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lexicographically-smallest-palindrome | Lexicographically Smallest Palindrome Solution in C++ | lexicographically-smallest-palindrome-so-27li | 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 | The_Kunal_Singh | NORMAL | 2023-05-26T04:08:09.611109+00:00 | 2023-05-26T04:08:09.611135+00:00 | 63 | 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. $$O(n)$$ -->\nO(n)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$... | 2 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | Easy Solution in Java | 8 ms - 100% beats | Fully Explained | easy-solution-in-java-8-ms-100-beats-ful-xgrl | \n\n# Approach\n\nTo solve the problem, we follow these steps:\n\n1. Convert the input string s into a character array arr for easy manipulation.\n2. Determine | akobirswe | NORMAL | 2023-05-23T16:02:42.630730+00:00 | 2023-07-03T20:32:26.659192+00:00 | 164 | false | \n\n# Approach\n\nTo solve the problem, we follow these steps:\n\n1. Convert the input string `s` into a character array `arr` for easy manipulation.\n2. Determine the length of the array and store it in the variable `n`.\n3. Iterate over the array from the beginning (`i = 0`) to the middle (`i < n / 2`).\n4. Check if ... | 2 | 0 | ['Array', 'Two Pointers', 'String', 'Java'] | 0 |
lexicographically-smallest-palindrome | [JavaScript] 2697. Lexicographically Smallest Palindrome | javascript-2697-lexicographically-smalle-feb1 | ---\n\nWeekly Contest 346 solutions:\n- Q1 - https://leetcode.com/problems/minimum-string-length-after-removing-substrings/solutions/3547566/javascript-2696-min | pgmreddy | NORMAL | 2023-05-21T06:43:02.488269+00:00 | 2023-05-21T06:47:26.528402+00:00 | 675 | false | ---\n\nWeekly Contest 346 solutions:\n- Q1 - https://leetcode.com/problems/minimum-string-length-after-removing-substrings/solutions/3547566/javascript-2696-minimum-string-length-after-removing-substrings/\n- Q2 - https://leetcode.com/problems/lexicographically-smallest-palindrome/solutions/3547563/javascript-2697-lexi... | 2 | 0 | ['JavaScript'] | 0 |
lexicographically-smallest-palindrome | Easily understandable C++ solution | easily-understandable-c-solution-by-heal-x56p | Intuition\nJust try to make the (i)th and (l-1-i)th characters same with the given condition (The lower value charater shoulb be taken) #where l is the length o | Healthy_UG_007 | NORMAL | 2023-05-21T05:58:34.036638+00:00 | 2023-05-21T05:58:34.036663+00:00 | 73 | false | # Intuition\nJust try to make the (i)th and (l-1-i)th characters same with the given condition (The lower value charater shoulb be taken) #where l is the length of th string\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nUse a for loop and iterate upto the middle and try to make t... | 2 | 0 | ['String', 'C++'] | 0 |
lexicographically-smallest-palindrome | | | ✅ Very Simple C++ Solution with 100% Time and Space Complexity ✅ | | | very-simple-c-solution-with-100-time-and-bmb4 | Intuition\nPlease UPVOTE if you LIKE the Solution and COMMENT your own Code and thoughts.\n\n# Approach\nThese Type of Questions are Solved using Two pointers,\ | Satyam_Singh_Nijwala | NORMAL | 2023-05-21T05:04:04.605848+00:00 | 2023-05-21T05:04:04.605887+00:00 | 218 | false | # Intuition\nPlease UPVOTE if you LIKE the Solution and COMMENT your own Code and thoughts.\n\n# Approach\nThese Type of Questions are Solved using Two pointers,\nOne from Starting index i=0 and other from end r=size of string -1.\n\n# Complexity\n- Time complexity:\n100%\n\n- Space complexity:\n100%\n\n# Code\n```\ncl... | 2 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | Easiest 2 Pointers Approach | Short code explained | easiest-2-pointers-approach-short-code-e-i3wj | Intuition\nWe need to make the string palindrome. Thus we need to compare first and last indices.\n\n# Approach\n- (step 1) Keep one index at beginning of strin | Jeetaksh | NORMAL | 2023-05-21T04:40:26.862455+00:00 | 2023-05-21T04:40:26.862489+00:00 | 122 | false | # Intuition\nWe need to make the string palindrome. Thus we need to compare first and last indices.\n\n# Approach\n- **(step 1)** Keep one index at beginning of string and other at the end.\n- **(case 1)** Update the first index to the character at last index if last index character is alphabetically smaller.\n- **(cas... | 2 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | Simplest of all solution | Super easy to understand | simplest-of-all-solution-super-easy-to-u-5vw8 | Just compare front and back letters.\n- If they are not equal then replace the larger letter with the smaller one.\n\n# Code\n\nclass Solution:\n def makeSma | younus-Sid | NORMAL | 2023-05-21T04:27:30.747284+00:00 | 2023-05-21T04:27:30.747320+00:00 | 659 | false | - Just compare front and back letters.\n- If they are not equal then replace the larger letter with the smaller one.\n\n# Code\n```\nclass Solution:\n def makeSmallestPalindrome(self, s: str) -> str:\n for i in range(int(len(s)/2)):\n if s[i] != s[len(s)-i-1]:\n if s[i] < s[len(s)-i-... | 2 | 0 | ['Python3'] | 0 |
lexicographically-smallest-palindrome | C++ || Easiest 4 lines of code | c-easiest-4-lines-of-code-by-mrigank_200-tjom | Here is my c++ code for this problem.\n\n# Complexity\n- Time complexity:O(n/2)\n\n- Space complexity:O(1)\n\n# Code\n\nclass Solution {\npublic:\n string ma | mrigank_2003 | NORMAL | 2023-05-21T04:25:23.190296+00:00 | 2023-05-21T04:25:23.190336+00:00 | 540 | false | Here is my c++ code for this problem.\n\n# Complexity\n- Time complexity:$$O(n/2)$$\n\n- Space complexity:$$O(1)$$\n\n# Code\n```\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n for(int i=0; i<s.size()/2; i++){\n if(s[i]!=s[s.size()-1-i]){\n (s[i]<s[s.size()-... | 2 | 0 | ['String', 'C++'] | 1 |
lexicographically-smallest-palindrome | EASIEST JAVA SOLUTION EVER | easiest-java-solution-ever-by-nikhil_rat-s9kv | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | nikhil_rathi93 | NORMAL | 2023-05-21T04:15:31.398198+00:00 | 2023-05-21T04:18:37.737871+00:00 | 553 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['C++', 'Java'] | 0 |
lexicographically-smallest-palindrome | JAVA Easy SOlution || 100 Faster | java-easy-solution-100-faster-by-mayur01-v5k7 | \n- Time complexity: O(n)\n\n- Space complexity: O(n)\n\n\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n \n char []ch=s.toC | mayur0106 | NORMAL | 2023-05-21T04:06:34.981335+00:00 | 2023-05-21T04:06:34.981366+00:00 | 280 | false | \n- Time complexity: O(n)\n\n- Space complexity: O(n)\n\n```\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n \n char []ch=s.toCharArray();\n \n int i=0;\n int j=s.length()-1;\n while(i<=j)\n {\n if(ch[i]<ch[j])ch[j]=ch[i];\n els... | 2 | 0 | ['Java'] | 0 |
lexicographically-smallest-palindrome | EASY SOLUTION USING LOOPS || BRUTE FORCE BEATING 100% || C++ & JAVA SOLUTION | easy-solution-using-loops-brute-force-be-mk6z | IntuitionTo convert the given string into a palindrome with the minimum number of operations, we need to ensure that the string reads the same forward and backw | arshi_bansal | NORMAL | 2025-03-15T10:11:40.223170+00:00 | 2025-03-15T10:11:40.223170+00:00 | 93 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
To convert the given string into a palindrome with the minimum number of operations, we need to ensure that the string reads the same forward and backward.
For each character s[i] in the first half, it should match its corresponding charact... | 1 | 0 | ['Two Pointers', 'String', 'Greedy', 'C++', 'Java'] | 0 |
lexicographically-smallest-palindrome | ☑️ Golang solution | golang-solution-by-codemonkey80s-jl5v | Code | CodeMonkey80s | NORMAL | 2025-01-28T00:25:41.604377+00:00 | 2025-01-28T00:25:41.604377+00:00 | 34 | false | # Code
```golang []
func makeSmallestPalindrome(s string) string {
output := []byte(s)
for i, j := 0, len(s)-1; i <= j; i, j = i+1, j-1 {
if s[i] == s[j] {
continue
}
output[i] = min(s[i], s[j])
output[j] = output[i]
}
return string(output)
}
``` | 1 | 0 | ['Go'] | 0 |
lexicographically-smallest-palindrome | Легчайшая | legchaishaia-by-danisdeveloper-zpte | Code | DanisDeveloper | NORMAL | 2025-01-03T11:18:39.656020+00:00 | 2025-01-03T11:18:39.656020+00:00 | 45 | false | # Code
```cpp []
class Solution {
public:
string makeSmallestPalindrome(string s) {
int n = s.length();
for(int i=0;i<n/2;++i){
char* left = &s[i];
char* right = &s[n - i - 1];
char diff = *right - *left;
if(diff == 0) continue;
if(diff > 0... | 1 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | Two pointers, in place. | two-pointers-in-place-by-michelusa-w6un | As we are ask for a palindrome, it is intuitive to track from
both:
start of string
and end of string
Because we are asked for minimum, apply minimum character | michelusa | NORMAL | 2024-12-29T23:56:14.704999+00:00 | 2024-12-29T23:56:14.704999+00:00 | 105 | false | As we are ask for a palindrome, it is intuitive to track from
both:
* start of string
* and end of string
Because we are asked for minimum, apply minimum character if the two compared elements are different.
O(N)
# Code
```cpp []
class Solution {
public:
string makeSmallestPalindrome(string s) {
for (siz... | 1 | 0 | ['Two Pointers', 'C++'] | 0 |
lexicographically-smallest-palindrome | Python Simple Code | python-simple-code-by-nagaraj08-4fxu | \n# Code\n\nclass Solution(object):\n def makeSmallestPalindrome(self, s):\n """\n :type s: str\n :rtype: str\n """\n i = | NAGARAJ08 | NORMAL | 2024-08-11T07:25:43.264159+00:00 | 2024-08-11T07:25:43.264188+00:00 | 263 | false | \n# Code\n```\nclass Solution(object):\n def makeSmallestPalindrome(self, s):\n """\n :type s: str\n :rtype: str\n """\n i = 0\n j = len(s)-1\n\n s = list(s)\n while i <j:\n\n if s[i]!=s[j]:\n if s[i]<s[j]:\n s[j] = ... | 1 | 0 | ['Python'] | 0 |
lexicographically-smallest-palindrome | Java | JavaScript | TypeScript | C++ | C# | Kotlin | Go Solution. | java-javascript-typescript-c-c-kotlin-go-o1va | Java []\npublic class Solution {\n\n public String makeSmallestPalindrome(String s) {\n char[] lexicographicallySmallestPalindrome = s.toCharArray();\ | LachezarTsK | NORMAL | 2024-07-15T16:32:12.168156+00:00 | 2024-07-15T16:37:25.007147+00:00 | 63 | false | ```Java []\npublic class Solution {\n\n public String makeSmallestPalindrome(String s) {\n char[] lexicographicallySmallestPalindrome = s.toCharArray();\n int left = 0;\n int right = lexicographicallySmallestPalindrome.length - 1;\n\n while (left < right) {\n if (lexicographica... | 1 | 0 | ['C++', 'Java', 'Go', 'TypeScript', 'Kotlin', 'JavaScript', 'C#'] | 0 |
lexicographically-smallest-palindrome | ✅ Easy C++ Solution | easy-c-solution-by-moheat-8gzk | Code\n\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int n = s.size();\n int left = 0;\n int right = n-1;\n | moheat | NORMAL | 2024-07-04T07:29:23.925560+00:00 | 2024-07-04T07:29:23.925595+00:00 | 225 | false | # Code\n```\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int n = s.size();\n int left = 0;\n int right = n-1;\n while(left < right)\n {\n if(s[left] != s[right])\n {\n char most = (s[left] < s[right]) ? s[left] : s[ri... | 1 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | [Python] O(N) using two pointers | python-on-using-two-pointers-by-pbelskiy-r01l | \nclass Solution:\n def makeSmallestPalindrome(self, s: str) -> str:\n a = list(s)\n\n left, right = 0, len(a) - 1\n\n while left < righ | pbelskiy | NORMAL | 2024-06-10T14:46:51.071769+00:00 | 2024-06-10T14:46:51.071807+00:00 | 40 | false | ```\nclass Solution:\n def makeSmallestPalindrome(self, s: str) -> str:\n a = list(s)\n\n left, right = 0, len(a) - 1\n\n while left < right:\n a[left] = a[right] = min(a[left], a[right])\n left += 1\n right -= 1\n \n return \'\'.join(a)\n``` | 1 | 0 | ['Two Pointers', 'Python'] | 0 |
lexicographically-smallest-palindrome | [Java] Easy 100% solution | java-easy-100-solution-by-ytchouar-zc45 | java\nclass Solution {\n public String makeSmallestPalindrome(final String s) {\n final char str[] = s.toCharArray();\n int i = 0, j = s.length | YTchouar | NORMAL | 2024-05-07T02:27:30.175283+00:00 | 2024-05-07T02:27:30.175310+00:00 | 228 | false | ```java\nclass Solution {\n public String makeSmallestPalindrome(final String s) {\n final char str[] = s.toCharArray();\n int i = 0, j = s.length() - 1;\n\n while(i < j) {\n str[i] = (char) Math.min(str[i], str[j]);\n str[j--] = str[i++];\n }\n\n return new S... | 1 | 0 | ['Java'] | 0 |
lexicographically-smallest-palindrome | Python solution, fast and intuitive | python-solution-fast-and-intuitive-by-se-kvm9 | 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 | Semako123 | NORMAL | 2024-03-26T19:00:00.441896+00:00 | 2024-03-26T19:00:00.441921+00:00 | 120 | 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 | ['Python3'] | 0 |
lexicographically-smallest-palindrome | Simple java code 5 ms beats 100 % | simple-java-code-5-ms-beats-100-by-arobh-obaf | \n# Complexity\n- \n\n# Code\n\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n int l = 0;\n int r = s.length() - 1;\n | Arobh | NORMAL | 2024-01-31T15:09:38.763123+00:00 | 2024-01-31T15:09:38.763153+00:00 | 8 | false | \n# Complexity\n- \n\n# Code\n```\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n int l = 0;\n int r = s.length() - 1;\n char[] arr = s.toCharArray();\n ... | 1 | 0 | ['Two Pointers', 'String', 'Java'] | 0 |
lexicographically-smallest-palindrome | java beats 100% | java-beats-100-by-susahin80-kw97 | Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\n\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n c | susahin80 | NORMAL | 2023-11-19T13:24:18.031494+00:00 | 2023-11-19T13:24:18.031518+00:00 | 56 | false | # Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\n```\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n char[] chars = s.toCharArray();\n int l = 0;\n int r = chars.length - 1;\n\n while (l < r) {\n int min = Math.min(chars[l], ... | 1 | 0 | ['Java'] | 0 |
lexicographically-smallest-palindrome | Super easy solution in JS. WOW. | super-easy-solution-in-js-wow-by-azamata-v05d | \n# Code\n\n/**\n * @param {string} s\n * @return {string}\n */\nvar makeSmallestPalindrome = function(s) {\n let str = s.split(\'\')\n for(let i = s.leng | AzamatAbduvohidov | NORMAL | 2023-08-12T05:15:02.754661+00:00 | 2023-08-12T05:15:02.754682+00:00 | 451 | false | \n# Code\n```\n/**\n * @param {string} s\n * @return {string}\n */\nvar makeSmallestPalindrome = function(s) {\n let str = s.split(\'\')\n for(let i = s.length - 1, j = 0; i >= Math.ceil(s.length / 2); i--, j++) {\n if(str[i] < str[j]) {\n str[j] = str[i]\n } else {\n str[i] = ... | 1 | 0 | ['JavaScript'] | 0 |
lexicographically-smallest-palindrome | C# easy | c-easy-by-ghmarek-qiug | Code\n\npublic class Solution {\n public string MakeSmallestPalindrome(string s) {\n \n char[] sArr = s.ToArray();\n\n for (int i = 0, j | GHMarek | NORMAL | 2023-06-27T15:01:03.500676+00:00 | 2023-06-27T15:01:03.500706+00:00 | 90 | false | # Code\n```\npublic class Solution {\n public string MakeSmallestPalindrome(string s) {\n \n char[] sArr = s.ToArray();\n\n for (int i = 0, j = s.Length - 1; j >= s.Length / 2; j--, i++)\n {\n if (s[i] != s[j])\n {\n if (s[i] < s[j])\n {... | 1 | 0 | ['C#'] | 0 |
lexicographically-smallest-palindrome | CPP || Easy Solution || 27 ms || 97%beats|| O(1) space || O(N) time | cpp-easy-solution-27-ms-97beats-o1-space-44ri | Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition is simple we have to make the string palindrome which is our first priori | aks1719 | NORMAL | 2023-05-30T17:03:35.279753+00:00 | 2023-05-30T17:04:08.777233+00:00 | 7 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition is simple we have to make the string palindrome which is our first priority so simply i can check that it is palindrom or not and second task is that it should be Lexicographically Smallest Palindrome so when the string pali... | 1 | 0 | ['String', 'C++'] | 0 |
lexicographically-smallest-palindrome | JAVA | One pass | java-one-pass-by-sourin_bruh-63e2 | Solution:\n\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n int n = s.length();\n StringBuilder sb = new StringBuilder() | sourin_bruh | NORMAL | 2023-05-23T17:32:25.872792+00:00 | 2023-05-23T17:32:25.872832+00:00 | 22 | false | # Solution:\n```\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n int n = s.length();\n StringBuilder sb = new StringBuilder();\n for (int i = 0; i < n; i++) {\n sb.append((char) Math.min(s.charAt(i), s.charAt(n - i - 1)));\n }\n return sb.toString(... | 1 | 0 | ['String', 'Java'] | 0 |
lexicographically-smallest-palindrome | python3; 214 ms; 16.5 MB | python3-214-ms-165-mb-by-nurmukhamed1-i7zh | 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 | Nurmukhamed1 | NORMAL | 2023-05-23T13:58:04.891049+00:00 | 2023-05-23T13:58:04.891097+00:00 | 858 | 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: 214 ms\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: 16.5 MB\n<!-- Add your space complexity here, e... | 1 | 0 | ['Python3'] | 1 |
lexicographically-smallest-palindrome | Replace with Minimum | replace-with-minimum-by-_kitish-n30f | Code\n\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int n = size(s);\n for(int i=0; i<=n/2; ++i){\n if(s | _kitish | NORMAL | 2023-05-21T19:58:03.065051+00:00 | 2023-05-21T19:58:03.065082+00:00 | 28 | false | # Code\n```\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int n = size(s);\n for(int i=0; i<=n/2; ++i){\n if(s[i] != s[n-i-1]){\n char e = min(s[i],s[n-i-1]);\n s[i] = e;\n s[n-i-1] = e;\n }\n }\n ... | 1 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | Two Pointer || Very easy to understand || Java | two-pointer-very-easy-to-understand-java-gvwl | \n# Approach\n Describe your approach to solving the problem. \n- Take two pointers, one at the 0th index and one at last index.\n- Check if the characters at b | sahilgupta8470 | NORMAL | 2023-05-21T10:45:40.904320+00:00 | 2023-05-21T10:45:40.904358+00:00 | 27 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Take two pointers, one at the 0th index and one at last index.\n- Check if the characters at both pointers are same or not.\n- If not same, then check which character is lexicographically smaller.\n- Change the bigger character to smaller characte... | 1 | 0 | ['Two Pointers', 'String', 'Java'] | 0 |
lexicographically-smallest-palindrome | C++ || EASY || SIMPLE | c-easy-simple-by-arya_ratan-1tfp | 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 | arya_ratan | NORMAL | 2023-05-21T09:36:39.540465+00:00 | 2023-05-21T09:36:39.540569+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)$$ --... | 1 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | Simple Easy Javascript Solution 100% beats 51.7 MB Memory | simple-easy-javascript-solution-100-beat-qz00 | Code\n\n/**\n * @param {string} s\n * @return {string}\n */\nvar makeSmallestPalindrome = function(s) {\nlet newString = ""\nfor(let i=0 ; i< s.length ; i ++){\ | dnt1997 | NORMAL | 2023-05-21T09:10:08.286742+00:00 | 2023-05-21T09:10:08.286784+00:00 | 154 | false | # Code\n```\n/**\n * @param {string} s\n * @return {string}\n */\nvar makeSmallestPalindrome = function(s) {\nlet newString = ""\nfor(let i=0 ; i< s.length ; i ++){\n let firstWord = s.charAt(i);\n let lastWord = s.charAt(s.length -i - 1);\n if(firstWord!==lastWord){\n if(firstWord.charCodeAt(0)>lastWor... | 1 | 0 | ['JavaScript'] | 0 |
lexicographically-smallest-palindrome | Clean PY | clean-py-by-bariscan97-btvj | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Bariscan97 | NORMAL | 2023-05-21T06:52:14.822036+00:00 | 2023-05-21T06:52:14.822076+00:00 | 559 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 1 | ['Array', 'String', 'String Matching', 'Python', 'Python3'] | 1 |
lexicographically-smallest-palindrome | c++ O(n) solution Two Pointers | c-on-solution-two-pointers-by-augus7-42o4 | \n Describe your first thoughts on how to solve this problem. \n\n# Approach: Two Pointers\n Describe your approach to solving the problem. \n\n# Complexity\n- | Augus7 | NORMAL | 2023-05-21T05:25:35.524682+00:00 | 2023-05-21T05:25:35.524721+00:00 | 192 | false | \n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach: Two Pointers\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$ \n<!-- Add your space complexity ... | 1 | 0 | ['Two Pointers', 'String', 'C++'] | 1 |
lexicographically-smallest-palindrome | easy and simple solution | easy-and-simple-solution-by-rahul_pinto-v0zm | \n# Complexity\n- Time complexity:$O(n/2)$\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:$O(1)$\n Add your space complexity here, e.g. O(n) | rahul_pinto | NORMAL | 2023-05-21T05:23:58.680280+00:00 | 2023-05-21T05:23:58.680332+00:00 | 9 | false | \n# Complexity\n- Time complexity:$O(n/2)$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$O(1)$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) \n {\n int first=0;\n int last=s.... | 1 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | Simple Solution || Easy To Understand | simple-solution-easy-to-understand-by-dc-8iez | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Dcoder_53 | NORMAL | 2023-05-21T04:30:53.975457+00:00 | 2023-05-25T07:07:36.858883+00:00 | 84 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:O(n)\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n ... | 1 | 0 | ['C++'] | 1 |
lexicographically-smallest-palindrome | Simple Two Pointer Solution | Java | simple-two-pointer-solution-java-by-raja-gvyl | Approach: Take two pointers one at 0 index other at last-1 at each iteration replace the large character with smaller one in stringbuilder return the string at | Rajat069 | NORMAL | 2023-05-21T04:24:59.268776+00:00 | 2023-05-21T04:24:59.268803+00:00 | 115 | false | **Approach:** Take two pointers one at 0 index other at last-1 at each iteration replace the large character with smaller one in stringbuilder return the string at last.\n```\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n StringBuilder sb = new StringBuilder(s);\n int f=0,l=s.le... | 1 | 0 | ['Java'] | 0 |
lexicographically-smallest-palindrome | Beats 100% Solutions | beats-100-solutions-by-deleted_user-vn95 | 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 | deleted_user | NORMAL | 2023-05-21T04:23:11.268618+00:00 | 2023-05-21T04:23:11.268661+00:00 | 171 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\n```\n/**\n * @param {string} s\n * @return {string}\n */\nvar makeSmallestPalindrome =... | 1 | 0 | ['JavaScript'] | 1 |
lexicographically-smallest-palindrome | Lexicographically Smallest Palindrome || C++ | lexicographically-smallest-palindrome-c-2feat | \n# Code\n\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int i=0;\n int j=s.size()-1;\n \n while(i<j){\n | krish_kakadiya | NORMAL | 2023-05-21T04:18:20.769805+00:00 | 2023-05-21T04:18:20.769847+00:00 | 367 | false | \n# Code\n```\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int i=0;\n int j=s.size()-1;\n \n while(i<j){\n if(s[i]!=s[j]){\n if(int(s[i])>int(s[j])) s[i]=s[j];\n else s[j]=s[i];\n }\n i++;\n ... | 1 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | Simple C++ Solution | simple-c-solution-by-roytanmay-5fia | Intuition\n Describe your first thoughts on how to solve this problem. \nTake the smallest from s[i] and s[n-i-1] for every index (0 <= i < n/2)\n\n# Complexit | roytanmay | NORMAL | 2023-05-21T04:13:16.982851+00:00 | 2023-05-21T04:13:16.982887+00:00 | 11 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTake the smallest from s[i] and s[n-i-1] for every index (0 <= i < n/2)\n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n\n# Code\n```\nclass Solution {\npublic:\n string makeSmall... | 1 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | very simple and easy to understand solution | very-simple-and-easy-to-understand-solut-1z2q | \n\n# Code\n\nclass Solution:\n def makeSmallestPalindrome(self, s: str) -> str:\n def helper(s) :\n if len(s) <2 :\n return | Vivek_Raj_ | NORMAL | 2023-05-21T04:09:49.544557+00:00 | 2023-05-21T04:09:49.544589+00:00 | 358 | false | \n\n# Code\n```\nclass Solution:\n def makeSmallestPalindrome(self, s: str) -> str:\n def helper(s) :\n if len(s) <2 :\n return s\n \n \n if ord(s[0]) <= ord(s[-1]) :\n return s[0] + helper(s[1:len(s)-1]) +s[0]\n else :\n ... | 1 | 1 | ['Python3'] | 1 |
lexicographically-smallest-palindrome | Check if you need to change character, if yes, change to smaller | check-if-you-need-to-change-character-if-1mkb | Intuition\nCheck for every position until the half of the string, check if elements are the same:\n - if yes, nothing to be done\n - if no, modify one of them b | salvadordali | NORMAL | 2023-05-21T04:04:33.281796+00:00 | 2023-05-21T04:24:38.147928+00:00 | 801 | false | # Intuition\nCheck for every position until the half of the string, check if elements are the same:\n - if yes, nothing to be done\n - if no, modify one of them by changing it to a smaller character\n\n\n\n# Complexity\n- Time complexity: $O(n)$\n- Space complexity: $O(n)$\n\n# Code\n\n```Rust []\nimpl Solution {\n pu... | 1 | 0 | ['Python3', 'Rust'] | 0 |
lexicographically-smallest-palindrome | Two pointer Approach only EASY TO UNDERSTAND | two-pointer-approach-only-easy-to-unders-oj0t | 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 | ajayadhikari | NORMAL | 2023-05-21T04:03:17.383166+00:00 | 2023-05-21T04:03:17.383225+00:00 | 20 | 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 | 1 | ['Python', 'C++', 'Java', 'Python3', 'MS SQL Server'] | 0 |
lexicographically-smallest-palindrome | Very simple and Easy Solution⚡🔥🔥 | very-simple-and-easy-solution-by-yashpad-gjaj | \n# Code\n\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int i=0;\n int j=s.size()-1;\n \n \n s | yashpadiyar4 | NORMAL | 2023-05-21T04:02:46.446588+00:00 | 2023-05-21T04:02:46.446623+00:00 | 149 | false | \n# Code\n```\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int i=0;\n int j=s.size()-1;\n \n \n string ans;\n while(i<j){\n if(s[i]!=s[j]){\n ans+=min(s[i],s[j]);\n }\n else{\n ans+=s[... | 1 | 0 | ['C++'] | 1 |
lexicographically-smallest-palindrome | Java || O(n) | java-on-by-nishant7372-4wkj | java []\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n int i=0;\n int j=s.length()-1;\n char[] arr = s.toCharArr | nishant7372 | NORMAL | 2023-05-21T04:02:41.640210+00:00 | 2023-05-21T04:02:41.640247+00:00 | 721 | false | ``` java []\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n int i=0;\n int j=s.length()-1;\n char[] arr = s.toCharArray();\n while(i<=j){\n if(arr[i]!=arr[j]){\n arr[i] = arr[j] = (char) Math.min(arr[i],arr[j]);\n }\n ... | 1 | 0 | ['Java'] | 0 |
lexicographically-smallest-palindrome | Java | java-by-neel_diyora-qgl8 | Code\n\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n char[] ch = s.toCharArray();\n int start = 0;\n int end = | anjan_diyora | NORMAL | 2023-05-21T04:02:33.206016+00:00 | 2023-05-21T04:02:33.206057+00:00 | 249 | false | # Code\n```\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n char[] ch = s.toCharArray();\n int start = 0;\n int end = s.length() - 1;\n \n while(start <= end) {\n if(ch[start] != ch[end]) {\n if(ch[end] > ch[start]) {\n ... | 1 | 0 | ['Java'] | 0 |
lexicographically-smallest-palindrome | 4 ms Beats 62.22% | 4-ms-beats-6222-by-bvc01654-n1z0 | IntuitionApproachComplexity
Time complexity:O(N)
Space complexity:O(1)
Code | bvc01654 | NORMAL | 2025-04-09T18:36:47.031966+00:00 | 2025-04-09T18:36:47.031966+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)$$ -->
- Space complexity:O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
... | 0 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | Lexicographically Smallest Palindrome | lexicographically-smallest-palindrome-by-fy9j | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | jyenduri | NORMAL | 2025-04-07T18:08:50.653600+00:00 | 2025-04-07T18:08:50.653600+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Go'] | 0 |
lexicographically-smallest-palindrome | Simple Java Code using 2 pointer | simple-java-code-using-2-pointer-by-kaly-o5k3 | IntuitionApproachComplexity
Time complexity: O(n)
Space complexity: O(n)
Code | kalyanram2053 | NORMAL | 2025-04-07T12:18:38.351341+00:00 | 2025-04-07T12:18:38.351341+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(n)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->... | 0 | 0 | ['Java'] | 0 |
lexicographically-smallest-palindrome | JAVA TWO POINTER APPROACH | java-two-pointer-approach-by-rohit-baba-cc2u | IntuitionApproachComplexity
Time complexity: O(n)
Space complexity: O(n)
Code | Rohit-Baba | NORMAL | 2025-04-05T06:39:22.363653+00:00 | 2025-04-05T06:39:22.363653+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(n)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->... | 0 | 0 | ['Two Pointers', 'Java'] | 0 |
lexicographically-smallest-palindrome | simple solution|| using two pointers | simple-solution-using-two-pointers-by-ko-aojd | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | kotla_jithendra | NORMAL | 2025-03-31T06:44:01.186389+00:00 | 2025-03-31T06:44:01.186389+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Two Pointers', 'String', 'Python3'] | 0 |
lexicographically-smallest-palindrome | 🔥✅✅ Dart Solution 📌📌 || with Explanation 👌👌 | dart-solution-with-explanation-by-nosar-435b | Solution
Two-Pointer Technique: We'll use two pointers starting from both ends of the string and move towards the center.
Character Comparison: At each step, we | NosaR | NORMAL | 2025-03-26T00:57:08.564786+00:00 | 2025-03-26T00:57:08.564786+00:00 | 2 | false | # Solution
1. **Two-Pointer Technique**: We'll use two pointers starting from both ends of the string and move towards the center.
2. **Character Comparison**: At each step, we compare the characters at both pointers:
- If they are different, we need to replace one of them to make them match.
- To ensure lexic... | 0 | 0 | ['Dart'] | 0 |
lexicographically-smallest-palindrome | Java, beats 83.93%, 6ms | java-beats-8393-6ms-by-nivet101-9muq | IntuitionIf we have two pointer moving inward from both ends when we encounter different characters, we choose the smaller of the two.Complexity
Time complexity | nivet101 | NORMAL | 2025-03-24T19:05:50.477591+00:00 | 2025-03-24T19:05:50.477591+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
If we have two pointer moving inward from both ends when we encounter different characters, we choose the smaller of the two.
# Complexity
- Time complexity:$$O(n)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity... | 0 | 0 | ['Java'] | 0 |
lexicographically-smallest-palindrome | MinChar solution | minchar-solution-by-ezpectus-i48a | What's a Palindrome?First, a palindrome is a word or phrase that reads the same forwards and backward. For example, "racecar" and "madam" are palindromes.The Pr | ezpectus | NORMAL | 2025-03-24T10:39:19.254762+00:00 | 2025-03-24T10:39:19.254762+00:00 | 2 | false |
What's a Palindrome?
First, a palindrome is a word or phrase that reads the same forwards and backward. For example, "racecar" and "madam" are palindromes.
The Problem:
You're given a string s. Your job is to change some of the characters in s so that it becomes a palindrome. But, you want to make the "smallest" pa... | 0 | 0 | ['Two Pointers', 'String', 'Greedy', 'C++', 'C#'] | 0 |
lexicographically-smallest-palindrome | JavaScript, beats 81.44%, 22ms | javascript-beats-8144-22ms-by-nivet101-g5st | IntuitionIf we have two pointer moving inward from both ends when we encounter different characters, we choose the smaller of the two.Complexity
Time complexity | nivet101 | NORMAL | 2025-03-24T02:32:54.977325+00:00 | 2025-03-24T02:32:54.977325+00:00 | 6 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
If we have two pointer moving inward from both ends when we encounter different characters, we choose the smaller of the two.
# Complexity
- Time complexity:$$O(n)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity... | 0 | 0 | ['JavaScript'] | 0 |
lexicographically-smallest-palindrome | Easy to understand solution in Java. Beats 84.11 % | easy-to-understand-solution-in-java-beat-qlok | Complexity
Time complexity:
O(n)
Space complexity:
O(n)
Code | Khamdam | NORMAL | 2025-03-21T15:42:33.499081+00:00 | 2025-03-21T15:42:33.499081+00:00 | 2 | false | # Complexity
- Time complexity:
O(n)
- Space complexity:
O(n)
# Code
```java []
class Solution {
public String makeSmallestPalindrome(String s) {
char[] chars = s.toCharArray();
int left = 0;
int right = chars.length - 1;
while (left < right) {
if (chars[left] < chars[r... | 0 | 0 | ['Two Pointers', 'String', 'Java'] | 0 |
lexicographically-smallest-palindrome | Optimized simple solution - beats 83.59%🔥 | optimized-simple-solution-beats-8359-by-n9lwm | IntuitionApproachComplexity
Time complexity: O(N)
Space complexity: O(N)
Code | cyrusjetson | NORMAL | 2025-03-18T11:15:51.510235+00:00 | 2025-03-18T11:15:51.510235+00:00 | 4 | 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)
- Space complexity: O(N)
# Code
```java []
class Solution {
public String makeSmallestPalindrome(String s) {
char[] t = ... | 0 | 0 | ['Java'] | 0 |
lexicographically-smallest-palindrome | C++ | 2 Pointer | c-2-pointer-by-kena7-lva2 | Code | kenA7 | NORMAL | 2025-03-17T13:40:04.263458+00:00 | 2025-03-17T13:40:04.263458+00:00 | 3 | false |
# Code
```cpp []
class Solution {
public:
string makeSmallestPalindrome(string s)
{
int i=0,j=s.size()-1;
while(i<j) {
if(s[i]!=s[j]) {
if(s[i]<s[j])
s[j]=s[i];
else
s[i]=s[j];
}
i++;
... | 0 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | C++ | Simple 0 ms 5 lines solution beats submitted solutions 100% in runtime and 90% in memory | c-simple-0-ms-5-lines-solution-beats-sub-qrwe | IntuitionReplace the twin characters with the smaller one if they are not equal.ApproachParse the string character by character till the mid of the string. Comp | thoka5pointsomeone | NORMAL | 2025-03-16T10:21:17.389066+00:00 | 2025-03-16T10:21:17.389066+00:00 | 3 | false | # Intuition
Replace the twin characters with the smaller one if they are not equal.
# Approach
Parse the string character by character till the mid of the string. Compare each character at a specific position from the beginning with the character at the same position from the end of the string. If the characters are n... | 0 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | Easy and Simple Solution to understand | easy-and-simple-solution-to-understand-b-jkgo | IntuitionA palindrome is a string that reads the same forward and backward. Our goal is to convert the given string into the lexicographically smallest palindro | gane0519 | NORMAL | 2025-03-15T09:39:02.376521+00:00 | 2025-03-15T09:39:02.376521+00:00 | 5 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
A palindrome is a string that reads the same forward and backward. Our goal is to convert the given string into the lexicographically smallest palindrome with minimal modifications.
To achieve this, we use a two-pointer approach:
1. Comp... | 0 | 0 | ['Two Pointers', 'String', 'Java'] | 0 |
lexicographically-smallest-palindrome | Simple straight forward two pointers | simple-straight-forward-two-pointers-by-hltto | Code | user9878Nx | NORMAL | 2025-03-13T17:18:37.468881+00:00 | 2025-03-13T17:18:37.468881+00:00 | 3 | false | # Code
```swift []
class Solution {
func makeSmallestPalindrome(_ s: String) -> String {
var chars = Array(s)
var i = 0
var j = chars.count - 1
while i < j {
if chars[i] < chars[j] {
chars[j] = chars[i]
} else {
... | 0 | 0 | ['Swift'] | 0 |
lexicographically-smallest-palindrome | byte performs better than rune | byte-performs-better-than-rune-by-wiiken-vqvd | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | wiikend | NORMAL | 2025-03-07T17:40:33.333889+00:00 | 2025-03-07T17:40:33.333889+00:00 | 4 | 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)$$ -->
# Code
`... | 0 | 0 | ['Go'] | 0 |
lexicographically-smallest-palindrome | easy to understand | easy-to-understand-by-budu123-p7fj | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | budu123 | NORMAL | 2025-03-04T20:07:52.547156+00:00 | 2025-03-04T20:07:52.547156+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | Easier way to solve this problem | easier-way-to-solve-this-problem-by-shia-lonv | ApproachFor all mirror-symmetric elements in the string, select the smaller one and use it as a replacement for both.
For example, given the string "abcd", we w | ShiaoC | NORMAL | 2025-03-02T04:06:03.263408+00:00 | 2025-03-02T04:06:03.263408+00:00 | 1 | false | # Approach
<!-- Describe your approach to solving the problem. -->
For all mirror-symmetric elements in the string, select the smaller one and use it as a replacement for both.
> For example, given the string "abcd", we will first compare 'a' and 'd', as they are closest to the beginning and end of the string. Since '... | 0 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | Lexicographically Smallest Palindrome Using Two-Pointer Approach | lexicographically-smallest-palindrome-us-hjd5 | IntuitionA palindrome is a string that reads the same forward and backward. To transform a given string into a lexicographically smallest palindrome, we need to | rabdya_767 | NORMAL | 2025-03-01T09:02:43.925248+00:00 | 2025-03-01T09:02:43.925248+00:00 | 5 | false | ---
# **Intuition**
A **palindrome** is a string that reads the same forward and backward. To transform a given string into a **lexicographically smallest palindrome**, we need to modify the characters while ensuring minimal changes.
A **lexicographically smaller** string means that, when comparing two characters... | 0 | 0 | ['JavaScript'] | 0 |
lexicographically-smallest-palindrome | Lexicographically Smallest Palindrome Transformation | lexicographically-smallest-palindrome-tr-6c8q | IntuitionThe goal is to transform the given string into the lexicographically smallest palindrome by modifying characters as needed. Since a palindrome reads th | rabdya_767 | NORMAL | 2025-03-01T08:59:02.136845+00:00 | 2025-03-01T08:59:02.136845+00:00 | 3 | false | ---
# **Intuition**
The goal is to transform the given string into the **lexicographically smallest palindrome** by modifying characters as needed. Since a palindrome reads the same forward and backward, we can compare corresponding characters from both ends and replace the larger one with the smaller one.
---
# *... | 0 | 0 | ['JavaScript'] | 0 |
lexicographically-smallest-palindrome | python code | python-code-by-vishalyadav28-pvlc | ApproachTwo pointer opposite directionComplexity
Time complexity:
O(N)
Space complexity:
O(N)
Code | vishalyadav28 | NORMAL | 2025-02-27T09:21:22.377650+00:00 | 2025-02-27T09:21:22.377650+00:00 | 5 | false | # Approach
Two pointer opposite direction
# Complexity
- Time complexity:
$$O(N)$$
- Space complexity:
$$O(N)$$
# Code
```python3 []
class Solution:
def makeSmallestPalindrome(self, s: str) -> str:
# sl = list(s)
# left=0
# right=len(sl)-1
# while left<right:
# if sl[le... | 0 | 0 | ['Two Pointers', 'String', 'Python3'] | 0 |
lexicographically-smallest-palindrome | java | java-by-shoreshan-7d7r | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | shoreshan | NORMAL | 2025-02-26T23:21:56.173821+00:00 | 2025-02-26T23:21:56.173821+00:00 | 2 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Java'] | 0 |
lexicographically-smallest-palindrome | 100% easy solution | 100-easy-solution-by-abhinav_gupta0007-8hf8 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Abhinav_Gupta0007 | NORMAL | 2025-02-23T06:07:57.062052+00:00 | 2025-02-23T06:07:57.062052+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | simple | simple-by-ryuji-lpde | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | ryuji | NORMAL | 2025-02-22T01:09:14.548209+00:00 | 2025-02-22T01:09:14.548209+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Rust'] | 0 |
lexicographically-smallest-palindrome | Very easy to understand-Two pointers | very-easy-to-understand-two-pointers-by-yrncg | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | lakshmikanth01 | NORMAL | 2025-02-21T20:15:42.548448+00:00 | 2025-02-21T20:15:42.548448+00:00 | 7 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Two Pointers', 'String', 'Python3'] | 0 |
lexicographically-smallest-palindrome | TRY TO YOUR BEST TO DO YOUR OWN:) | try-to-your-best-to-do-your-own-by-srina-zn11 | Code | Srinath_Y | NORMAL | 2025-02-20T07:39:29.399280+00:00 | 2025-02-20T07:39:29.399280+00:00 | 2 | false |
# Code
```python []
class Solution(object):
def makeSmallestPalindrome(self, s):
s=list(s)
n=len(s)
for i in range(n/2):
if s[i]!=s[n-1-i]:
if ord(s[i])<ord(s[n-1-i]):
s[n-1-i]=s[i]
else:
s[i]=s[n-1-i]
... | 0 | 0 | ['String', 'Python'] | 0 |
count-number-of-nice-subarrays | [Java/C++/Python] Sliding Window, O(1) Space | javacpython-sliding-window-o1-space-by-l-6tpq | Solution 1: atMost\nHave you read this? 992. Subarrays with K Different Integers\nExactly K times = at most K times - at most K - 1 times\n\n\n# Complexity\nTim | lee215 | NORMAL | 2019-11-03T04:04:38.512895+00:00 | 2020-06-15T07:40:04.379933+00:00 | 92,408 | false | # **Solution 1: atMost**\nHave you read this? [992. Subarrays with K Different Integers](https://leetcode.com/problems/subarrays-with-k-different-integers/discuss/523136/JavaC%2B%2BPython-Sliding-Window)\nExactly `K` times = at most `K` times - at most `K - 1` times\n<br>\n\n# **Complexity**\nTime `O(N)` for one pass\n... | 590 | 10 | [] | 64 |
count-number-of-nice-subarrays | C++: Visual explanation. O(1) space. Two pointers | c-visual-explanation-o1-space-two-pointe-xbcd | Algorithm\nThe best way to explain this approach is to look at the example:\nLet\'s say k = 2 and we have the following array:\n\n\n\nWe only going to work with | andnik | NORMAL | 2020-02-13T21:38:11.823200+00:00 | 2020-02-19T06:59:46.412712+00:00 | 25,796 | false | ##### Algorithm\nThe best way to explain this approach is to look at the example:\nLet\'s say **k = 2** and we have the following array:\n\n\n\nWe only going to work with numbers `1` and `2` because we are only interested in if number is odd or even... | 455 | 3 | ['C'] | 33 |
count-number-of-nice-subarrays | Deque with Picture | deque-with-picture-by-votrubac-p9dz | Intuition\nThis is similar to 992. Subarrays with K Different Integers.\n\nWhen we find k odd numbers, we have one nice subarray, plus an additional subarray fo | votrubac | NORMAL | 2019-11-03T05:38:38.885686+00:00 | 2022-08-04T15:58:59.519948+00:00 | 17,815 | false | #### Intuition\nThis is similar to [992. Subarrays with K Different Integers](https://leetcode.com/problems/subarrays-with-k-different-integers/discuss/235235/C%2B%2BJava-with-picture-prefixed-sliding-window).\n\nWhen we find `k` odd numbers, we have one nice subarray, plus an additional subarray for each even number p... | 210 | 5 | ['C', 'Java'] | 25 |
count-number-of-nice-subarrays | ✅Beats 100% -Explained with [ Video ] -C++/Java/Python/JS -Prefix Sum -Interview Solution | beats-100-explained-with-video-cjavapyth-7ett | \n\n# YouTube Video Explanation:\n\n### To watch the video please click on "Watch On Youtube" option available the left bottom corner of the thumbnail.\n\n **I | lancertech6 | NORMAL | 2024-06-22T00:56:21.994495+00:00 | 2024-06-22T12:41:41.216007+00:00 | 33,583 | false | \n\n# YouTube Video Explanation:\n\n### To watch the video please click on "Watch On Youtube" option available the left bottom corner of the thumbnail.\n\n<!-- **If you want a video ... | 161 | 2 | ['Array', 'Hash Table', 'Math', 'Sliding Window', 'Python', 'C++', 'Java', 'JavaScript'] | 14 |
count-number-of-nice-subarrays | Subarray Sum Equals K | subarray-sum-equals-k-by-herbert5812-a5a7 | If you transform the input array into binary, then the problem becomes the \'Subarray Sum Equals K\' problem. You can think of k odd numbers means sum of then i | herbert5812 | NORMAL | 2019-11-03T04:37:21.643479+00:00 | 2019-11-03T04:54:11.287397+00:00 | 10,762 | false | If you transform the input array into binary, then the problem becomes the \'Subarray Sum Equals K\' problem. You can think of k odd numbers means sum of then is k.\n\n```\nclass Solution {\npublic:\n int numberOfSubarrays(vector<int>& nums, int k) {\n unordered_map<int, int> m;\n const int n = nums.si... | 151 | 4 | ['C'] | 16 |
count-number-of-nice-subarrays | [Java] PrefixSum 1pass 10line 7ms | java-prefixsum-1pass-10line-7ms-by-wangy-nsuo | At index i, if current odd numbers from the beginning is M,\nand we checked there was N previous index with (M - K) oddnum, then we got N subarrays\nres += N\n\ | wangyxwyx | NORMAL | 2019-11-03T04:53:28.283787+00:00 | 2019-11-03T05:12:53.656412+00:00 | 6,804 | false | At index i, if current odd numbers from the beginning is M,\nand we checked there was N previous index with (M - K) oddnum, then we got N subarrays\nres += N\n\n\n```\n public int numberOfSubarrays(int[] nums, int k) {\n int cur = 0, ans = 0;\n Map<Integer, Integer> map = new HashMap<>();\n map.... | 80 | 1 | [] | 11 |
count-number-of-nice-subarrays | [Prefix Sum & Sliding Window Tutorial] Count Number of Nice Subarrays | prefix-sum-sliding-window-tutorial-count-vll6 | This post includes the solution for both using Prexis Sum and Sliding Window.\n\nTopic : Prefix Sum\n\n### Prefix Sum:\nPrefix sum, also known as cumulative su | never_get_piped | NORMAL | 2024-06-22T00:20:57.777541+00:00 | 2024-06-27T07:11:51.997435+00:00 | 14,886 | false | **This post includes the solution for both using Prexis Sum and Sliding Window.**\n\n**Topic** : Prefix Sum\n\n### Prefix Sum:\nPrefix sum, also known as cumulative sum, is a technique used in computer science and mathematics to efficiently calculate the sum of a sequence of numbers. The prefix sum of a sequence `a[0]... | 53 | 1 | ['Array', 'Hash Table', 'C', 'PHP', 'Prefix Sum', 'Java', 'Go', 'Python3', 'JavaScript'] | 15 |
count-number-of-nice-subarrays | [Java/Python 3] 1 pass Sliding Window O(n) time O(1) space w/ brief explanation. | javapython-3-1-pass-sliding-window-on-ti-sy4m | Whenever the count of odd numbers reach k, for each high boundary of the sliding window, we have indexOfLeftMostOddInWin - lowBound options for the low boundary | rock | NORMAL | 2019-11-03T04:01:23.903614+00:00 | 2020-10-17T12:33:12.091462+00:00 | 7,952 | false | 1. Whenever the count of odd numbers reach `k`, for each high boundary of the sliding window, we have `indexOfLeftMostOddInWin - lowBound` options for the low boundary, where `indexOfLeftMostOddInWin` is the index of the leftmost odd number within the window, and `lowBound` is the index of the low boundary exclusively;... | 48 | 4 | [] | 10 |
count-number-of-nice-subarrays | Python - Two pointer | python-two-pointer-by-harshhx-kyjr | \nclass Solution:\n def numberOfSubarrays(self, nums: List[int], k: int) -> int:\n right ,left = 0,0\n ans = 0 \n odd_cnt = 0\n a | harshhx | NORMAL | 2021-06-11T19:01:12.047999+00:00 | 2021-06-11T19:01:12.048029+00:00 | 5,163 | false | ```\nclass Solution:\n def numberOfSubarrays(self, nums: List[int], k: int) -> int:\n right ,left = 0,0\n ans = 0 \n odd_cnt = 0\n ans = 0\n cur_sub_cnt = 0\n for right in range(len(nums)):\n \n if nums[right]%2 == 1:\n odd_cnt += 1\n ... | 41 | 0 | ['Two Pointers', 'Python', 'Python3'] | 9 |
count-number-of-nice-subarrays | easy peasy python solution with explanation | easy-peasy-python-solution-with-explanat-x0e6 | \t# Just keep count of the current odd number.\n\t# Look in the dictionary if we can find (currendOds - k), \n\t# if it exisits that means I can get an subarray | lostworld21 | NORMAL | 2019-11-03T18:31:51.327405+00:00 | 2019-11-04T15:36:06.568313+00:00 | 5,465 | false | \t# Just keep count of the current odd number.\n\t# Look in the dictionary if we can find (currendOds - k), \n\t# if it exisits that means I can get an subarray with k odds.\n\t# Also keep count of number of different types of odds too,\n\t# because for K =1 , [2,2,1] is a valid list, so does, [2,1] and [1].\n\t\n ... | 36 | 1 | ['Python', 'Python3'] | 8 |
count-number-of-nice-subarrays | C++ || Two Pointers || Clean Code | c-two-pointers-clean-code-by-jennifer-kf21 | \nclass Solution {\npublic:\n int numberOfSubarrays(vector<int>& nums, int k) {\n int n = nums.size();\n int ans = 0,odd = 0,cnt = 0;\n | Jennifer__ | NORMAL | 2022-08-25T10:50:59.088068+00:00 | 2022-08-25T10:50:59.088100+00:00 | 4,258 | false | ```\nclass Solution {\npublic:\n int numberOfSubarrays(vector<int>& nums, int k) {\n int n = nums.size();\n int ans = 0,odd = 0,cnt = 0;\n int l = 0,r = 0;\n while(r<n)\n {\n if(nums[r]%2 != 0)\n {\n odd++;\n cnt = 0;\n ... | 32 | 0 | ['Two Pointers', 'C', 'Sliding Window'] | 3 |
count-number-of-nice-subarrays | JAVA | Sliding Window | java-sliding-window-by-aman0786khan-2dt8 | \nclass Solution {\n public int numberOfSubarrays(int[] nums, int k) {\n int oddcount=0;\n int res=0;\n int i=0;\n int count=0;\n | aman0786khan | NORMAL | 2021-07-04T10:09:41.445090+00:00 | 2021-07-04T10:09:41.445122+00:00 | 2,351 | false | ```\nclass Solution {\n public int numberOfSubarrays(int[] nums, int k) {\n int oddcount=0;\n int res=0;\n int i=0;\n int count=0;\n for(int j=0;j<nums.length;j++){\n if(nums[j]%2==1){\n oddcount++;\n count=0;\n }\n whi... | 30 | 0 | [] | 4 |
count-number-of-nice-subarrays | java || sliding window || two pinter | java-sliding-window-two-pinter-by-ankurj-3bhh | \nclass Solution {\n public int numberOfSubarrays(int[] nums, int k) {\n int i=0;\n int j=0;\n int oddCount=0;\n int count=0;\n | ANKURJANA-1 | NORMAL | 2022-08-29T13:06:11.568062+00:00 | 2022-08-29T13:06:11.568103+00:00 | 4,309 | false | ```\nclass Solution {\n public int numberOfSubarrays(int[] nums, int k) {\n int i=0;\n int j=0;\n int oddCount=0;\n int count=0;\n int temp=0;\n \n while(j<nums.length){\n if(nums[j]%2==1){\n oddCount++;\n temp=0;\n ... | 27 | 0 | ['Sliding Window', 'Java'] | 6 |
count-number-of-nice-subarrays | Prefix sum+Sliding window vs at most k odds||35ms Beats 99.99% | prefix-sumsliding-window-vs-at-most-k-od-bz7i | Intuition\n Describe your first thoughts on how to solve this problem. \nThe concept is to use sliding window. With the help of prefix sum, it made an acceptibl | anwendeng | NORMAL | 2024-06-22T02:09:04.411473+00:00 | 2024-06-22T06:00:05.777780+00:00 | 4,366 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe concept is to use sliding window. With the help of prefix sum, it made an acceptible solution.\n\n2nd approach usse at most k odds argument which is applied to solve the hard question [992. Subarrays with K Different Integers](https:/... | 26 | 5 | ['Array', 'Sliding Window', 'Prefix Sum', 'C++'] | 5 |
count-number-of-nice-subarrays | Detailed Explanation | One Pass, O(1) Space - C++, Python, Java | detailed-explanation-one-pass-o1-space-c-odm5 | Approach\n Describe your approach to solving the problem. \nOur goal is to count the number of sub arrays that have exactly k odd numbers. Let\'s take a look at | not_yl3 | NORMAL | 2024-06-22T00:31:44.387493+00:00 | 2024-06-22T02:18:27.734937+00:00 | 6,024 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nOur goal is to count the number of sub arrays that have exactly `k` odd numbers. Let\'s take a look at Example 3 where `k = 2` (but I added an extra 1 to the end):\n\n`nums = [2,2,2,1,2,2,1,2,2,2,1]`\n\nOur first nice subarray starts at index 7 since ... | 26 | 0 | ['Array', 'Math', 'C', 'Sliding Window', 'C++', 'Java', 'Python3'] | 7 |
count-number-of-nice-subarrays | C++ || 3 APPROACHES || SOLUTION | c-3-approaches-solution-by-_himanshu_12-b977 | \nclass Solution {\npublic:\n int numberOfSubarrays(vector<int>& nums, int k) {\n int i=0;\n int j=0;\n int count=0;\n int result | _himanshu_12 | NORMAL | 2022-09-11T06:48:10.429934+00:00 | 2022-09-11T06:48:10.429968+00:00 | 5,382 | false | ```\nclass Solution {\npublic:\n int numberOfSubarrays(vector<int>& nums, int k) {\n int i=0;\n int j=0;\n int count=0;\n int result=0;\n int first_occur=0;\n vector<int>occur(nums.size(),0);\n int index=0;\n int current=0;\n // Variable window size prob... | 26 | 2 | ['C', 'Sliding Window', 'Prefix Sum', 'C++'] | 5 |
count-number-of-nice-subarrays | C++| 🚀 ✅ Sliding Window | 🚀 ✅ With Explaination | 🚀 ✅ Easy to Understand | c-sliding-window-with-explaination-easy-kuve5 | Intuition:\r\nThe problem requires finding the number of subarrays having exactly k odd integers. We can solve the problem using sliding window technique where | devanshupatel | NORMAL | 2023-04-11T15:12:40.847989+00:00 | 2023-04-14T14:15:44.875906+00:00 | 6,576 | false | # Intuition:\r\nThe problem requires finding the number of subarrays having exactly k odd integers. We can solve the problem using sliding window technique where we maintain a window of contiguous subarray and slide it from left to right. While sliding the window, we keep track of the number of odd integers inside the ... | 24 | 0 | ['Sliding Window', 'C++'] | 4 |
count-number-of-nice-subarrays | Easy C++ Solution || T.C = O(N) || Sliding Window | easy-c-solution-tc-on-sliding-window-by-k1yie | Intuition\r\nTo solve the problem of counting the number of subarrays with exactly k odd numbers, we can use a sliding window approach. This method allows us to | kumar_kshitij | NORMAL | 2024-06-22T00:07:55.227272+00:00 | 2024-06-22T00:07:55.227288+00:00 | 5,418 | false | # Intuition\r\nTo solve the problem of counting the number of subarrays with exactly `k` odd numbers, we can use a sliding window approach. This method allows us to efficiently manage the window of elements in the array and count the number of valid subarrays without needing to recompute for each possible subarray.\r\n... | 21 | 0 | ['Sliding Window', 'C++'] | 6 |
count-number-of-nice-subarrays | C++ Sliding Window Solution O(1) Space | c-sliding-window-solution-o1-space-by-ro-oj38 | Similar to https://leetcode.com/problems/subarrays-with-k-different-integers/\n\npublic:\n int numarr(vector<int>&nums,int k){\n int ans=0;\n i | rom111 | NORMAL | 2020-09-02T09:28:59.954373+00:00 | 2020-09-02T09:28:59.954413+00:00 | 3,036 | false | Similar to https://leetcode.com/problems/subarrays-with-k-different-integers/\n```\npublic:\n int numarr(vector<int>&nums,int k){\n int ans=0;\n int count=0;\n int i=0,j=0,n=nums.size();\n while(j<n){\n if(nums[j]%2==1){\n count++;\n }\n if(... | 21 | 1 | [] | 2 |
count-number-of-nice-subarrays | nice subarrays || c++ || easy | nice-subarrays-c-easy-by-sheetaljoshi-nkhd | //exactly similar to the problem:subarrays sum equal to given sum(k)\n// we just need to convert odds with 1 and even with 0\n//and find the given sub arrays ha | sheetaljoshi | NORMAL | 2021-11-28T06:52:27.358079+00:00 | 2021-11-28T07:23:36.312548+00:00 | 2,051 | false | //exactly similar to the problem:subarrays sum equal to given sum(k)\n// we just need to convert odds with 1 and even with 0\n//and find the given sub arrays having sum k\nUPVOTE IF YOU GET:)\n```\nclass Solution {\npublic:\n \n int numberOfSubarrays(vector<int>& a, int k) {\n int ans=0,sum=0,n=a.size();\n... | 17 | 1 | ['C', 'C++'] | 4 |
count-number-of-nice-subarrays | C++ Prefix State Map / Two Pointers / Sliding Window | c-prefix-state-map-two-pointers-sliding-31ara | See my latest update in repo LeetCode\n\n## Solution 1. Prefix State Map\n\nUse a map m to store the mapping from the count of odd numbers cnt to the first inde | lzl124631x | NORMAL | 2021-10-11T04:34:14.057512+00:00 | 2021-10-11T04:34:14.057565+00:00 | 2,491 | false | See my latest update in repo [LeetCode](https://github.com/lzl124631x/LeetCode)\n\n## Solution 1. Prefix State Map\n\nUse a map `m` to store the mapping from the count of odd numbers `cnt` to the first index in the array that has `cnt` numbers in front of it and including itself.\n\nWhen `cnt >= k`, we add `m[cnt - k +... | 17 | 0 | [] | 1 |
count-number-of-nice-subarrays | Python Solution Prefix Sum | python-solution-prefix-sum-by-yuyingji-xy2h | \ndef numberOfSubarrays(self, nums, k):\n """\n :type nums: List[int]\n :type k: int\n :rtype: int\n """\n res = 0\n | yuyingji | NORMAL | 2019-11-03T04:03:16.509603+00:00 | 2019-11-03T18:38:39.962951+00:00 | 3,257 | false | ```\ndef numberOfSubarrays(self, nums, k):\n """\n :type nums: List[int]\n :type k: int\n :rtype: int\n """\n res = 0\n count = 0\n prefix = {}\n prefix[0] = 1\n for i in range(len(nums)):\n \n if nums[i] % 2 != 0:\n ... | 17 | 2 | [] | 7 |
count-number-of-nice-subarrays | 💹Easy 3 Approaches 🔰📊|| Beats 98% || 🚀 Hashing and Two Pointers 🎯|| In Depth | easy-3-approaches-beats-98-hashing-and-t-mf1e | \r\n# Intuition:\r\nThe problem is essentially about finding subarrays with exactly \'k\' odd numbers. \r\nWe can use a hashmap to count the occurrences of pref | laggerk | NORMAL | 2024-06-22T07:13:49.400576+00:00 | 2024-06-22T07:13:49.400621+00:00 | 2,836 | false | \r\n# Intuition:\r\nThe problem is essentially about finding subarrays with exactly \'k\' odd numbers. \r\nWe can use a hashmap to count the occurrences of prefix sums, which helps us keep track \r\nof the number of odd numbers encountered so far.\r\n \r\n# Approach 1:\r\n1. Initialize a hashmap to store the count of p... | 16 | 1 | ['Array', 'Hash Table', 'Math', 'Sliding Window', 'Python', 'C++', 'Java', 'Python3'] | 3 |
count-number-of-nice-subarrays | JAVA || Picture + Detail Explanation || prefix sum + HashMap || Easy Solution | java-picture-detail-explanation-prefix-s-gpyl | \n\n In this problem , First we will change the odd digit from 1 and even with zero in given array.\n then we use the same approach which we used in the leetcod | ayushx | NORMAL | 2021-10-20T20:45:56.983692+00:00 | 2021-11-02T18:23:57.601451+00:00 | 2,988 | false | \n\n* In this problem , First we will change the odd digit from 1 and even with zero in given array.\n* then we use the same approach which we used in the leetcode problem 560. If you want detail explanation th... | 15 | 2 | ['Sliding Window', 'Prefix Sum', 'Java'] | 5 |
count-number-of-nice-subarrays | Java Solution | java-solution-by-credit_card-lcab | Replace all odd numbers with 1 and even with zeroes\n\nNow the problem becomes\nhttps://leetcode.com/problems/subarray-sum-equals-k/\n\nFind number of sub arrys | credit_card | NORMAL | 2020-09-06T14:15:24.362308+00:00 | 2020-10-04T14:53:12.707737+00:00 | 1,295 | false | Replace all odd numbers with 1 and even with zeroes\n\nNow the problem becomes\n[https://leetcode.com/problems/subarray-sum-equals-k/](https://leetcode.com/problems/subarray-sum-equals-k/)\n\nFind number of sub arrys with sum = K\n```\npublic int numberOfSubarrays(int[] nums, int k) {\n //Replace all odd by 1 a... | 15 | 0 | [] | 3 |
count-number-of-nice-subarrays | ✅ One Line Solution | one-line-solution-by-mikposp-gv5a | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)Code #1.1 - One LineTime complexity: O(n). Space com | MikPosp | NORMAL | 2024-06-22T09:18:33.601994+00:00 | 2025-02-11T10:52:26.574263+00:00 | 1,323 | false | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)
# Code #1.1 - One Line
Time complexity: $$O(n)$$. Space complexity: $$O(n)$$.
```
class Solution:
def numberOfSubarrays(self, a: List[int], k: int) -> int:
return (z:=Counter([q:=0])) and sum(z.upda... | 14 | 0 | ['Array', 'Hash Table', 'Math', 'Prefix Sum', 'Python', 'Python3'] | 1 |
count-number-of-nice-subarrays | Sliding window | sliding-window-by-anil-budamakuntla-kvga | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ANIL-BUDAMAKUNTLA | NORMAL | 2024-06-22T00:33:52.290783+00:00 | 2024-06-22T00:33:52.290808+00:00 | 1,237 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:o(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:o(1)\n<!-- Add your space complexity here, e.g. $$O... | 14 | 0 | ['Array', 'Sliding Window', 'C++'] | 5 |
count-number-of-nice-subarrays | C++ Sliding Window Solution + Probability Rule O(N) | c-sliding-window-solution-probability-ru-v1vg | Runtime: 112 ms, faster than 93.17% of C++ online submissions for Count Number of Nice Subarrays.\nMemory Usage: 67.5 MB, less than 91.98% of C++ online submiss | ahsan83 | NORMAL | 2021-07-11T15:57:15.687238+00:00 | 2021-07-11T17:26:57.770515+00:00 | 1,880 | false | Runtime: 112 ms, faster than 93.17% of C++ online submissions for Count Number of Nice Subarrays.\nMemory Usage: 67.5 MB, less than 91.98% of C++ online submissions for Count Number of Nice Subarrays.\n\n```\nInorder to solve the problem we can easily find the main subarray containing K odd numbers using Sliding\nWindo... | 14 | 0 | ['Array', 'C', 'Sliding Window', 'Probability and Statistics'] | 3 |
count-number-of-nice-subarrays | C++ : Sliding Window || Two pointers || O(n) Space | c-sliding-window-two-pointers-on-space-b-n2j4 | This question is slightly based on sliding window, however one needs proper idea of whats going on.\nWe actually just need to maintain the condition of (Oddnumb | nivedita_chatterjee_021 | NORMAL | 2022-02-13T14:44:46.111529+00:00 | 2022-02-13T14:44:46.111556+00:00 | 2,206 | false | This question is slightly based on sliding window, however one needs proper idea of whats going on.\nWe actually just need to maintain the condition of (Oddnumbers==k), once we get it equal.\nTill the time our odd number count is less than k, we just simply keep updating **"end"**.\nOnce our oddnum count is equal to k,... | 13 | 0 | ['Two Pointers', 'C', 'Sliding Window', 'C++'] | 2 |
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