')\n {\n str[i] = replaceChar(str, i);\n }\n }\n return String.valueOf(str);\n... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python Easy Solution - 90+% Faster | python-easy-solution-90-faster-by-gsbc95-yc96 | \nclass Solution:\n def modifyString(self, s: str) -> str:\n prev = \'\'\n for index,value in enumerate(s):\n if value == \'?\':\n | gsbc95 | NORMAL | 2021-08-19T20:36:43.063699+00:00 | 2021-08-19T20:39:15.357553+00:00 | 109 | false | ```\nclass Solution:\n def modifyString(self, s: str) -> str:\n prev = \'\'\n for index,value in enumerate(s):\n if value == \'?\':\n if index+1 < len(s):\n for i in range(97,123):\n if chr(i) !=s[index+1] and chr(i) != prev:\n ... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python3, fast and simple | python3-fast-and-simple-by-mihailp-uh9r | \nclass Solution:\n def modifyString(self, s: str) -> str:\n ans = []\n\n for i, ch in enumerate(s):\n if ch != "?":\n | MihailP | NORMAL | 2021-07-29T16:12:15.705703+00:00 | 2021-07-29T16:12:15.705754+00:00 | 303 | false | ```\nclass Solution:\n def modifyString(self, s: str) -> str:\n ans = []\n\n for i, ch in enumerate(s):\n if ch != "?":\n ans.append(ch)\n else:\n prev_ch = ans[-1] if i > 0 else ""\n next_ch = s[i + 1] if i < len(s) - 1 else ""\n ... | 1 | 1 | ['Python', 'Python3'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | simple C++ | simple-c-by-rap_tors-j1hx | \nstring modifyString(string s) {\n for(int i = 0; i < s.size(); i++) {\n if(s[i] == \'?\') {\n for(char ch = \'a\'; ch <= \'c\ | rap_tors | NORMAL | 2021-07-04T17:22:28.492226+00:00 | 2021-07-04T17:22:28.492268+00:00 | 169 | false | ```\nstring modifyString(string s) {\n for(int i = 0; i < s.size(); i++) {\n if(s[i] == \'?\') {\n for(char ch = \'a\'; ch <= \'c\'; ch++) {\n if(i > 0 and s[i-1] == ch) continue;\n if(i < s.size() - 1 and s[i + 1] == ch) continue;\n ... | 1 | 0 | ['C', 'C++'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Easy to understand JavaScript solution (reduce) | easy-to-understand-javascript-solution-r-ayo1 | \tvar modifyString = function(s) {\n\t\tconst getChar = (num, left, right) => {\n\t\t\tconst char = String.fromCharCode(num);\n\t\t\treturn char === left || cha | tzuyi0817 | NORMAL | 2021-06-26T07:10:00.799116+00:00 | 2021-06-26T07:10:00.799148+00:00 | 130 | false | \tvar modifyString = function(s) {\n\t\tconst getChar = (num, left, right) => {\n\t\t\tconst char = String.fromCharCode(num);\n\t\t\treturn char === left || char === right ? getChar(++num, left, right) : char;\n\t\t};\n\n\t\treturn [...s].reduce((acc, curr, index) => {\n\t\t\tif (curr === \'?\') {\n\t\t\t\tconst checkL... | 1 | 0 | ['JavaScript'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | [Java] Straightforward Clean Code | java-straightforward-clean-code-by-mllej-w0tm | \nclass Solution {\n public String modifyString(String s) {\n StringBuilder str = new StringBuilder("a"+s+"z");\n for (int i = 0; i < s.length( | mllejuly | NORMAL | 2021-06-13T08:50:57.540351+00:00 | 2021-09-25T09:26:35.579753+00:00 | 177 | false | ```\nclass Solution {\n public String modifyString(String s) {\n StringBuilder str = new StringBuilder("a"+s+"z");\n for (int i = 0; i < s.length(); i++) { \n if (s.charAt(i) == \'?\') {\n if (Math.min(str.charAt(i),str.charAt(i+2))>\'a\') {\n str.setCharAt(... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | java easy 1 ms, faster than 100.00% | java-easy-1-ms-faster-than-10000-by-tany-88zq | public String modifyString(String s) {\n StringBuilder sb = new StringBuilder();\n for(int i = 0; i < s.length(); i++){\n if(s.charAt(i | tanyasinghal92 | NORMAL | 2021-06-13T07:29:20.307644+00:00 | 2021-06-13T07:29:20.307681+00:00 | 198 | false | public String modifyString(String s) {\n StringBuilder sb = new StringBuilder();\n for(int i = 0; i < s.length(); i++){\n if(s.charAt(i) == \'?\'){\n char ch = \'a\';\n while((i != 0 && sb.charAt(i-1) == ch) || (i != s.length() - 1 && s.charAt(i+1) ... | 1 | 2 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python simple single pass solution, Time: O(n) | python-simple-single-pass-solution-time-otc4o | \nclass Solution:\n def modifyString(self, s: str) -> str:\n output = ""\n \n for index in range(len(s)):\n if s[index] != \' | kaushal087 | NORMAL | 2021-06-06T13:58:38.673070+00:00 | 2021-06-06T13:58:38.673115+00:00 | 121 | false | ```\nclass Solution:\n def modifyString(self, s: str) -> str:\n output = ""\n \n for index in range(len(s)):\n if s[index] != \'?\':\n output += s[index] \n continue\n\n not_allowed = set()\n \n if index - 1 >= 0:\n ... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python O(n) time | O(n) Space | python-on-time-on-space-by-wintersoldier-inxt | \n# Time = O(n)\n# Space = O(n)\nclass Solution:\n def modifyString(self, s: str) -> str:\n d = {}\n s = list(s)\n\n for i in range(len( | wintersoldier | NORMAL | 2021-06-05T21:25:42.118683+00:00 | 2021-06-05T21:25:42.118713+00:00 | 92 | false | ```\n# Time = O(n)\n# Space = O(n)\nclass Solution:\n def modifyString(self, s: str) -> str:\n d = {}\n s = list(s)\n\n for i in range(len(s)):\n if s[i] != \'?\':\n d[s[i]] = i \n continue \n \n for j in range(1, 26+1):\n ... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | c# solution | c-solution-by-obpb4ae4twxy-gbl0 | \npublic class Solution\n{\n public string ModifyString(string s) \n {\n var c = 0;\n \n var result = new StringBuilder();\n f | obpb4ae4twxy | NORMAL | 2021-06-05T16:32:45.359448+00:00 | 2021-06-05T16:34:15.173057+00:00 | 57 | false | ```\npublic class Solution\n{\n public string ModifyString(string s) \n {\n var c = 0;\n \n var result = new StringBuilder();\n for (int i = 0; i < s.Length; i++)\n {\n if (s[i] != \'?\')\n {\n result.Append(s[i]);\n continue;\... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | simple circular C++ 100% faster | simple-circular-c-100-faster-by-dzzhdzzh-4hzw | \nclass Solution {\npublic:\n string modifyString(string s) {\n int j = 0;\n for(int i = 0; i < s.size(); ++i){\n char c = s[i];\n | dzzhdzzh | NORMAL | 2021-06-04T07:04:52.273862+00:00 | 2021-06-04T07:04:52.273909+00:00 | 59 | false | ```\nclass Solution {\npublic:\n string modifyString(string s) {\n int j = 0;\n for(int i = 0; i < s.size(); ++i){\n char c = s[i];\n if(c == \'?\'){\n char l = i > 0 ? s[i-1] : \'*\';\n char r = i < s.size() - 1 ? s[i + 1] : \'*\';\n c... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | 100 % :: Python | Maintain a pool of characters | 100-python-maintain-a-pool-of-characters-p7hv | \nclass Solution:\n def modifyString(self, s: str) -> str:\n pool = set([chr(i) for i in range(97, 123)])\n s = list(s)\n for i in range | tuhinnn_py | NORMAL | 2021-05-20T16:21:33.702338+00:00 | 2021-05-20T16:21:33.702391+00:00 | 78 | false | ```\nclass Solution:\n def modifyString(self, s: str) -> str:\n pool = set([chr(i) for i in range(97, 123)])\n s = list(s)\n for i in range(len(s)):\n if s[i] == \'?\':\n pool -= set([s[i - 1] if i >= 1 else None, s[i + 1] if i <= len(s) - 2 else None])\n ... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | [C++] 0ms Concise Solution | 5 lines | Beats 100% Time | c-0ms-concise-solution-5-lines-beats-100-uci6 | \n string modifyString(string s) {\n for(int i = 0; i < s.length(); i++){\n if(s[i] == \'?\'){\n if((i == 0 || s[i - 1] != \ | sherlasd | NORMAL | 2021-05-07T16:22:31.384082+00:00 | 2021-05-07T16:22:31.384114+00:00 | 208 | false | ```\n string modifyString(string s) {\n for(int i = 0; i < s.length(); i++){\n if(s[i] == \'?\'){\n if((i == 0 || s[i - 1] != \'a\') && (i == s.length() - 1 || s[i + 1] != \'a\')) s[i] = \'a\'; \n else if((i == 0 || s[i - 1] != \'b\') && (i == s.length() - 1 || s[i + ... | 1 | 0 | ['C', 'C++'] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | 100% faster || C++ Solution || Clean Code | 100-faster-c-solution-clean-code-by-kann-hao7 | \nclass Solution {\npublic:\n string modifyString(string s) {\n int n = s.size();\n for(int i = 0; i < n; i++)\n if(s[i] == \'?\')\n | kannu_priya | NORMAL | 2021-04-24T09:58:08.409446+00:00 | 2021-04-24T09:58:08.409477+00:00 | 136 | false | ```\nclass Solution {\npublic:\n string modifyString(string s) {\n int n = s.size();\n for(int i = 0; i < n; i++)\n if(s[i] == \'?\')\n for(int c = \'a\'; c <= \'z\'; c++)\n {\n if(i-1 >= 0 && c == s[i-1]) continue;\n if(i+1... | 1 | 0 | ['C', 'C++'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Very Simple C++ Solution | 100% Faster 0ms | very-simple-c-solution-100-faster-0ms-by-g9i6 | \nclass Solution {\npublic:\n \n char fun(char cp, char cn)\n {\n for(int i=0; i<26; i++)\n {\n char c = i +\'a\';\n | imanshul | NORMAL | 2021-04-21T09:11:22.842666+00:00 | 2021-04-21T09:11:22.842700+00:00 | 56 | false | ```\nclass Solution {\npublic:\n \n char fun(char cp, char cn)\n {\n for(int i=0; i<26; i++)\n {\n char c = i +\'a\';\n if(cp!=c && cn!=c)\n {\n return c;\n }\n }\n return 0;\n }\n string modifyString(string s) {\n ... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | simple C++ solution | simple-c-solution-by-tuhin12-qe3o | \nclass Solution {\npublic:\n string modifyString(string s) \n {\n int len=s.length();\n char ch;\n \n for(int i=0;i<len;i++)\ | Tuhin12 | NORMAL | 2021-04-16T15:26:44.668879+00:00 | 2021-04-16T15:26:44.668917+00:00 | 66 | false | ```\nclass Solution {\npublic:\n string modifyString(string s) \n {\n int len=s.length();\n char ch;\n \n for(int i=0;i<len;i++)\n {\n ch=\'a\';\n if(s[i]==\'?\')\n {\n if(i==0)\n {\n while(ch==s[i... | 1 | 1 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Javascript using charCode and .slice | javascript-using-charcode-and-slice-by-k-um9a | \nvar modifyString = function(s) {\n for(let i = 0; i < s.length; i++){\n if(s[i] === \'?\'){\n let charCode = 97\n const charCo | kziechmann | NORMAL | 2021-04-14T05:12:03.037124+00:00 | 2021-04-14T05:12:03.037154+00:00 | 47 | false | ```\nvar modifyString = function(s) {\n for(let i = 0; i < s.length; i++){\n if(s[i] === \'?\'){\n let charCode = 97\n const charCodeLeft = s.charCodeAt(i-1)\n const charCodeRight = s.charCodeAt(i+1)\n while(charCodeLeft === charCode || charCodeRight === charCode){\... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Elegant Python Solution | elegant-python-solution-by-true-detectiv-buur | \nclass Solution:\n def modifyString(self, s: str) -> str:\n alphabet = set(string.ascii_lowercase)\n \n res = \'\'\n for i in ra | true-detective | NORMAL | 2021-04-06T14:27:16.042049+00:00 | 2021-04-06T14:28:32.998979+00:00 | 133 | false | ```\nclass Solution:\n def modifyString(self, s: str) -> str:\n alphabet = set(string.ascii_lowercase)\n \n res = \'\'\n for i in range(len(s)):\n if s[i] != \'?\':\n res += s[i]\n continue\n \n last_char = res[-1] if i > 0 el... | 1 | 1 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Java solution faster than 100% | java-solution-faster-than-100-by-bhavyas-wfo4 | We dont want consecutive characters and multiple solutions are possible, so all we need to check is whether the character we replace is not the same as its left | bhavyaspatel | NORMAL | 2021-03-29T19:24:50.359444+00:00 | 2021-03-29T19:24:50.359489+00:00 | 179 | false | We dont want consecutive characters and multiple solutions are possible, so all we need to check is whether the character we replace is not the same as its left/right neighbor.\nWe can simply solve this problem by checking with 3 alphabets : \'a\', \'b\' and \'c\'\nIf left is null, check 2nd character\nelse if right is... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | [C++] Concise Solution Using string::find | c-concise-solution-using-stringfind-by-k-rgvq | Search for every ? in the string, assigning it either a, b, or c depending on the adjacent values.\n\nstring modifyString(string s) {\n\tfor (int i = s.find(\'? | KasraCode | NORMAL | 2021-03-23T18:18:14.373828+00:00 | 2021-03-23T18:18:14.373857+00:00 | 38 | false | Search for every `?` in the string, assigning it either `a`, `b`, or `c` depending on the adjacent values.\n```\nstring modifyString(string s) {\n\tfor (int i = s.find(\'?\'); i != string::npos; i = s.find(\'?\', i + 1)) {\n\t\tchar prev = i > 0 ? s[i - 1] : 0, next = i < s.size() - 1 ? s[i + 1] : 0;\n\t\ts[i] = (prev ... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python O(n) | python-on-by-shacoz-l3z0 | \nclass Solution:\n def modifyString(self, s: str) -> str:\n a_to_z = "abcdefghijklmnopqrstuvwxyz"\n ans = ""\n for i in range (0,len(s) | ShacoZ | NORMAL | 2021-03-03T06:40:47.457914+00:00 | 2021-03-03T06:40:47.457950+00:00 | 109 | false | ```\nclass Solution:\n def modifyString(self, s: str) -> str:\n a_to_z = "abcdefghijklmnopqrstuvwxyz"\n ans = ""\n for i in range (0,len(s)):\n if s[i] =="?":\n left, right ="?","?"\n if i-1 >=0 :\n left = ans[i-1]\n if i... | 1 | 0 | [] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python O(n) solution | python-on-solution-by-stuti2009-k38u | We can simply start with converting string in to list of characters as string is immutable in python. Then iterate over each element in list if it is "?" check | stuti2009 | NORMAL | 2021-03-03T05:02:04.846391+00:00 | 2021-03-03T05:04:37.450648+00:00 | 154 | false | We can simply start with converting string in to list of characters as string is immutable in python. Then iterate over each element in list if it is "?" check previous and next value and replace it with any character apart from neighbours.\n \n `class Solution:\n def modifyString(self, s: str) -> str:\n ... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python3 solution, beat 97% | python3-solution-beat-97-by-hotrabbit05-f0m8 | Python\nclass Solution:\n def modifyString(self, s: str) -> str:\n chars = \'abcdefghijklmnopqrstuvwxyz\'\n s_list = list(s)\n for i in | hotrabbit05 | NORMAL | 2021-02-15T03:30:37.850335+00:00 | 2021-02-15T03:30:37.850437+00:00 | 119 | false | ```Python\nclass Solution:\n def modifyString(self, s: str) -> str:\n chars = \'abcdefghijklmnopqrstuvwxyz\'\n s_list = list(s)\n for i in range(len(s)):\n if s_list[i] != \'?\':\n continue\n pre, post = \'\', \'\'\n if i - 1 >= 0:\n ... | 1 | 0 | [] | 1 |
replace-all-s-to-avoid-consecutive-repeating-characters | Python Solution | python-solution-by-sunnychugh-nsmp | ```\nclass Solution(object):\n def modifyString(self, s):\n """\n :type s: str\n :rtype: str\n """\n \n import rand | sunnychugh | NORMAL | 2021-02-03T00:13:57.711480+00:00 | 2021-02-03T00:13:57.711519+00:00 | 133 | false | ```\nclass Solution(object):\n def modifyString(self, s):\n """\n :type s: str\n :rtype: str\n """\n \n import random\n import string\n def gen_char():\n char = random.choice(string.ascii_lowercase)\n return char\n \n s = lis... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | [JAVA] Simple solution, O(n) | java-simple-solution-on-by-yurokusa-u7qc | \nclass Solution {\n public String modifyString(String s) {\n var word = s.toCharArray();\n for (int i = 0; i < word.length; i++) {\n | yurokusa | NORMAL | 2021-02-02T21:40:32.419381+00:00 | 2021-02-02T21:40:46.057249+00:00 | 398 | false | ```\nclass Solution {\n public String modifyString(String s) {\n var word = s.toCharArray();\n for (int i = 0; i < word.length; i++) {\n if (word[i] == \'?\') {\n word[i] = \'a\';\n var left = i == 0 ? \'z\' : word[i - 1];\n var right = i == word.... | 1 | 0 | ['Java'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | easy java random solution | easy-java-random-solution-by-darkrwe-3ntv | \nclass Solution {\n Map<Integer, Character> map = new HashMap();\n Random rand = new Random();\n \n public String modifyString(String s) {\n | darkrwe | NORMAL | 2021-01-06T12:12:55.530037+00:00 | 2021-01-06T12:13:27.258334+00:00 | 222 | false | ```\nclass Solution {\n Map<Integer, Character> map = new HashMap();\n Random rand = new Random();\n \n public String modifyString(String s) {\n char[] ch = s.toCharArray();\n map.put(0, \'a\');\n map.put(1, \'b\');\n map.put(2, \'c\');\n map.put(3, \'d\');\n map.pu... | 1 | 1 | ['Java'] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | [Java] Straightforward solution, beats 100% | java-straightforward-solution-beats-100-y0son | Time Complexity: O(n)\nSpace Complexity: O(n)\n```\nclass Solution {\n public String modifyString(String s) {\n StringBuilder sb = new StringBuilder() | Miss_S | NORMAL | 2020-12-25T17:47:52.175883+00:00 | 2020-12-25T17:49:17.662787+00:00 | 154 | false | Time Complexity: O(n)\nSpace Complexity: O(n)\n```\nclass Solution {\n public String modifyString(String s) {\n StringBuilder sb = new StringBuilder();\n for(int i=0;i<s.length();i++){\n if(s.charAt(i)!=\'?\'){\n sb.append(s.charAt(i));\n }else{\n sb.... | 1 | 0 | [] | 0 |
replace-all-s-to-avoid-consecutive-repeating-characters | python3 simple solution | python3-simple-solution-by-julianayo-le97 | \nclass Solution:\n def modifyString(self, s: str) -> str:\n if s == \'?\':\n return \'a\'\n s_set = set(s)\n available_chars | JulianaTsv | NORMAL | 2020-12-13T09:26:24.593259+00:00 | 2020-12-13T09:51:57.842396+00:00 | 480 | false | ```\nclass Solution:\n def modifyString(self, s: str) -> str:\n if s == \'?\':\n return \'a\'\n s_set = set(s)\n available_chars = set()\n for char in s:\n if char == \'?\':\n if not available_chars:\n available_chars = set(\'abcdefg... | 1 | 0 | ['Python3'] | 0 |
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