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minimum-window-substring | ✅☑Beats 97% Users || [C++/Java/Python/JavaScript] || EXPLAINED🔥 | beats-97-users-cjavapythonjavascript-exp-g9af | PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n---\n\n\n# Approaches\n(Also explained in the code)\n\n1. Problem Objective:\n\n - The goal is to find the minimum win | MarkSPhilip31 | NORMAL | 2024-02-04T00:35:17.748122+00:00 | 2024-02-04T00:35:17.748153+00:00 | 33,043 | false | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n---\n\n\n# Approaches\n(Also explained in the code)\n\n1. **Problem Objective:**\n\n - The goal is to find the minimum window in string `s`... | 99 | 1 | ['Hash Table', 'String', 'Sliding Window', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 11 |
minimum-window-substring | [C++] Short Sliding Window Solution with Explanation | c-short-sliding-window-solution-with-exp-dg4w | \n string minWindow(string s, string t) {\n unordered_map<char, int> letters; //unordered map for storing the characters in t that we need to check fo | hunarbatra | NORMAL | 2020-01-17T18:09:43.980793+00:00 | 2020-01-17T18:09:43.980828+00:00 | 20,266 | false | ```\n string minWindow(string s, string t) {\n unordered_map<char, int> letters; //unordered map for storing the characters in t that we need to check for in s\n for(auto c : t) letters[c]++; \n int count = 0; //counts number of t\'s letters in current window\n int low = 0, min_length = I... | 97 | 8 | ['C', 'Sliding Window', 'C++'] | 15 |
minimum-window-substring | Python | My advice | python-my-advice-by-dev-josh-wp3w | My advice for solving this problem is to:\n Understand the intuition and what to do at a high level\n Try to implement your own solution WITHOUT copying anyone | dev-josh | NORMAL | 2021-02-01T16:21:32.655771+00:00 | 2021-02-01T16:21:32.655817+00:00 | 6,294 | false | My advice for solving this problem is to:\n* Understand the intuition and what to do at a high level\n* Try to implement your own solution WITHOUT copying anyone elses\n* This is how you will learn\n* You will remember high level concepts, but never line for line code\n\nIntuition:\n* Two pointers, left and right\n* Bo... | 76 | 0 | ['Python', 'Python3'] | 12 |
minimum-window-substring | ✅🔥Sliding Window Approach with Explanation - C++/Java/Python | sliding-window-approach-with-explanation-tgop | \n# Intuition\nIn this problem, we need to find the minimum window substring of string s that contains all characters from string t. We can use a sliding window | dhruba-datta | NORMAL | 2022-03-28T15:40:08.454696+00:00 | 2023-12-04T17:10:34.912935+00:00 | 10,133 | false | \n# Intuition\nIn this problem, we need to find the minimum window substring of string `s` that contains all characters from string `t`. We can use a sliding window approach to find the minimum window substring efficiently.\n\n# Approach 01\n1. Create an unordered_map `mp` to store the count of characters in string `t`... | 75 | 1 | ['Two Pointers', 'Sliding Window', 'Python', 'C++', 'Java', 'Python3'] | 10 |
minimum-window-substring | Share my neat java solution | share-my-neat-java-solution-by-tlj77-kd5z | public String minWindow(String S, String T) {\n if(S==null||S.isEmpty()||T==null||T.isEmpty()) return "";\n int i=0, j=0;\n int[] Tmap=new | tlj77 | NORMAL | 2015-04-22T02:58:02+00:00 | 2018-09-29T18:05:17.067477+00:00 | 33,284 | false | public String minWindow(String S, String T) {\n if(S==null||S.isEmpty()||T==null||T.isEmpty()) return "";\n int i=0, j=0;\n int[] Tmap=new int[256];\n int[] Smap=new int[256];\n for(int k=0; k< T.length(); k++){\n Tmap[T.charAt(k)]++;\n }\n int found=0;\n ... | 72 | 8 | ['Java'] | 11 |
minimum-window-substring | C++ Solution 68ms explanation | c-solution-68ms-explanation-by-leodicap9-a275 | \nWe are going to use a two pointer approach to solve this.\n\nThe idea here is that \n1. We will store the characters of t in a map lets say mapt.\n2. We will | leodicap99 | NORMAL | 2020-05-07T05:31:08.846128+00:00 | 2020-05-07T05:31:08.846160+00:00 | 6,558 | false | ```\nWe are going to use a two pointer approach to solve this.\n\nThe idea here is that \n1. We will store the characters of t in a map lets say mapt.\n2. We will have two pointers l and r.\n3. Whille we traverse through s we check if the character is found in mapt If so we will store the character into another hash ma... | 53 | 2 | ['C', 'C++'] | 10 |
minimum-window-substring | Simple Java Solution | Sliding Window | simple-java-solution-sliding-window-by-g-f3s1 | Kindly upvote, if it helps you!\n```\n public String minWindow(String s, String t) {\n String result = "";\n if(s.length() < t.length())\n | guptap151 | NORMAL | 2022-01-26T14:28:19.416146+00:00 | 2022-01-26T18:06:32.834936+00:00 | 4,902 | false | Kindly upvote, if it helps you!\n```\n public String minWindow(String s, String t) {\n String result = "";\n if(s.length() < t.length())\n return result;\n int minWindow = Integer.MAX_VALUE;\n\t\t//We will use two variables \'have\' & \'need\' to keep a track whether the characters \n\t\... | 45 | 1 | ['Sliding Window', 'Java'] | 6 |
minimum-window-substring | Java 4ms bit 97.6% | java-4ms-bit-976-by-huang593-liwd | Basically, there are two pointers for windows sliding. One for exploiting new matched substring, other pointer for expiring previous substring.\n\n public St | huang593 | NORMAL | 2016-03-30T03:10:24+00:00 | 2018-10-02T07:05:13.712462+00:00 | 12,670 | false | Basically, there are two pointers for windows sliding. One for exploiting new matched substring, other pointer for expiring previous substring.\n\n public String minWindow(String s, String t) {\n char[] s_array = s.toCharArray();\n char[] t_array = t.toCharArray();\n int[] map = new ... | 45 | 1 | [] | 4 |
minimum-window-substring | Sliding Window | Beats 100% | Java | Python | C++ | JS | Go | Rust | sliding-window-beats-100-java-python-c-j-57n6 | \n\n### INTUITION\n\n\nWe\u2019re given two strings: s, the large string, and t, the smaller one. Our job is to find the smallest substring of s that contains a | kartikdevsharma_ | NORMAL | 2024-07-16T10:30:58.919549+00:00 | 2024-09-11T11:45:39.812527+00:00 | 7,533 | false | \n\n### INTUITION\n\n\nWe\u2019re given two strings: `s`, the large string, and `t`, the smaller one. Our job is to find the **smallest substring** of `s` that contains **all the characters** of `t`, and crucially, we need to consider duplicates. So, if `t` contains two `A`s, then any substring we pick from `s` needs t... | 42 | 0 | ['Hash Table', 'String', 'Sliding Window', 'C++', 'Java', 'Go', 'Python3', 'Rust', 'JavaScript'] | 6 |
minimum-window-substring | C++/JAVA/Python | 🚀 ✅ Sliding Window | ✅ Fully Explained | ✅ Hash Table | ✅ String | cjavapython-sliding-window-fully-explain-ex0g | Intuition:\nThe problem asks to find the minimum window in s that contains all the characters of t. One way to approach this problem is to use a sliding window | devanshupatel | NORMAL | 2023-04-14T18:05:53.116403+00:00 | 2023-04-24T20:23:28.926976+00:00 | 9,064 | false | # Intuition:\nThe problem asks to find the minimum window in s that contains all the characters of t. One way to approach this problem is to use a sliding window technique. We can maintain a window that starts from the beginning of s and moves forward until it contains all the characters of t. Once we have such a windo... | 38 | 0 | ['Hash Table', 'Sliding Window', 'Python', 'C++', 'Java'] | 4 |
minimum-window-substring | c++ || Advance || Sliding window || Fast | c-advance-sliding-window-fast-by-iam_sin-ckn3 | Please upvote, your one upvote makes me happy\n\n^____^\n\nHere is the code\n\n\nclass Solution {\npublic:\n\n string minWindow(string str, string pat) {\n | Iam_SinghSunny | NORMAL | 2022-10-22T02:50:23.038629+00:00 | 2023-05-13T03:02:30.827669+00:00 | 6,945 | false | **Please upvote, your one upvote makes me happy**\n\n^____^\n\n**Here is the code**\n\n```\nclass Solution {\npublic:\n\n string minWindow(string str, string pat) {\n \n int len1 = str.length();\n int len2 = pat.length();\n \n const int no_of_chars = 256;\n\n if (len1 < len2) {\n ... | 38 | 1 | ['C', 'Sliding Window'] | 3 |
minimum-window-substring | Simple Python Solution Beats 99% with Detailed Explanation | simple-python-solution-beats-99-with-det-hp5w | The key idea is how you update the tracking variables. There are four essentials variables to track: 1. remaing length of a match 2. The position of previous ma | Brownian_motion | NORMAL | 2018-09-15T20:12:14.414718+00:00 | 2018-10-09T05:15:57.640374+00:00 | 5,924 | false | The key idea is how you update the tracking variables. There are four essentials variables to track: 1. remaing length of a match 2. The position of previous matched first element 3. start_position of returned answer 4. end_position of returned answer. \n\nAnd a dictionary to count the occurrence of characters is usefu... | 36 | 0 | [] | 2 |
minimum-window-substring | Java | TC: O(S+T) | SC: O(T) | Space-optimized Sliding Window using Two Pointers | java-tc-ost-sc-ot-space-optimized-slidin-kzl2 | java\n/**\n * Space-optimized Sliding Window using Two Pointers\n *\n * Time Complexity: O(S + T)\n *\n * Space Complexity: O(T)\n *\n * S = length of String s. | NarutoBaryonMode | NORMAL | 2021-10-01T06:46:46.340722+00:00 | 2021-10-07T07:56:11.751803+00:00 | 5,710 | false | ```java\n/**\n * Space-optimized Sliding Window using Two Pointers\n *\n * Time Complexity: O(S + T)\n *\n * Space Complexity: O(T)\n *\n * S = length of String s. T = length of String t\n */\nclass Solution1 {\n public String minWindow(String s, String t) {\n if (s == null || t == null) {\n throw ... | 35 | 0 | ['Two Pointers', 'String', 'Sliding Window', 'Java'] | 3 |
minimum-window-substring | [Python3] Sliding window O(N+M) | python3-sliding-window-onm-by-yourick-us52 | \n# Approach\nThis problem follows the Sliding Window pattern and has a lot of similarities with 567 Permutation in a String with one difference. In this proble | yourick | NORMAL | 2023-11-09T17:11:24.818215+00:00 | 2024-06-27T19:59:25.410276+00:00 | 2,608 | false | \n# Approach\nThis problem follows the Sliding Window pattern and has a lot of similarities with [567 Permutation in a String](https://leetcode.com/problems/permutation-in-string/description/) with one difference. In this problem, we need to find a substring having all characters of the pattern which means that the req... | 34 | 0 | ['Sliding Window', 'Python', 'Python3'] | 5 |
minimum-window-substring | Three O(N) concise implemetation according to leetcode oj discuss | three-on-concise-implemetation-according-ci60 | // according to http://leetcode.com/2010/11/finding-minimum-window-in-s-which.html\n // finds the first window that satisfies the constraint\n // then con | shichaotan | NORMAL | 2014-12-31T19:13:25+00:00 | 2014-12-31T19:13:25+00:00 | 9,563 | false | // according to http://leetcode.com/2010/11/finding-minimum-window-in-s-which.html\n // finds the first window that satisfies the constraint\n // then continue maintaining the constraint throughout\n // time complexity O(2N)\n string minWindow(string S, string T) {\n int m = S.size(), n = T.size(... | 33 | 3 | [] | 2 |
minimum-window-substring | 💡JavaScript Solution w/ Detailed Comments | javascript-solution-w-detailed-comments-jxrzs | javascript\nvar minWindowSlidingWindow = function (s, t) {\n\t// `right` is -1 since every loop, we start by expanding the right boundary\n\t// setting this to | aminick | NORMAL | 2019-10-23T02:27:27.114250+00:00 | 2019-10-23T02:45:04.825380+00:00 | 6,620 | false | ``` javascript\nvar minWindowSlidingWindow = function (s, t) {\n\t// `right` is -1 since every loop, we start by expanding the right boundary\n\t// setting this to -1 ensures that we will check the first char on the first time\n let min = "", left = 0, right = -1;\n let map = {};\n\t\n\t// this creates a map for ... | 32 | 0 | ['JavaScript'] | 6 |
minimum-window-substring | Accepted Python solution using hashtable | accepted-python-solution-using-hashtable-0lno | class Solution:\n # @return a string\n def minWindow(self, S, T):\n indices = {}\n for char in T:\n indices[c | xiaoying10101 | NORMAL | 2015-01-03T04:14:51+00:00 | 2018-10-26T04:13:16.498421+00:00 | 9,739 | false | class Solution:\n # @return a string\n def minWindow(self, S, T):\n indices = {}\n for char in T:\n indices[char] = []\n miss = list(T)\n start = 0\n end = len(S)\n for i in range(len(S)):\n if S[i] in T:\n... | 31 | 1 | [] | 4 |
minimum-window-substring | Python easy to understand | python-easy-to-understand-by-clarketm-pgo7 | python\ndef min_window(S: str, T: str) -> str:\n """\n Minimum Window Substring\n\n :param str S:\n :param str T:\n :return str:\n """\n Tc | clarketm | NORMAL | 2019-05-05T01:41:24.979072+00:00 | 2019-05-05T01:41:24.979115+00:00 | 5,634 | false | ```python\ndef min_window(S: str, T: str) -> str:\n """\n Minimum Window Substring\n\n :param str S:\n :param str T:\n :return str:\n """\n Tc = Counter(T)\n Sc = Counter()\n\n best_i = -sys.maxsize\n best_j = sys.maxsize\n\n i = 0\n\n for j, char in enumerate(S):\n Sc[char] +... | 29 | 2 | ['Python'] | 9 |
minimum-window-substring | EXPLAINED | Easy to understand code with comments | 3ms | explained-easy-to-understand-code-with-c-s72w | \nclass Solution {\n public String minWindow(String s, String t) {\n int[] count = new int[128];\n\n // Count the characters in t\n for | heisenbergknocks | NORMAL | 2020-11-13T08:02:12.571412+00:00 | 2022-10-25T08:54:03.558312+00:00 | 2,799 | false | ```\nclass Solution {\n public String minWindow(String s, String t) {\n int[] count = new int[128];\n\n // Count the characters in t\n for (char ch : t.toCharArray()) count[ch]++;\n\n char[] sourceStr = s.toCharArray();\n String windowString = "";\n int windowLeft = 0, windo... | 28 | 0 | ['Sliding Window', 'Java'] | 7 |
minimum-window-substring | [Python] O(n+m) sliding window, explained | python-onm-sliding-window-explained-by-d-0lmr | The idea of sliding window with 2 pointers: we create counter cnt_t is frequencies how many time s we need to take each symbol, for example for abca we have a:2 | dbabichev | NORMAL | 2021-08-15T07:52:22.792580+00:00 | 2021-08-15T07:52:22.792624+00:00 | 1,442 | false | The idea of sliding window with 2 pointers: we create counter `cnt_t` is frequencies how many time s we need to take each symbol, for example for `abca` we have `a:2, b:1, c:1`. We create also `cnt_s` as empty counter, which will keep information about frequencies is current window `[beg, end)` - not that we do not inc... | 27 | 3 | ['Sliding Window'] | 3 |
minimum-window-substring | Super Easy Java Solution or or 100% faster or or Easy to understand | super-easy-java-solution-or-or-100-faste-63p3 | Looking for Contribution in Hacktoberfest\n## You are welcomed to contribute in my Repos:-\n# GITHUB LINK --> Yaduttam95\n# All PRs are getting accepted...\n\n# | Yaduttam_Pareek | NORMAL | 2022-10-22T01:41:11.734701+00:00 | 2022-10-22T01:41:11.734744+00:00 | 2,732 | false | # Looking for Contribution in Hacktoberfest\n## You are welcomed to contribute in my Repos:-\n# GITHUB LINK --> [Yaduttam95](https://github.com/Yaduttam95)\n# All PRs are getting accepted...\n\n# Please upvote if Helpful\n```\nclass Solution {\n public String minWindow(String s, String t) {\n \n \n ... | 26 | 0 | [] | 3 |
minimum-window-substring | ✅ Minimum Window Substring || Using Map w/ Explanation || C++ | Python | Java | minimum-window-substring-using-map-w-exp-4tlx | IDEA\nThe solution is a bit intuitive. We keep expanding the window by moving the right pointer. When the window has all the desired characters, we contract (if | Maango16 | NORMAL | 2021-08-15T07:58:11.255661+00:00 | 2021-08-15T07:58:11.255715+00:00 | 1,917 | false | **IDEA**\nThe solution is a bit intuitive. We keep expanding the window by moving the right pointer. When the window has all the desired characters, we contract (if possible) and save the smallest window till now.\n`The answer is the smallest desirable window.`\n\n**EXAMPLE**\nFor eg. `S = "ABAACBAB" T = "ABC"`. Then o... | 26 | 2 | [] | 9 |
minimum-window-substring | Sliding Windows||Hash tables->Freq Array||0ms Beats 100% | sliding-windowshash-tables-freq-array0ms-4tes | Intuition\n Describe your first thoughts on how to solve this problem. \n2 kinds of solutions. the main idea is the the sliding windows. but for implementation | anwendeng | NORMAL | 2024-02-04T00:46:24.152403+00:00 | 2024-02-04T06:54:34.247401+00:00 | 6,243 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n2 kinds of solutions. the main idea is the the sliding windows. but for implementation it needs the info of frequencies of alphabets.\n1 is just using hash tables (map or unordered_map) which is slow & done many months ago.\nOther is usi... | 24 | 0 | ['Array', 'Hash Table', 'String', 'Bit Manipulation', 'C', 'Sliding Window', 'C++', 'Python3'] | 8 |
minimum-window-substring | A visualized first principles approach for better understanding | a-visualized-first-principles-approach-f-xfis | Heres a visualization techinque that might help you get a better idea of the solution. \n\nImagine the strings (henceforth S and T) as a long chain of character | principled_man | NORMAL | 2020-07-20T22:58:48.675684+00:00 | 2020-07-21T16:42:00.081153+00:00 | 1,141 | false | Heres a visualization techinque that might help you get a better idea of the solution. \n\nImagine the strings (henceforth **S** and **T**) as a long chain of characters\n\nFirst we will build a tool that will help us extract information about the target string **T**\n\nWe will extract 2 pieces of info from **T** - \nf... | 24 | 0 | ['Go'] | 3 |
minimum-window-substring | Here is my C++ code with line by line Explanation as comments | here-is-my-c-code-with-line-by-line-expl-7zrv | Please Upvote if you find it useful\n\n\nstring minWindow(string s, string t) {\n \n //This result variable will store the string which we will return\n | yvrjprshr | NORMAL | 2021-10-15T07:24:10.931891+00:00 | 2021-10-15T07:24:10.931931+00:00 | 3,128 | false | Please **Upvote** if you find it useful\n\n```\nstring minWindow(string s, string t) {\n \n //This result variable will store the string which we will return\n string result;\n\n //It will check if any of the two string is empty and return empty string result\n if(s.empty() || t.empty()){\n return... | 22 | 1 | ['Two Pointers', 'C', 'Sliding Window'] | 2 |
minimum-window-substring | ✅ One Line Solution | one-line-solution-by-mikposp-ptcs | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)\n\n# Code #1.1 - Oneliner\nTime complexity: O(n). S | MikPosp | NORMAL | 2024-02-04T12:00:50.308006+00:00 | 2024-02-04T21:23:19.893242+00:00 | 4,474 | false | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)\n\n# Code #1.1 - Oneliner\nTime complexity: $$O(n)$$. Space complexity: $$O(n)$$.\n```\nclass Solution:\n def minWindow(self, s: str, t: str) -> str:\n return (g:={\'s\':s,\'t\':t,\'a\':0,\'c\':Counter... | 21 | 0 | ['Hash Table', 'String', 'Sliding Window', 'Counting', 'Python', 'Python3'] | 10 |
minimum-window-substring | O(N) JAVA Sliding Window solution with explanation | on-java-sliding-window-solution-with-exp-t6ci | Sliding Window Solution:\n\n 1) Spread the right pointer until it satisfies the requirement \n 2) Shrink the left pointer to get the minimum range \n | cheng_zhang | NORMAL | 2015-10-06T03:26:07+00:00 | 2015-10-06T03:26:07+00:00 | 6,765 | false | **Sliding Window Solution:**\n\n 1) Spread the right pointer until it satisfies the requirement \n 2) Shrink the left pointer to get the minimum range \n 3) Keep the above steps.\n\n**Time complexity = O(2n) = O(n)**\n\nThere're 2 loops: for loop of i, while loop of j. As j only steps forward, never steps bac... | 21 | 4 | [] | 7 |
minimum-window-substring | [JavaScript] Simple Sliding Window + Hash Map Solution | javascript-simple-sliding-window-hash-ma-7r40 | Time Complexity: O(S + T) where S and T are the respective lengths of strings s and t\nSpace Complexity: O(S + T)\n\nIntuition\n\nThis is actually very similar | ryangdev | NORMAL | 2022-07-18T23:48:27.856387+00:00 | 2022-07-18T23:48:27.856420+00:00 | 2,733 | false | Time Complexity: O(S + T) where S and T are the respective lengths of strings `s` and `t`\nSpace Complexity: O(S + T)\n\n**Intuition**\n\nThis is actually very similar to the [Permutation in String](https://leetcode.com/problems/permutation-in-string/) problem. The difference is that we don\'t match based on string le... | 20 | 0 | ['Sliding Window', 'JavaScript'] | 3 |
minimum-window-substring | Java 100% 2ms | Clean Code w/ Video Explanation | java-100-2ms-clean-code-w-video-explanat-j6g2 | Please Upvote if you find the explanation helpful\n\nVideo Explanation\nMinimum Window Substring | YouTube\n\nJava Solution\n\n//2ms\n\nclass Solution {\n pu | sagnik_20 | NORMAL | 2022-10-22T02:37:39.148739+00:00 | 2022-10-22T02:37:39.148777+00:00 | 2,844 | false | *Please **Upvote** if you find the explanation helpful*\n\n**Video Explanation**\n[Minimum Window Substring | YouTube](https://www.youtube.com/watch?v=vVSEJuN6BHA&feature=youtu.be)\n\n**Java Solution**\n```\n//2ms\n\nclass Solution {\n public String minWindow(String s, String t) {\n int [] map = new int[128];... | 18 | 1 | ['Two Pointers', 'Sliding Window', 'Java'] | 6 |
minimum-window-substring | easy peasy python [comments] solution | easy-peasy-python-comments-solution-by-l-ieto | \tdef minWindow(self, s: str, t: str) -> str:\n ln_s = len(s)\n ln_t = len(t)\n if ln_s == 0 or ln_t == 0 or ln_t > ln_s:\n retu | lostworld21 | NORMAL | 2019-09-25T03:17:23.733874+00:00 | 2019-09-25T03:17:23.733931+00:00 | 5,029 | false | \tdef minWindow(self, s: str, t: str) -> str:\n ln_s = len(s)\n ln_t = len(t)\n if ln_s == 0 or ln_t == 0 or ln_t > ln_s:\n return ""\n dct = {}\n for ch in t:\n dct[ch] = dct.get(ch, 0) + 1\n \n i = j = 0\n minWindow = ln_s + 1\n outp... | 18 | 2 | ['Sliding Window', 'Python', 'Python3'] | 2 |
minimum-window-substring | Java | Beats 100 % | java-beats-100-by-beingbmc12-r2l1 | java\nclass Solution {\n public String minWindow(String s, String t) {\n if(s.isEmpty()) return "";\n int[] need = new int[128];\n for(c | beingbmc12 | NORMAL | 2019-06-09T07:06:31.863273+00:00 | 2019-06-09T07:06:31.863306+00:00 | 1,488 | false | ```java\nclass Solution {\n public String minWindow(String s, String t) {\n if(s.isEmpty()) return "";\n int[] need = new int[128];\n for(char c : t.toCharArray()) need[c]++;\n char[] a = s.toCharArray();\n int r = 0, l = 0, missing = t.length(), i = 0, j = 0;\n while(r < s.... | 18 | 0 | [] | 5 |
minimum-window-substring | Python, easy to understand, no weird tricks | python-easy-to-understand-no-weird-trick-rww4 | \nimport collections\nclass Solution(object):\n def minWindow(self, s, t):\n """\n :type s: str\n :type t: str\n :rtype: str\n | frankli92 | NORMAL | 2019-03-26T20:56:14.265147+00:00 | 2019-03-26T20:56:14.265204+00:00 | 1,086 | false | ```\nimport collections\nclass Solution(object):\n def minWindow(self, s, t):\n """\n :type s: str\n :type t: str\n :rtype: str\n """\n if len(s) < len(t):\n return ""\n \n hashmap = collections.Counter(t)\n counter = len(t)\n min_windo... | 18 | 0 | [] | 5 |
minimum-window-substring | Simple JavaScript solution with comments | simple-javascript-solution-with-comments-ssxg | function minWindow(s, t) {\n var ans = '';\n \n // 1. process hashmap\n var map = {};\n t.split('').forEach(ch => map[ch] = ( | linfongi | NORMAL | 2016-05-16T03:39:09+00:00 | 2016-05-16T03:39:09+00:00 | 4,167 | false | function minWindow(s, t) {\n var ans = '';\n \n // 1. process hashmap\n var map = {};\n t.split('').forEach(ch => map[ch] = (map[ch] || 0) + 1);\n var count = Object.keys(map).length;\n \n // 2. traverse s to find boundaries\n // both l & r are inclusiv... | 18 | 0 | [] | 9 |
minimum-window-substring | C++ - Easiest Beginner Friendly Sol || Sliding Window | c-easiest-beginner-friendly-sol-sliding-f1qv2 | Intuition of this Problem:\nReference - https://leetcode.com/problems/minimum-window-substring/solutions/26808/here-is-a-10-line-template-that-can-solve-most-su | singhabhinash | NORMAL | 2023-02-05T09:21:51.345562+00:00 | 2023-02-05T09:21:51.345594+00:00 | 2,501 | false | # Intuition of this Problem:\nReference - https://leetcode.com/problems/minimum-window-substring/solutions/26808/here-is-a-10-line-template-that-can-solve-most-substring-problems/?orderBy=hot\n<!-- Describe your first thoughts on how to solve this problem. -->\n**NOTE - PLEASE READ APPROACH FIRST THEN SEE THE CODE. YOU... | 17 | 0 | ['Hash Table', 'String', 'Sliding Window', 'C++'] | 2 |
minimum-window-substring | [C++] [Sliding Window] - Simple and easy to understand with explanation | O(n) time complexity | c-sliding-window-simple-and-easy-to-unde-p3vv | Make two frequency count array and store the frequency of t initially\n2. If character match with the characters of t, increment the count variable\n3. if count | morning_coder | NORMAL | 2020-10-11T09:21:10.403538+00:00 | 2020-10-11T09:25:22.376006+00:00 | 4,905 | false | 1. Make two frequency count array and store the frequency of t initially\n2. If character match with the characters of t, increment the count variable\n3. if count==length of t (t_len) => we found a window with all characters of t\n4. Minimize the window size by removing extra characters from the start of window (using... | 17 | 2 | ['C', 'Sliding Window', 'C++'] | 4 |
minimum-window-substring | C++ || Sliding Window || Map || String || Easy Explanation || Simple | c-sliding-window-map-string-easy-explana-iewz | Approach \n- Sliding Window Technique!!\n- Map for storing Frequency\n\n# Complexity\n- Time complexity: O(m+n)\n Add your time complexity here, e.g. O(n) \n\n- | yashgaherwar2002 | NORMAL | 2022-10-22T05:39:05.234493+00:00 | 2022-10-22T05:39:05.234538+00:00 | 2,541 | false | # Approach \n- Sliding Window Technique!!\n- Map for storing Frequency\n\n# Complexity\n- Time complexity: O(m+n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n\n // Sliding Window ... | 15 | 0 | ['String', 'Ordered Map', 'Sliding Window', 'C++'] | 3 |
minimum-window-substring | [Python] O(N) and O(T) with explanation | python-on-and-ot-with-explanation-by-unk-n3b0 | \nTime Complexity: O(N)\nSpace Complexity: O(T)\n\n\n\nfrom collections import Counter\nclass Solution:\n def minWindow(self, s: str, t: str) -> str:\n\t\t# | Unkn0wnPerson | NORMAL | 2021-01-08T06:43:17.396055+00:00 | 2021-01-08T08:07:58.167722+00:00 | 1,693 | false | ```\nTime Complexity: O(N)\nSpace Complexity: O(T)\n```\n\n```\nfrom collections import Counter\nclass Solution:\n def minWindow(self, s: str, t: str) -> str:\n\t\t# Count number of occurrences of each character from target string.\n t=Counter(t) \n\t\t\n #This will store number of unique characters ... | 15 | 0 | ['Sliding Window', 'Python3'] | 2 |
minimum-window-substring | ⭐✅ LOGICAL COMMENTS || EASY C++ CODE || Including logical comments | logical-comments-easy-c-code-including-l-9odz | Easy to understand code with complete comments:\n\n\nclass Solution {\npublic:\n string minWindow(string s, string t) {\n unordered_map<char, int>mp;\ | SachinSahu | NORMAL | 2022-06-22T05:24:18.500542+00:00 | 2022-11-08T13:54:19.178848+00:00 | 909 | false | # Easy to understand code with complete comments:\n\n```\nclass Solution {\npublic:\n string minWindow(string s, string t) {\n unordered_map<char, int>mp;\n \n // keeping track of common char b/w s ans t\n for(auto i : t)\n mp[i]++;\n \n int start = 0,end =0; //... | 14 | 0 | ['Two Pointers', 'C', 'Sliding Window', 'C++'] | 1 |
minimum-window-substring | C++ 😊 ADITYA VERMA SOLUTION 💖 | c-aditya-verma-solution-by-sayan_11_mait-zoqn | \nclass Solution\n{\npublic:\n string minWindow(string s, string t)\n {\n unordered_map<char, int> mp;\n for (int i = 0; i < t.size(); i++)\ | sayan_11_maitra | NORMAL | 2022-01-14T16:41:51.422347+00:00 | 2022-09-17T10:08:58.417616+00:00 | 999 | false | ```\nclass Solution\n{\npublic:\n string minWindow(string s, string t)\n {\n unordered_map<char, int> mp;\n for (int i = 0; i < t.size(); i++)\n {\n mp[t[i]]++;\n }\n int count = mp.size();\n string ans;\n\n int mini = INT_MAX;\n\n int i = 0;\n ... | 14 | 1 | ['C'] | 4 |
minimum-window-substring | C++ - 2 approaches for hashmap - way too much commenting and explanation | c-2-approaches-for-hashmap-way-too-much-9ptw2 | The idea of sliding window is this:\n1. Start two pointers - begin and end, both at 0.\n2. Keep begin constant. Move end (i.e., do end++) till the condition is | kiyosaanb | NORMAL | 2020-09-14T17:34:46.595961+00:00 | 2020-09-14T17:35:25.490167+00:00 | 1,932 | false | **The idea of sliding window is this:**\n1. Start two pointers - begin and end, both at 0.\n2. Keep begin constant. Move end (i.e., do end++) till the condition is satisfied.\n3. Once the condition is satisfied, move the begin pointer (do begin++) for as long as the condition is still satisfied. Update the minLength an... | 14 | 1 | ['C', 'C++'] | 5 |
minimum-window-substring | O( n )✅ | C++ (Step by step explanation)✅ | o-n-c-step-by-step-explanation-by-monste-o5g2 | Intuition\nThe problem requires finding the minimum window in string \'s\' that contains all characters from string \'t\'. We can use a sliding window approach | monster0Freason | NORMAL | 2023-10-27T17:43:52.444678+00:00 | 2023-10-27T17:43:52.444710+00:00 | 1,673 | false | # Intuition\nThe problem requires finding the minimum window in string \'s\' that contains all characters from string \'t\'. We can use a sliding window approach to solve this problem efficiently.\n\n# Approach\n1. First, we handle the edge case: if \'t\' is an empty string, we return an empty string as the result.\n\n... | 13 | 0 | ['Sliding Window', 'C++'] | 1 |
minimum-window-substring | ✅ [Python/Rust/C++] fast & concise using two-pointer sliding window (with detailed comments) | pythonrustc-fast-concise-using-two-point-3djq | \u2705 IF YOU LIKE THIS SOLUTION, PLEASE UPVOTE.\n*\nThis solution employs a two-pointer sliding window approach to search for the minimum substring. Time compl | stanislav-iablokov | NORMAL | 2022-10-22T11:49:45.391794+00:00 | 2022-10-23T11:15:05.703121+00:00 | 1,639 | false | **\u2705 IF YOU LIKE THIS SOLUTION, PLEASE UPVOTE.**\n****\nThis solution employs a two-pointer sliding window approach to search for the minimum substring. Time complexity is linear: **O(n+m)**. Space complexity is linear: **O(n)**. \n\n| Language | Runtime | Memory |\n|---|---|---|\n| [**Python**](https://leetcode.co... | 13 | 1 | ['C', 'Python', 'Rust'] | 1 |
minimum-window-substring | JavaScript - 96.66 % better - used char Array instead if Hash Map | javascript-9666-better-used-char-array-i-u0sl | ```\nvar minWindow = function(s, t) {\n \n let arr = new Array(128).fill(0); // Ascii charSet array to store count\n let result = [-Infinity, Infinity] | varungowde | NORMAL | 2020-03-28T07:50:01.342088+00:00 | 2020-03-28T07:50:01.342126+00:00 | 3,396 | false | ```\nvar minWindow = function(s, t) {\n \n let arr = new Array(128).fill(0); // Ascii charSet array to store count\n let result = [-Infinity, Infinity] // result not yet known\n let missing = t.length; // missing words initially\n \n for(let i=0; i < t.length; i++){ // increase the count in arr\n ... | 13 | 0 | ['Array', 'Sliding Window', 'JavaScript'] | 2 |
minimum-window-substring | Simple Python two-pointer solution | simple-python-two-pointer-solution-by-ot-on6e | Please see and vote for my solutions for these similar problems.\n1208. Get Equal Substrings Within Budget\n3. Longest Substring Without Repeating Characters\n1 | otoc | NORMAL | 2019-07-27T07:03:54.464836+00:00 | 2019-09-29T04:19:38.236871+00:00 | 1,345 | false | Please see and vote for my solutions for these similar problems.\n[1208. Get Equal Substrings Within Budget](https://leetcode.com/problems/get-equal-substrings-within-budget/discuss/392901/Simple-Python-moving-window)\n[3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-w... | 13 | 0 | [] | 1 |
minimum-window-substring | Simple 2 pointer approach || Beats 100% 🔥🔥 || Python & CPP || T.C. : O(n) & S.C. : O(1) | simple-2-pointer-approach-beats-100-pyth-l7z8 | IntuitionThe problem is about finding the smallest substring of s that contains all the characters of string t (including duplicates). The task can be efficient | devilshiv_07 | NORMAL | 2025-01-25T19:03:52.111077+00:00 | 2025-02-08T11:42:12.211457+00:00 | 2,060 | false | 
---
# Intuition
The problem is about finding the smallest substring of s that contains all the characters of string t (including duplicates). The task can be efficiently solved usi... | 12 | 0 | ['Hash Table', 'Two Pointers', 'Sliding Window', 'C++', 'Python3'] | 1 |
minimum-window-substring | C++ - Easiest Beginner Friendly Sol || Sliding window | c-easiest-beginner-friendly-sol-sliding-t54i5 | Intuition of this Problem:\nReference - https://leetcode.com/problems/minimum-window-substring/solutions/26808/here-is-a-10-line-template-that-can-solve-most-su | singhabhinash | NORMAL | 2023-02-04T13:34:44.302376+00:00 | 2023-02-04T13:34:44.302403+00:00 | 2,963 | false | # Intuition of this Problem:\nReference - https://leetcode.com/problems/minimum-window-substring/solutions/26808/here-is-a-10-line-template-that-can-solve-most-substring-problems/?orderBy=hot\n<!-- Describe your first thoughts on how to solve this problem. -->\n**NOTE - PLEASE READ APPROACH FIRST THEN SEE THE CODE. YOU... | 12 | 0 | ['Hash Table', 'Sliding Window', 'C++', 'Java', 'Python3'] | 3 |
minimum-window-substring | Simple c++ solution | Sliding Window | Hashmap | Explained clearly with comments | simple-c-solution-sliding-window-hashmap-wz1x | \n\n string minWindow(string s, string t) {\n // Initialize minl as INT_MAX\n int minl = INT_MAX;\n \n // Map to keep count of all c | maitreyeepaliwal | NORMAL | 2021-07-24T19:49:59.753700+00:00 | 2021-07-24T19:49:59.753745+00:00 | 1,329 | false | \n```\n string minWindow(string s, string t) {\n // Initialize minl as INT_MAX\n int minl = INT_MAX;\n \n // Map to keep count of all characters of t \n unordered_map <int, int> mp;\n for(auto ch: t) mp[ch]++;\n \n // Sliding Window Approach\n // Let c be t... | 12 | 0 | ['Two Pointers', 'C', 'Sliding Window'] | 1 |
minimum-window-substring | Sliding Window Thinking Process | sliding-window-thinking-process-by-grace-4ti5 | For example\n\ne.g. S = "ADOBECODEBANC", T = "ABC"\n ADOBEC \n\t BECODEBA \n CODEBA\n BANC\nThe substrings abov | gracemeng | NORMAL | 2018-06-23T13:13:00.025631+00:00 | 2018-06-23T13:13:00.025631+00:00 | 1,064 | false | > For example\n```\ne.g. S = "ADOBECODEBANC", T = "ABC"\n ADOBEC \n\t BECODEBA \n CODEBA\n BANC\nThe substrings above are candidates for the result.\n```\n> In Brute Force, \n```\nfor left in [0: sLen - 1]\n for right in [left*: sLen - 1]\n if (canCover()) \n\t upda... | 12 | 1 | [] | 3 |
minimum-window-substring | My 12ms simple C++ code (O(1) space, O(N) time) | my-12ms-simple-c-code-o1-space-on-time-b-rslz | Just used an array dict to count the occurence of the letters in t. To distinguish the letters that are not in t, we initialize dict with -slen and for those le | lejas | NORMAL | 2015-08-05T03:24:49+00:00 | 2015-08-05T03:24:49+00:00 | 3,697 | false | Just used an array dict to count the occurence of the letters in t. To distinguish the letters that are not in t, we initialize dict with -slen and for those letters that are not in t, their corresponding elements in dict will be -slen. For example, t="abc", then dict['a']= dict['b']=dict['c']=1, while the others, suc... | 12 | 1 | [] | 1 |
minimum-window-substring | Java, SlidingWindow, HashMap, TwoPointer Solution | java-slidingwindow-hashmap-twopointer-so-3hi4 | plz... upvote! if you find my solution helpful.\n\nStatus: Accepted*\nRuntime: 4 ms\nMemory Usage: 43.9 MB\nAll test cases passed.\n\n | kumar-rinku0 | NORMAL | 2022-07-25T09:50:36.379176+00:00 | 2022-07-25T09:52:12.730683+00:00 | 2,272 | false | ***plz... upvote! if you find my solution helpful.***\n\nStatus: **Accepted***\nRuntime: 4 ms\nMemory Usage: 43.9 MB\nAll test cases passed.\n\n[<iframe src="https://leetcode.com/playground/jpJPCeKg/shared" frameBorder="0" width="800" height="600"></iframe>](http://) | 11 | 0 | ['Sliding Window', 'Java'] | 3 |
minimum-window-substring | O(n) 5ms Java Solution Beats 93.18% | on-5ms-java-solution-beats-9318-by-longs-gr7t | This solution adopts the idea described in this [LeetCode article][1]. It explains this O(n) solution very well. Because of that article, I kept the comments si | longstation | NORMAL | 2016-03-04T23:53:48+00:00 | 2016-03-04T23:53:48+00:00 | 5,275 | false | This solution adopts the idea described in this [LeetCode article][1]. It explains this O(n) solution very well. Because of that article, I kept the comments simple. I highly suggest you to read it before trying this code.\n\n public class Solution {\n public String minWindow(String s, String t) {\n char... | 11 | 0 | ['Java'] | 4 |
minimum-window-substring | [Animated Video] Sliding Window Template + Visualization | animated-video-sliding-window-template-v-q06l | CodeInMotion ResourcesEvery Leetcode Pattern You Need To Knowhttps://www.blog.codeinmotion.io/p/leetcode-patternsBlind 75 Animated Playlisthttps://www.youtube.c | codeinmotion | NORMAL | 2025-01-03T22:03:54.784433+00:00 | 2025-01-03T22:03:54.784433+00:00 | 624 | false | 
# *CodeInMotion Resources*
## Every Leetcode Pattern You Need To Know
https://www.blog.codeinmotion.io/p/leetcode-patterns
## Blind 75 Animated Playlist
https:/... | 10 | 0 | ['Hash Table', 'String', 'Sliding Window', 'Python', 'C++', 'Java', 'Go', 'Python3', 'JavaScript', 'C#'] | 0 |
minimum-window-substring | C++ STL SOLUTION | | 2 APPROACHES | | EASY TO UNDERSTAND | c-stl-solution-2-approaches-easy-to-unde-338v | APPROACH 1\n\n## Approach\nSteps:\n1. We first create a hashmap for the characters of the string \'t\'.\n\n\nfor (auto value:t){\n m[value]++;\n}\n\n\n2. The | ALANT535 | NORMAL | 2024-02-04T07:36:02.503088+00:00 | 2024-02-04T07:36:02.503120+00:00 | 1,493 | false | # APPROACH 1\n\n## Approach\nSteps:\n1. We first create a hashmap for the characters of the string \'t\'.\n\n```\nfor (auto value:t){\n m[value]++;\n}\n```\n\n2. Then we use two pointers, left and right to iterate through the string \'s\'. For each index, we do the following-\n\n\n2.1. If s[i] is inside m, the hashm... | 10 | 0 | ['Hash Table', 'String', 'Sliding Window', 'C++'] | 6 |
minimum-window-substring | ✔️ Python's Simple and Easy to Understand Solution Using Sliding Window | 99% Faster 🔥 | pythons-simple-and-easy-to-understand-so-wvv7 | \uD83D\uDD3C IF YOU FIND THIS POST HELPFUL PLEASE UPVOTE \uD83D\uDC4D\n\n\nclass Solution:\n def minWindow(self, s: str, t: str) -> str:\n lookup = Co | pniraj657 | NORMAL | 2022-10-22T05:27:11.294594+00:00 | 2022-10-31T05:28:42.546285+00:00 | 2,337 | false | **\uD83D\uDD3C IF YOU FIND THIS POST HELPFUL PLEASE UPVOTE \uD83D\uDC4D**\n\n```\nclass Solution:\n def minWindow(self, s: str, t: str) -> str:\n lookup = Counter(t)\n mx = float("inf")\n output = ""\n S = len(s)\n start, end = 0, 0\n count = len(lookup)\n \n w... | 10 | 0 | ['Sliding Window', 'Python', 'Python3'] | 2 |
minimum-window-substring | ✔️ 100% Fastest Swift Solution | 100-fastest-swift-solution-by-sergeylesc-cfcp | \nclass Solution {\n func minWindow(_ s: String, _ t: String) -> String {\n guard s.count >= t.count else { return "" }\n \n let sChars | sergeyleschev | NORMAL | 2022-04-06T05:43:40.603423+00:00 | 2022-04-06T05:43:40.603468+00:00 | 1,271 | false | ```\nclass Solution {\n func minWindow(_ s: String, _ t: String) -> String {\n guard s.count >= t.count else { return "" }\n \n let sChars = Array(s)\n let tChars = Array(t)\n \n let indexs = validIndexs(sChars, tChars)\n guard s.count >= t.count else { return "" }\n ... | 10 | 0 | ['Swift'] | 3 |
minimum-window-substring | C++|| Sliding Window|| Using arrays | c-sliding-window-using-arrays-by-hiteshc-650u | Runtime: 8 ms\n\n\nclass Solution {\npublic:\n string minWindow(string s, string t) {\n \n //frequency mapping\n int FP[256]={0};\n | hiteshcmonga | NORMAL | 2021-06-09T15:28:27.804483+00:00 | 2021-06-09T15:29:01.647947+00:00 | 720 | false | Runtime: 8 ms\n\n```\nclass Solution {\npublic:\n string minWindow(string s, string t) {\n \n //frequency mapping\n int FP[256]={0};\n int FS[256]={0};\n int cnt=0;\n int start=0; //left contraction\n int start_idx=-1 ;//for best window\n int min_so_far=INT_MAX... | 10 | 3 | ['C', 'Sliding Window'] | 0 |
minimum-window-substring | Minimum window substring | minimum-window-substring-by-r9n-zyb4 | \n\nusing System;\nusing System.Collections.Generic;\n\npublic class Solution\n{\n public string MinWindow(string s, string t)\n {\n if (s.Length = | r9n | NORMAL | 2024-08-15T01:48:27.810142+00:00 | 2024-08-15T01:48:27.810165+00:00 | 261 | false | \n```\nusing System;\nusing System.Collections.Generic;\n\npublic class Solution\n{\n public string MinWindow(string s, string t)\n {\n if (s.Length == 0 || t.Length == 0)\n return "";\n\n var required = new Dictionary<char, int>();\n var window = new Dictionary<char, int>();\n ... | 9 | 0 | ['C#'] | 0 |
minimum-window-substring | Sliding Window 🪟 | Python 🐍 | CLEAR EXPLANATION 📒 | sliding-window-python-clear-explanation-iumlu | Efficiency\n\n\n\n# Approach\nSliding Window algorithm: Create 2 pointers to identify the start and end points of the output string.\n1. First, we create a Hash | kcp_1410 | NORMAL | 2024-02-04T19:30:55.437579+00:00 | 2024-02-04T19:31:35.777700+00:00 | 149 | false | # Efficiency\n\n\n\n# Approach\n**Sliding Window algorithm**: Create 2 pointers to identify the start and end points of the output string.\n1. First, we create a HashMap `t_chars` and count the occurrences ... | 9 | 0 | ['Hash Table', 'String', 'Sliding Window', 'Python3'] | 2 |
minimum-window-substring | Beats 100% 🚀➡️Detailed Explanation 💯➡️[Java/C++/Python3/JavaScript] | beats-100-detailed-explanation-javacpyth-yh81 | Intuition\n1. Window Creation with Two Pointers:\n - Use two pointers, one for the left end of the window and another for the right end, to create a window o | Shivansu_7 | NORMAL | 2024-02-04T09:52:18.108292+00:00 | 2024-02-04T09:52:18.108325+00:00 | 3,014 | false | # Intuition\n1. **Window Creation with Two Pointers:**\n - Use two pointers, one for the left end of the window and another for the right end, to create a window of letters in string `s`.\n - This window should aim to contain all the characters present in string `t`.\n2. **Expanding the Window:**\n - Start by ... | 9 | 0 | ['Sliding Window', 'C++', 'Java', 'Python3', 'JavaScript'] | 6 |
minimum-window-substring | 🚀🚀 Beats 100% | Optimized Approach 🔥🔥 | Fully Explained 💎 | beats-100-optimized-approach-fully-expla-xrvg | Intuition \uD83E\uDD14:\nWe are given two strings, s and t, and we need to find the minimum window substring in s that contains all characters of t. To approach | The_Eternal_Soul | NORMAL | 2024-02-04T01:29:13.091631+00:00 | 2024-02-04T01:29:13.091660+00:00 | 2,351 | false | ### Intuition \uD83E\uDD14:\nWe are given two strings, `s` and `t`, and we need to find the minimum window substring in `s` that contains all characters of `t`. To approach this problem, we\'ll use two hash maps to keep track of characters in `s` and `t`, and then we\'ll use a sliding window technique to find the minim... | 9 | 1 | ['Hash Table', 'String', 'C', 'Sliding Window', 'Python', 'C++', 'Java', 'JavaScript'] | 3 |
minimum-window-substring | SUPER EASY SOLUTION | super-easy-solution-by-himanshu__mehra-jfc9 | Intuition\n Describe your first thoughts on how to solve this problem. \nhttps://www.youtube.com/watch?v=_t8kq_RpiPU\n\n# Code\n\nclass Solution {\npublic:\n | himanshu__mehra__ | NORMAL | 2023-07-11T18:00:27.491009+00:00 | 2023-07-11T18:00:27.491029+00:00 | 1,811 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nhttps://www.youtube.com/watch?v=_t8kq_RpiPU\n\n# Code\n```\nclass Solution {\npublic:\n string minWindow(string s, string t) {\n unordered_map<char,int> hast_s;\n unordered_map<char,int> hast_t;\n if(s.length()<t.l... | 9 | 0 | ['Hash Table', 'Sliding Window', 'C++'] | 1 |
minimum-window-substring | Brute force to Optimal | Sliding Window | C++ | brute-force-to-optimal-sliding-window-c-l91j6 | Brute Force\n\nclass Solution {\n bool check(unordered_map<char, int> &mp, unordered_map<char, int> &m) {\n if(m.size() < mp.size()) return false;\n | TusharBhart | NORMAL | 2022-10-22T05:19:36.454471+00:00 | 2022-10-22T05:19:36.454511+00:00 | 2,074 | false | # Brute Force\n```\nclass Solution {\n bool check(unordered_map<char, int> &mp, unordered_map<char, int> &m) {\n if(m.size() < mp.size()) return false;\n for(auto i : m) {\n if(i.second < mp[i.first]) return false;\n }\n return true;\n }\npublic:\n string minWindow(string... | 9 | 0 | ['Hash Table', 'Sliding Window', 'C++'] | 2 |
minimum-window-substring | Simple Python O(N) Sliding Window Explained with Example | simple-python-on-sliding-window-explaine-n2wn | O(N) Sliding Window Python Solution with Counter\n\ndef minWindow(self, s: str, t: str) -> str:\n\ttCounter = Counter(t) # counter for t to check with\n\twindow | bloomh | NORMAL | 2022-10-22T03:54:39.488692+00:00 | 2022-10-23T04:38:53.947778+00:00 | 2,228 | false | **O(N) Sliding Window Python Solution with Counter**\n```\ndef minWindow(self, s: str, t: str) -> str:\n\ttCounter = Counter(t) # counter for t to check with\n\twindow = Counter() # sliding window\n\tans = "" # answer\n\tlast = 0 # last index in our window\n\tfor i,char in enumerate(s):\n\t\twindow[char] = window.get(c... | 9 | 0 | ['Sliding Window', 'Python'] | 3 |
minimum-window-substring | The most easy sliding window solution | the-most-easy-sliding-window-solution-by-7xtz | \nstring minWindow(string s, string t) {\n string res = "";\n if(s.size() < t.size()) return res;\n unordered_map<char, int> umap;\n | codingGuy2016 | NORMAL | 2022-05-11T18:04:52.697989+00:00 | 2022-05-11T18:04:52.698090+00:00 | 913 | false | ```\nstring minWindow(string s, string t) {\n string res = "";\n if(s.size() < t.size()) return res;\n unordered_map<char, int> umap;\n for(auto it: t) umap[it]++;\n int i=0, j=0;\n int length = INT_MAX, count = umap.size();\n while(j<s.size()){\n if(umap.find... | 9 | 0 | ['C', 'Sliding Window', 'C++'] | 2 |
minimum-window-substring | [Python 3] From Brute Force to Sliding Window - 3 solutions - O(N) | python-3-from-brute-force-to-sliding-win-8wx9 | Solution 1: Brute force\npython\nclass Solution:\n def minWindow(self, s: str, t: str) -> str:\n def isCover(cnt1, cnt2): # return True if all charact | hiepit | NORMAL | 2021-08-16T16:09:46.569921+00:00 | 2023-08-21T13:51:12.776208+00:00 | 459 | false | **Solution 1: Brute force**\n```python\nclass Solution:\n def minWindow(self, s: str, t: str) -> str:\n def isCover(cnt1, cnt2): # return True if all characters in cnt2 is covered by cnt1\n for k, v in cnt2.items():\n if cnt1[k] < v:\n return False\n ret... | 9 | 1 | ['Sliding Window'] | 0 |
minimum-window-substring | [C++] Sliding Window || Easy To Understand | c-sliding-window-easy-to-understand-by-m-k9dp | IDEA\nWe keep expanding the window by moving the right pointer. When the window has all the desired characters, we contract (if possible) and save the smallest | Maango16 | NORMAL | 2021-08-15T08:52:17.480608+00:00 | 2021-08-15T08:52:17.480634+00:00 | 559 | false | **IDEA**\nWe keep expanding the window by moving the right pointer. When the window has all the desired characters, we contract (if possible) and save the smallest window till now.\n`The answer is the smallest desirable window.`\n\u200B\n**EXAMPLE**\nFor eg. `S = "ABAACBAB" T = "ABC"`. Then our answer window is `"ACB"`... | 9 | 4 | [] | 2 |
minimum-window-substring | Share my neat and easy-understand python solution in O(n) | share-my-neat-and-easy-understand-python-vprw | class Solution(object):\n '''\n b -- begin index of window; e -- end index of window.\n example: S="acbbaca", T="aba"\n \n 0 | mach7 | NORMAL | 2015-10-14T23:03:52+00:00 | 2015-10-14T23:03:52+00:00 | 2,662 | false | class Solution(object):\n '''\n b -- begin index of window; e -- end index of window.\n example: S="acbbaca", T="aba"\n \n 0 1 2 3 4 5 6\n a c b b a c a\n \n initially, b=e=0\n a:2\n b:1\n pc=2 (positive count. pc>0 indicates that there ar... | 9 | 0 | ['Python'] | 1 |
minimum-window-substring | The fast 7ms O(N) Java solution use only one array without map | the-fast-7ms-on-java-solution-use-only-o-tni3 | public String minWindow(String s, String t) {\n if (t.length() <= 0 || s.length() < t.length()) return "";\n int start = 0, end = 0, i = 0, j = 0, | alpenliebe | NORMAL | 2016-01-13T21:26:57+00:00 | 2016-01-13T21:26:57+00:00 | 3,471 | false | public String minWindow(String s, String t) {\n if (t.length() <= 0 || s.length() < t.length()) return "";\n int start = 0, end = 0, i = 0, j = 0, count = t.length(), min = s.length()+1;\n int[] table = new int[256];\n \n for(int k = 0; k<count; k++){\n char c = t.charA... | 9 | 2 | ['Array', 'Dynamic Programming', 'Java'] | 1 |
minimum-window-substring | C++ post referred to top voted post | c-post-referred-to-top-voted-post-by-rai-bakx | First, thanks the post from @vinceyuan\n\nHere I just try explain it carefully so to help some beginners to better understand the inspiring ideas behind the sol | rainbowsecret | NORMAL | 2016-02-20T11:06:22+00:00 | 2016-02-20T11:06:22+00:00 | 3,525 | false | First, thanks the post from @vinceyuan\n\nHere I just try explain it carefully so to help some beginners to better understand the inspiring ideas behind the solution.\n\nPreprocessing step:\n\nstore the condition variable in the unordered_map\n\nWhile loop step:\n\n check sub-conditions, update global variable\n\... | 9 | 1 | [] | 6 |
minimum-window-substring | Minimum Window Substring - Solution Explanation and Code (Video Solution Available) | minimum-window-substring-solution-explan-2t72 | Video SolutionIntuitionThe problem involves finding the minimum window in the string s that contains all the characters of string t. My first thought was to use | CodeCrack7 | NORMAL | 2025-01-25T03:40:16.974122+00:00 | 2025-01-25T03:40:16.974122+00:00 | 979 | false | # Video Solution
[https://youtu.be/OoLytq4DRro?si=agfFSpEcxAEablx_]()
# Intuition
The problem involves finding the minimum window in the string `s` that contains all the characters of string `t`. My first thought was to use a sliding window approach with a frequency map to track the required characters and their counts... | 8 | 0 | ['Java'] | 1 |
minimum-window-substring | 🔥🔥🔥 Easy Sliding Window Solution | Explained 🔥🔥🔥 | easy-sliding-window-solution-explained-b-0u59 | Intuition\nThe main problem is finding the smallest substring in s that contains all the characters in t. Use the sliding window concept to solve this problem.\ | pratamandiko | NORMAL | 2024-02-04T02:43:51.157837+00:00 | 2024-02-04T02:43:51.157866+00:00 | 1,406 | false | # Intuition\nThe main problem is finding the **smallest** substring in **s** that contains all the characters in **t**. Use the sliding window concept to solve this problem.\n\n# Approach\n**Character Frequency t:** We start by counting the frequency of each character in **t** using a hashmap.\n\n**Sliding Window Expan... | 8 | 0 | ['Hash Table', 'String', 'Sliding Window', 'Python', 'C++', 'Java', 'Go', 'Python3', 'JavaScript'] | 0 |
minimum-window-substring | [Python] ✅ || Easy Mind Map Diagram || Two Pointer + Sliding Window [Dynamic Size] || Space O(m+n) | python-easy-mind-map-diagram-two-pointer-r7jt | Intuition\nThe "Minimum Window Substring" problem requires finding the smallest substring in s that contains all the characters of t. Initially, the challenge s | chitralpatil | NORMAL | 2024-01-08T19:57:07.765945+00:00 | 2024-01-08T19:57:07.765972+00:00 | 944 | false | # Intuition\nThe "Minimum Window Substring" problem requires finding the smallest substring in `s` that contains all the characters of `t`. Initially, the challenge seems to lie in efficiently checking if a substring contains all required characters, and then minimizing this substring. The key is realizing the dynamic ... | 8 | 0 | ['Hash Table', 'Two Pointers', 'String', 'Sliding Window', 'Python3'] | 0 |
minimum-window-substring | 🙎C++ || Easy || Daily LeetCoding Challenge (22nd October 2022) | c-easy-daily-leetcoding-challenge-22nd-o-dfrh | Easy to Understand\nIf understand the sol^n then Please Upvote\n\nclass Solution {\npublic:\n string minWindow(string s, string t) {\n if (s.size() < | spyder_master | NORMAL | 2022-10-22T04:36:54.298965+00:00 | 2022-10-22T04:36:54.298993+00:00 | 5,674 | false | # Easy to Understand\n**If understand the sol^n then Please Upvote**\n```\nclass Solution {\npublic:\n string minWindow(string s, string t) {\n if (s.size() < t.size() or s.empty()) {\n return "";\n }\n \n int i = 0, j = 0;\n int start = -1, len = INT_MAX;\n std::... | 8 | 0 | ['Hash Table', 'C', 'Sliding Window'] | 2 |
minimum-window-substring | [Java] Sliding window | Aditya Verma | java-sliding-window-aditya-verma-by-naga-0gxp | \nclass Solution {\n public String minWindow(String s, String t) {\n \n Map<Character, Integer> map = new HashMap<>();\n for(char ch: t. | nagato19 | NORMAL | 2022-08-14T19:15:15.550855+00:00 | 2022-08-14T19:15:15.550907+00:00 | 807 | false | ```\nclass Solution {\n public String minWindow(String s, String t) {\n \n Map<Character, Integer> map = new HashMap<>();\n for(char ch: t.toCharArray())\n map.put(ch, map.getOrDefault(ch, 0) + 1);\n \n int count = map.size(), ansLength = Integer.MAX_VALUE;\n Stri... | 8 | 0 | ['Sliding Window', 'Java'] | 0 |
minimum-window-substring | 🔥 JAVA 2️⃣-pointer sliding window || Simple Explanation 👀😄 | java-2-pointer-sliding-window-simple-exp-jkc8 | \uD83D\uDCCC 2 pointer Sliding window with 2 Hashmaps \uD83D\uDC49\uD83C\uDFFB \'Acquired\' and \'Required\' \n\n ## \u2705 At any point of time we will find | Yash_kr | NORMAL | 2022-06-13T09:55:10.298019+00:00 | 2022-06-22T05:02:02.859113+00:00 | 882 | false | # \uD83D\uDCCC 2 pointer Sliding window with 2 Hashmaps \uD83D\uDC49\uD83C\uDFFB \'Acquired\' and \'Required\' \n\n ## \u2705 At any point of time we will find : Minimum window substring starting from ith index (i.e. first pointer )\n \n \u270D\uD83C\uDFFB **Notations** : \n1. acquired -> Hashmap containing detai... | 8 | 0 | ['Two Pointers', 'Sliding Window', 'Java'] | 0 |
minimum-window-substring | Sliding window C++ solution | sliding-window-c-solution-by-workday-rzkr | Using a sliding window to increase the right boundary to include more elements. Once the sliding window includes all elements in the target array, try to increa | workday | NORMAL | 2022-03-15T04:12:48.371258+00:00 | 2022-03-15T04:12:48.371327+00:00 | 542 | false | Using a sliding window to increase the right boundary to include more elements. Once the sliding window includes all elements in the target array, try to increase the left boundary to decrease the size of the window until the window doesn\'t include all elements in the target array. Repeat the previous steps till the r... | 8 | 0 | ['C', 'Sliding Window'] | 2 |
minimum-window-substring | JS optimized solution with explanations O(t+s) time O(t+s) space | js-optimized-solution-with-explanations-c6554 | Note\n- Should be pretty straight forward if you did these problems before: 567-permutation-in-string, 438. Find All Anagrams in a String\n- Do those first.\n- | inxela | NORMAL | 2022-02-17T00:04:44.457157+00:00 | 2022-02-17T17:25:31.038455+00:00 | 1,497 | false | # Note\n- Should be pretty straight forward if you did these problems before: [567-permutation-in-string](https://leetcode.com/problems/permutation-in-string/), [438. Find All Anagrams in a String](https://leetcode.com/problems/find-all-anagrams-in-a-string/)\n- Do those first.\n- Here is the idea:\n1. create a hashmap... | 8 | 0 | ['Two Pointers', 'Sliding Window', 'JavaScript'] | 0 |
minimum-window-substring | Aditya Verma Approach || Easy Hashmap with Sliding Window || Minimum Window Substring | aditya-verma-approach-easy-hashmap-with-h77av | \nif (t.length() > s.length()) {\n return "";\n}\n\nif (s == t)\n return s;\nunordered_map < char, int > m;\n\nfor (auto it: t) {\n m[it]++;\n}\n\nint cnt = | akshansh773 | NORMAL | 2021-10-01T06:39:55.185780+00:00 | 2023-03-20T06:22:11.087501+00:00 | 1,025 | false | ```\nif (t.length() > s.length()) {\n return "";\n}\n\nif (s == t)\n return s;\nunordered_map < char, int > m;\n\nfor (auto it: t) {\n m[it]++;\n}\n\nint cnt = m.size();\nint i = 0;\nint j = 0;\nint ans = INT_MAX;\nint start = -1;\nint end = -1;\nwhile (j < s.length()) {\n if (m.find(s[j]) != m.end()) {\n m[s[j]... | 8 | 0 | ['C++'] | 1 |
minimum-window-substring | C++|| Commented || Easy to Understand || Sliding Window|| Map | c-commented-easy-to-understand-sliding-w-i7a9 | \nclass Solution {\npublic:\n string minWindow(string s, string t)\n {\n unordered_map<char, int> mapt; // For frequency of characters of string t\ | pragyatewary24 | NORMAL | 2021-08-15T18:22:13.661135+00:00 | 2021-08-15T18:22:13.661179+00:00 | 815 | false | ```\nclass Solution {\npublic:\n string minWindow(string s, string t)\n {\n unordered_map<char, int> mapt; // For frequency of characters of string t\n for(auto ch : t)\n {\n mapt[ch]++;\n }\n unordered_map<char, int> maps; // For frequency of characters of string s\n... | 8 | 0 | ['C', 'Sliding Window'] | 2 |
minimum-window-substring | Easy to understand C++ sliding window technique | easy-to-understand-c-sliding-window-tech-090u | \nclass Solution {\npublic:\n string minWindow(string s, string t) {\n int mm = s.size(), n = t.size();\n unordered_map<char, int> m;\n | tejas702 | NORMAL | 2021-08-15T14:56:23.037338+00:00 | 2021-08-15T14:57:01.438480+00:00 | 588 | false | ```\nclass Solution {\npublic:\n string minWindow(string s, string t) {\n int mm = s.size(), n = t.size();\n unordered_map<char, int> m;\n for(int i=0;i<n;i++)m[t[i]]++;\n int l = 0, r = 0;\n int res = INT_MAX;\n string st = "";\n string ryan = s.substr(l,r-l+1);\n ... | 8 | 6 | ['C', 'Sliding Window'] | 0 |
minimum-window-substring | Template for sliding window problems | template-for-sliding-window-problems-by-ex624 | Inspired by the top post, I create my own template for sliding window problems. I feel it is more generic and is good for all the sliding window problems I stud | yu6 | NORMAL | 2020-08-26T01:22:23.091984+00:00 | 2021-01-03T19:27:33.267840+00:00 | 678 | false | Inspired by the top post, I create my own template for sliding window problems. I feel it is more generic and is good for all the sliding window problems I studied so far.\n\n**The template**\n```\nwhile(r<n) {\n\t//process r\n\twhile(r cannot move) {\n\t\t//process l to free r\n\t\tl++;\n\t}\n\t//process r\n\tr++;\n}\... | 8 | 0 | [] | 1 |
minimum-window-substring | Python 20 lines AC O(n) solution | python-20-lines-ac-on-solution-by-round1-bcxy | Idea is pretty simple, keep two pointers left and right.\n\n If s[left:right] has all chars in T, calculate distance and keep answer, then move left pointer.\n\ | round1lc | NORMAL | 2015-05-12T06:48:28+00:00 | 2015-05-12T06:48:28+00:00 | 4,858 | false | Idea is pretty simple, keep two pointers left and right.\n\n If s[left:right] has all chars in T, calculate distance and keep answer, then move left pointer.\n\n If s[left:right] doesn't have all chars in T, move right pointer.\n\n\n class Solution:\n # @param {string} s\n # @param {string} t\n # @return {s... | 8 | 0 | ['Python'] | 3 |
minimum-window-substring | Sliding Window Explained || Beginner-friendly | sliding-window-explained-beginner-friend-ojis | IntuitionAs we need to check for substring, window would be the best approach to handle this. The only issue was deciding when to shrink. Once we find all the e | Bulba_Bulbasar | NORMAL | 2025-01-26T08:23:14.219517+00:00 | 2025-01-26T08:30:42.968145+00:00 | 1,107 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
As we need to check for substring, window would be the best approach to handle this. The only issue was deciding when to shrink. Once we find all the elements we will shrink the window from left till the next element and then look for the m... | 7 | 0 | ['Hash Table', 'Two Pointers', 'Sliding Window', 'Java'] | 1 |
minimum-window-substring | ✅HARD MADE EASY | SLIDING WINDOW | JAVA | Explained🔥🔥🔥 | hard-made-easy-sliding-window-java-expla-3mx4 | Problem: Find the minimum window substring in string S which will contain all the characters in string T in inclusion and order.\n\nApproach:\n\nWe employ a sli | AdnanNShaikh | NORMAL | 2024-08-16T16:41:58.405132+00:00 | 2024-08-16T16:41:58.405173+00:00 | 730 | false | **Problem**: Find the minimum window substring in string S which will contain all the characters in string T in inclusion and order.\n\n**Approach**:\n\nWe employ a sliding window technique, where we maintain a window of characters in string S.\nA HashMap is used to keep track of the characters in string T and their re... | 7 | 0 | ['Java'] | 1 |
minimum-window-substring | Easy to Understand, C++ solution Sliding Window's Most Important Question | easy-to-understand-c-solution-sliding-wi-8z8q | Intuition\nAs we see the problems of String and Array, in which we have to optimise the solution and we have to perform sequential operations on those problems. | sribhargav1345 | NORMAL | 2024-02-04T06:26:29.617144+00:00 | 2024-02-04T06:26:29.617174+00:00 | 1,288 | false | # Intuition\nAs we see the problems of String and Array, in which we have to optimise the solution and we have to perform sequential operations on those problems. We need to think of **Sliding Window**\n\n# Approach\nStore the elements of string \'t\' in a map, and compare that with the elements of string \'s\', and if... | 7 | 0 | ['Ordered Map', 'Sliding Window', 'C++'] | 1 |
minimum-window-substring | ✔️ ✔️Beat 97.97%|| Sliding Window Approach with Clear Comments and Explanation | beat-9797-sliding-window-approach-with-c-ww6s | Please upvote if you find the solution helpful.\n\n\n## Explanation\n\n1. Initialization:\n - Create a character frequency map mp1 for string t.\n - Initial | Inam_28_06 | NORMAL | 2024-02-04T01:10:30.818487+00:00 | 2024-02-04T03:37:43.827510+00:00 | 1,311 | false | Please **upvote** if you find the solution helpful.\n\n\n## Explanation\n\n1. **Initialization**:\n - Create a character frequency map `mp1` for string `t`.\n - Initialize sliding window pointers `i` and `j... | 7 | 0 | ['C', 'Sliding Window', 'Python', 'Java'] | 0 |
minimum-window-substring | O( n )✅ | Python (Step by step explanation)✅ | o-n-python-step-by-step-explanation-by-m-q7j2 | Intuition\nThe problem requires finding the minimum window in string \'s\' that contains all characters from string \'t\'. We can use a sliding window approach | monster0Freason | NORMAL | 2023-10-27T07:34:54.036360+00:00 | 2023-10-27T07:34:54.036381+00:00 | 485 | false | # Intuition\nThe problem requires finding the minimum window in string \'s\' that contains all characters from string \'t\'. We can use a sliding window approach to solve this problem efficiently.\n\n# Approach\n1. First, we handle the edge case: if \'t\' is an empty string, we return an empty string as the result.\n\n... | 7 | 0 | ['Sliding Window', 'Python3'] | 0 |
apply-operations-to-an-array | 🚀 Beats 100% | In-Place Processing + Zero Shifting | Apply Operations on Array | beats-100-in-place-processing-zero-shift-p3w7 | Youtube🚀 Beats 100% | In-Place Processing + Zero Shifting | Apply Operations on Array🔼 Please Upvote🔼 Please Upvote🔼 Please Upvote🔼 Please Upvote💡 If this helpe | rahulvijayan2291 | NORMAL | 2025-03-01T03:38:56.513541+00:00 | 2025-03-01T03:38:56.513541+00:00 | 18,214 | false | # Youtube
https://youtu.be/81UhXYJOz1g
---
# 🚀 Beats 100% | In-Place Processing + Zero Shifting | Apply Operations on Array
---
# **🔼 Please Upvote**
# **🔼 Please Upvote**
# **🔼 Please Upvote... | 74 | 3 | ['Array', 'C', 'C++', 'Java', 'TypeScript', 'Python3', 'JavaScript', 'C#'] | 6 |
apply-operations-to-an-array | ✅✅✅ C++ & Java || O(n) time & O(1) space || Very Simple and Easy to Understand | c-solution-on-time-o1-space-very-simple-2l5hy | Approach:
Simply change the value of nums[i] to 2 times & set nums[i+1] to zero when nums[i] == nums[i+1].
Then take a pointer and keep on accumulating non zero | kreakEmp | NORMAL | 2022-11-06T04:00:38.429018+00:00 | 2025-03-01T07:13:05.365924+00:00 | 6,232 | false |
# Approach:
- Simply change the value of nums[i] to 2 times & set nums[i+1] to zero when nums[i] == nums[i+1].
- Then take a pointer and keep on accumulating non zero value at the front.
- Set all remaining values to zero, utill pointer is less then size of the array.
```cpp []
vector<int> applyOperations(vector<int... | 65 | 9 | ['C++', 'Java'] | 11 |
apply-operations-to-an-array | One Pass [C++/Java/Python3] | one-pass-cjavapython3-by-xxvvpp-za84 | Just do Operation and Swapping of non-zeroes at front of array simultaneously.\n# C++\n vector applyOperations(vector& A) {\n for (int i = 0, j = 0; i | xxvvpp | NORMAL | 2022-11-06T04:01:46.657909+00:00 | 2022-11-06T08:15:39.924890+00:00 | 2,977 | false | Just do `Operation` and `Swapping of non-zeroes at front` of array simultaneously.\n# C++\n vector<int> applyOperations(vector<int>& A) {\n for (int i = 0, j = 0; i < size(A); ++i){\n if (i + 1 < size(A) and A[i] == A[i + 1]){\n A[i] *= 2;\n A[i + 1] = 0;\n ... | 56 | 2 | ['C'] | 5 |
apply-operations-to-an-array | ✅ Two Pointers | Python | C++ | Java | C# | JS | Go | Swift | PHP | Rust | Kotlin | Dart | Ruby | two-pointers-python-by-otabek_kholmirzae-kumt | IntuitionThe problem involves two distinct operations: first, we need to perform a series of comparisons and modifications on adjacent elements, and then we nee | otabek_kholmirzaev | NORMAL | 2025-03-01T08:44:49.113410+00:00 | 2025-03-01T10:30:19.084090+00:00 | 6,009 | false | # Intuition
**The problem involves two distinct operations:** first, we need to perform a series of comparisons and modifications on adjacent elements, and then we need to move all zeros to the end of the array. A natural way to approach this is to follow the instructions step by step.
# Approach
1. **Apply Operations... | 49 | 0 | ['Swift', 'Python', 'C++', 'Java', 'Go', 'Python3', 'Rust', 'Kotlin', 'JavaScript', 'C#'] | 3 |
apply-operations-to-an-array | ✅ [Python/C++] use the NOT trick to move zeros (explained) | pythonc-use-the-not-trick-to-move-zeros-e54cc | \u2705 IF YOU LIKE THIS SOLUTION, PLEASE UPVOTE.\n*\nThis solution employs a brute force approach to perform operations along with (stable-)sorting using a *log | stanislav-iablokov | NORMAL | 2022-11-06T04:00:48.478990+00:00 | 2022-11-06T05:19:33.054526+00:00 | 2,892 | false | **\u2705 IF YOU LIKE THIS SOLUTION, PLEASE UPVOTE.**\n****\nThis solution employs a brute force approach to perform operations along with (stable-)sorting using a **logical not** trick to move zeros to the end of array. Time complexity is linear: **O(N\\*logN)**. Space complexity is linear: **O(N)**.\n\n\n**Python.**\n... | 37 | 1 | [] | 10 |
apply-operations-to-an-array | ✅2 Method's ||🌟JAVA||🧑💻 BEGINNER FREINDLY||C++||Python3 | 2-methods-java-beginner-freindlycpython3-gdqh | Approach 1: Two-Pass1.Apply Operations and Shift Zeros in a Single Pass:
. Use a pointer index to track the position where the next non-zero element should be | Varma5247 | NORMAL | 2025-03-01T00:41:28.990563+00:00 | 2025-03-01T05:42:01.649157+00:00 | 2,847 | false |
# Approach 1: Two-Pass
<!-- Describe your approach to solving the problem. -->
**1.Apply Operations and Shift Zeros in a Single Pass:**
. Use a pointer ```index``` to track the position where the next non-zero element should be placed.
.Iterate through the array:
.If the current element (```nums[i]```) is equal ... | 22 | 0 | ['Array', 'Two Pointers', 'Python', 'C++', 'Java', 'TypeScript', 'Python3', 'JavaScript'] | 1 |
apply-operations-to-an-array | 2 pointers O(1) space vs 1 loop||C++ Py3 beats 100% | 2-passes-2-pointers-o1-spacec-beats-100-gs9qz | Intuition1st pass is for applying operations
2nd pass use 2 pointers to reuse the array nums for the answer
Both C++ & Python are made.a 1 loop solution is also | anwendeng | NORMAL | 2025-03-01T00:17:41.882187+00:00 | 2025-03-01T02:56:39.007387+00:00 | 2,249 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
1st pass is for applying operations
2nd pass use 2 pointers to reuse the array nums for the answer
Both C++ & Python are made.
a 1 loop solution is also made.
# Approach
<!-- Describe your approach to solving the problem. -->
1. Use a loop... | 17 | 0 | ['Two Pointers', 'C++', 'Python3'] | 5 |
apply-operations-to-an-array | ✅✅🔥BEATS 96.37% SIMPLEST PROGRAM | JAVA 🔥 Python 🔥 C++ 🔥 JS 🔥 | Simplest Program | O(n) & O(1) | beats-9637-simplest-program-java-python-zotsv | IntuitionThis method modifies the given array based on certain operations:
If two consecutive elements are equal, double the first one and set the second one to | peEAWW4Y33 | NORMAL | 2025-03-01T12:12:23.310810+00:00 | 2025-03-01T12:12:23.310810+00:00 | 596 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
This method modifies the given array based on certain operations:
- If two consecutive elements are equal, double the first one and set the second one to 0.
- Shift all non-zero elements to the left while maintaining their relative order.
... | 15 | 0 | ['Two Pointers', 'Python', 'C++', 'Java', 'JavaScript'] | 0 |
apply-operations-to-an-array | Python 3 || 6 lines, iteration || T/S: 99% / 52% | python-3-6-lines-iteration-ts-99-52-by-s-bsbl | [https://leetcode.com/problems/apply-operations-to-an-array/submissions/1166861606/?submissionId=1166861050)updateI could be wrong, but I think that time comple | Spaulding_ | NORMAL | 2022-11-06T07:05:15.846795+00:00 | 2025-03-01T01:28:10.872483+00:00 | 945 | false | ```Python3 []
class Solution:
def applyOperations(self, nums: List[int]) -> List[int]:
for i in range(len(nums)-1):
if nums[i] == nums[i+1]:
nums[i]*= 2
nums[i + 1] = 0 # <-- performing the "operation."
nums.sort(key=lambda x: x == 0) # Sorti... | 14 | 0 | ['C++', 'Java', 'Python3'] | 5 |
apply-operations-to-an-array | ✅c++|| Simple solution|| Move Zeroes problem extension || O(1) space | c-simple-solution-move-zeroes-problem-ex-dgec | The idea is super simple go through all n - 1 elements and see if condition satisifies\nthen the problem is move all zeroes to end \nwe solve it using two point | dinesh55 | NORMAL | 2022-11-06T04:02:37.108295+00:00 | 2022-11-06T06:49:57.678610+00:00 | 941 | false | The idea is super simple go through all n - 1 elements and see if condition satisifies\nthen the problem is move all zeroes to end \nwe solve it using two pointers \n\nLink to move zeroes to end problem :- <b>[Move Zeroes Problem](https://leetcode.com/problems/move-zeroes/)</b>\n\n```c++\nclass Solution {\npublic:\n ... | 11 | 1 | ['C'] | 0 |
apply-operations-to-an-array | ✅Simple Java Solutions in O(N) || ✅Runtime 0ms || ✅ Beats 100% | simple-java-solutions-in-on-runtime-0ms-t5v1y | \n# Intuition\nDescribe your first thoughts on how to solve this problem. \n\n\n\nThis code defines a `Solution` class with a method called `applyOperations`, | ahmedna126 | NORMAL | 2023-09-27T16:28:12.413227+00:00 | 2023-11-07T11:36:10.517206+00:00 | 398 | false | <!-- \n# Intuition\nDescribe your first thoughts on how to solve this problem. \n\n\n\nThis code defines a `Solution` class with a method called `applyOperations`, which takes an integer array `nums` as input and returns another integer array as output. The goal of this code is to perform certain operations on the inp... | 10 | 0 | ['Java'] | 0 |
apply-operations-to-an-array | ✅C++|| Simple Traversal || Easy Solution | c-simple-traversal-easy-solution-by-indr-i7k6 | \nclass Solution {\npublic:\n vector<int> applyOperations(vector<int>& nums) {\n int n=nums.size();\n\n vector<int>ans;\n \n for( | indresh149 | NORMAL | 2022-11-06T04:04:07.403275+00:00 | 2022-11-06T04:04:07.403318+00:00 | 1,264 | false | ```\nclass Solution {\npublic:\n vector<int> applyOperations(vector<int>& nums) {\n int n=nums.size();\n\n vector<int>ans;\n \n for(int i=0;i<n-1;i++)\n {\n if(nums[i]==nums[i+1])\n {\n nums[i]=2*nums[i];\n \n nums[... | 10 | 0 | ['C'] | 1 |
apply-operations-to-an-array | JavaScript easy and fast solution (93.71% faster) | javascript-easy-and-fast-solution-9371-f-8oag | \n\n\n\n* only sort the zeros: nums.sort((a,b)=> !a - !b)\n\n\n# Code\n\n/**\n * @param {number[]} nums\n * @return {number[]}\n */\nvar applyOperations = func | fahim_ash | NORMAL | 2022-11-14T08:34:31.993881+00:00 | 2022-11-14T08:35:15.490547+00:00 | 496 | false | \n\n\n\n* only sort the zeros: nums.sort((a,b)=> !a - !b)\n\n\n# Code\n```\n/**\n * @param {number[]} nums\n * @return {number[]}\n */\nvar applyOperations = function(nums) {\n for (let i=0;i<nums.length;i++){\n if (nums[i]==nums[i+1]){\n [nums[i],nums[i+1]]=[nums[i]*2,0];\n }\n }return ... | 9 | 0 | ['JavaScript'] | 0 |
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