question_slug stringlengths 3 77 | title stringlengths 1 183 | slug stringlengths 12 45 | summary stringlengths 1 160 ⌀ | author stringlengths 2 30 | certification stringclasses 2
values | created_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | updated_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | hit_count int64 0 10.6M | has_video bool 2
classes | content stringlengths 4 576k | upvotes int64 0 11.5k | downvotes int64 0 358 | tags stringlengths 2 193 | comments int64 0 2.56k |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
minimum-bit-flips-to-convert-number | xor bitset->1 line||beats 100% | xor-bitset-1-linebeats-100-by-anwendeng-ynjf | Intuition\n Describe your first thoughts on how to solve this problem. \nEasy question.\n1-line solution is provided.\n# Approach\n Describe your approach to so | anwendeng | NORMAL | 2024-09-11T01:02:33.893466+00:00 | 2024-09-11T11:45:23.436705+00:00 | 2,811 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nEasy question.\n1-line solution is provided.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nxx=x^y;\nzz=x&y&xx;//in fact zz=0\n\nxb.count()+zb.count() is the answer\n# Complexity\n- Time complexity:\n<!-- Add your t... | 24 | 1 | ['Bit Manipulation', 'C++'] | 8 |
minimum-bit-flips-to-convert-number | Java - XOR and counting bits - beats 100% | java-xor-and-counting-bits-beats-100-by-cuup8 | We need to count the number of corresponding bits of start and goal that are different.\nxor-ing start and goal will result in a new number with binary represen | abanoubcs | NORMAL | 2022-04-02T16:17:07.965214+00:00 | 2022-04-02T22:43:00.069103+00:00 | 3,141 | false | We need to count the number of corresponding bits of start and goal that are different.\nxor-ing start and goal will result in a new number with binary representation of 0 where the corresponding bits of start and goal are equal and 1 where the corresponding bits are different.\n\nFor example: 10 and 7 \n10 = 1010\n 7... | 24 | 0 | ['Java'] | 8 |
minimum-bit-flips-to-convert-number | Simple Two lines code using XOR | simple-two-lines-code-using-xor-by-user0-rxy0 | Intuition\n# XOR Operation (^):\n\nThe XOR operation returns a bit that is 1 where the bits of start and goal differ, and 0 where they are the same.\nBy calcula | user0197ub | NORMAL | 2024-09-11T01:11:51.171047+00:00 | 2024-09-11T01:11:51.171066+00:00 | 5,602 | false | # Intuition\n# XOR Operation (^):\n\nThe XOR operation returns a bit that is 1 where the bits of start and goal differ, and 0 where they are the same.\nBy calculating xor_value = start ^ goal, you create a number where each 1 represents a bit that differs between start and goal.\n# Counting the Number of Differing Bits... | 21 | 2 | ['Bit Manipulation', 'C++', 'Python3'] | 9 |
minimum-bit-flips-to-convert-number | C++/Java/JavaScript || Easy Solution || Beats 100% || Bit Manipulation | cjavajavascript-easy-solution-beats-100-rlgau | Intuition\n - Question -> Here given Two numbers. If both no. are represented in Binary then how many minimum bits are required to change in one no. so it conve | shivang21007 | NORMAL | 2023-05-11T19:57:43.918196+00:00 | 2023-05-11T19:57:43.918241+00:00 | 2,059 | false | # Intuition\n - Question -> Here given Two numbers. If both no. are represented in Binary then how many minimum bits are required to change in one no. so it convert into second no. \n\n*example* -> \n```\n 10 = 1 0 (1) 0\n 7 = 0 1 (1) 1\n```\nhere, 3 bits are different in both no.s and need to be ... | 18 | 0 | ['Bit Manipulation', 'C++', 'Java', 'JavaScript'] | 3 |
minimum-bit-flips-to-convert-number | O(1) | 0 - ms beats 100 % | Java | Python | Go | C++ | Rust | JavaScript | o1-0-ms-beats-100-java-python-go-c-rust-jlb2c | \n### Code 1\n\n\n\njava []\npublic class Solution {\n public int minBitFlips(int start, int goal) {\n return Integer.bitCount(start ^ goal);\n }\n | kartikdevsharma_ | NORMAL | 2024-09-11T04:33:07.520883+00:00 | 2024-09-11T14:26:09.368333+00:00 | 2,493 | false | \n### Code 1\n\n\n\n```java []\npublic class Solution {\n public int minBitFlips(int start, int goal) {\n return Integer.bitCount(start ^ goal);\n }\n}\n\n```\n\n\n```cpp []\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n return __builtin_popcount(start ^ goal);\n }\n};\n... | 14 | 0 | ['Bit Manipulation', 'C++', 'Java', 'Go', 'Python3', 'Rust', 'JavaScript'] | 5 |
minimum-bit-flips-to-convert-number | c++ | Easy O(N) solution | Basic Maths | c-easy-on-solution-basic-maths-by-yash2a-qd3s | Please upvote if you find this solution helpful\nCode:\n\nclass Solution {\npublic:\n //we just check whether binary bit is equal or not \n //if it is we | Yash2arma | NORMAL | 2022-04-05T08:37:52.408073+00:00 | 2022-04-05T08:40:26.193274+00:00 | 1,540 | false | **Please upvote if you find this solution helpful**\n**Code:**\n```\nclass Solution {\npublic:\n //we just check whether binary bit is equal or not \n //if it is we do nothing otherwise we flips the bit and increase the count\n int minBitFlips(int start, int goal) \n { \n int flips=0;\n\t\t\n\t\t//... | 12 | 0 | ['Math', 'C', 'C++'] | 3 |
minimum-bit-flips-to-convert-number | __builtin_popcount(start ^ goal) | builtin_popcountstart-goal-by-votrubac-1w89 | C++\ncpp\nint minBitFlips(int start, int goal) {\n return __builtin_popcount(start ^ goal);\n}\n | votrubac | NORMAL | 2022-04-03T19:19:24.871537+00:00 | 2022-04-03T19:19:24.871572+00:00 | 904 | false | **C++**\n```cpp\nint minBitFlips(int start, int goal) {\n return __builtin_popcount(start ^ goal);\n}\n``` | 12 | 0 | ['C'] | 5 |
minimum-bit-flips-to-convert-number | ✔ python | easy | interview thinking | python-easy-interview-thinking-by-peb-dx0h | The given question requires us to count the total number of flips we need to do inorder to make start -> goal.\nEx: \n7 - 0111\n10 - 1010\nHere, we only flip t | peb | NORMAL | 2022-07-09T21:34:30.613365+00:00 | 2022-12-04T15:22:16.507182+00:00 | 1,923 | false | The given question requires us to count the total number of flips we need to do inorder to make `start -> goal`.\nEx: \n7 - 0111\n10 - 1010\nHere, we only flip the bits which are different in both **start** and **goal**, i.e. `01` or `10`. And, what helps us to find if the bits are different? **XOR**. \nNow, we count ... | 9 | 0 | ['Bit Manipulation', 'Python', 'Python3'] | 3 |
minimum-bit-flips-to-convert-number | Java solution with explaination | java-solution-with-explaination-by-shubh-xjzd | Intuition\nWe want to check how many bits need to flipped in the input number to get the output number. So we would need to check the bits which are different i | shubhampawar16298 | NORMAL | 2023-01-07T17:56:26.053480+00:00 | 2023-01-07T17:56:26.053529+00:00 | 1,828 | false | # Intuition\nWe want to check how many bits need to flipped in the input number to get the output number. So we would need to check the bits which are different in both numbers (No point in counting bits which are same in both) and count them.\n\n# Approach\nE.g. \n10 = 1010\n7 = 0111\nso different bits are, from righ... | 8 | 0 | ['Java'] | 1 |
minimum-bit-flips-to-convert-number | Simple Java Code ☠️ | simple-java-code-by-abhinandannaik1717-em31 | Code\njava []\nclass Solution {\n public int minBitFlips(int start, int goal) {\n String str1 = Integer.toBinaryString(start);\n String str2 = | abhinandannaik1717 | NORMAL | 2024-09-11T07:03:08.513339+00:00 | 2024-09-11T07:03:08.513373+00:00 | 584 | false | # Code\n```java []\nclass Solution {\n public int minBitFlips(int start, int goal) {\n String str1 = Integer.toBinaryString(start);\n String str2 = Integer.toBinaryString(goal);\n int n1 = str1.length();\n int n2 = str2.length();\n int count = 0,i=n1-1,j=n2-1;\n while(i>=0 &... | 7 | 0 | ['String', 'Java'] | 0 |
minimum-bit-flips-to-convert-number | 💢☠💫Easiest👾Faster✅💯 Lesser🧠 🎯 C++✅Python3🐍✅Java✅C✅Python🐍✅C#✅💥🔥💫Explained☠💥🔥 Beats 100 | easiestfaster-lesser-cpython3javacpython-mmf1 | Intuition\n\n Describe your first thoughts on how to solve this problem. \njavascript []\n/**\n * @param {number} start\n * @param {number} goal\n * @return {nu | Edwards310 | NORMAL | 2024-09-11T05:29:06.114252+00:00 | 2024-09-11T05:29:06.114289+00:00 | 160 | false | # Intuition\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n```javascript []\n/**\n * @param {number} start\n * @param {number} goal\n * @return {number}\n */\nvar minBitFli... | 7 | 0 | ['Bit Manipulation', 'C', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 0 |
minimum-bit-flips-to-convert-number | not bit manipulation just plain string conversion | not-bit-manipulation-just-plain-string-c-hz20 | Intuition\n Describe your first thoughts on how to solve this problem. \nwe convert the bigger number to string and convert the smaller number to string and we | srinivas_bodduru | NORMAL | 2024-09-11T02:19:04.147008+00:00 | 2024-09-11T02:19:04.147035+00:00 | 1,330 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nwe convert the bigger number to string and convert the smaller number to string and we compare the two strings and increment the counter where ever we find differences \n# Approach\n<!-- Describe your approach to solving the problem. -->\... | 7 | 1 | ['Bit Manipulation', 'C', 'Python', 'C++', 'TypeScript', 'Python3', 'JavaScript'] | 4 |
minimum-bit-flips-to-convert-number | 🔥Beats 100%🔥✅Without Inbuilt Function Super Easy (C++/Java/Python) Bit Manipulation Solution✅ | beats-100without-inbuilt-function-super-gxso5 | Intuition\nThe intuition is to iteratively checks each bit of the two numbers using a loop, which continues until the larger of start and goal becomes zero (ind | suyogshete04 | NORMAL | 2024-03-22T03:37:51.196328+00:00 | 2024-03-22T03:41:09.421277+00:00 | 1,108 | false | # Intuition\nThe intuition is to iteratively checks each bit of the two numbers using a loop, which continues until the larger of `start` and `goal` becomes zero (indicating all bits have been checked). Within each iteration, it applies a bitwise AND with 1 to both numbers to extract their least significant bits, compa... | 7 | 0 | ['Bit Manipulation', 'C++', 'Java', 'Python3'] | 1 |
minimum-bit-flips-to-convert-number | Python Solution | One liner | python-solution-one-liner-by-tejpratap1-ju4n | Just xor so we get 1 at places where bits are different and then count those bits.\n\nclass Solution:\n def minBitFlips(self, start: int, goal: int) -> int:\ | TejPratap1 | NORMAL | 2022-04-02T16:20:21.723192+00:00 | 2022-04-02T16:20:21.723223+00:00 | 911 | false | **Just xor so we get 1 at places where bits are different and then count those bits.**\n```\nclass Solution:\n def minBitFlips(self, start: int, goal: int) -> int:\n return (bin(start^goal).count("1"))\n``` | 7 | 0 | ['Bit Manipulation', 'Python'] | 3 |
minimum-bit-flips-to-convert-number | 👍One liner solution✅As fast as wind💨! Shortest🔥Fastest🔥Easiest🔥Full explanation🌟 | one-liner-solutionas-fast-as-wind-shorte-n1bm | \n### Explanation of the Problem\n\nThe problem asks us to determine the minimum number of "bit flips" required to convert one number, start, into another numbe | aijcoder | NORMAL | 2024-09-11T09:08:59.201923+00:00 | 2024-09-11T09:11:20.980378+00:00 | 1,090 | false | \n### Explanation of the Problem\n\nThe problem asks us to determine the minimum number of "bit flips" required to convert one number, `start`, into another number, `goal`. A "bit flip" means changing a bit in a binary number from 0 to 1, or from 1 to 0.\n\n#### Binary Representation of Numbers\nEach integer can be rep... | 6 | 0 | ['Bit Manipulation', 'C++', 'Java', 'Python3', 'JavaScript'] | 2 |
minimum-bit-flips-to-convert-number | Flip Different Bits Only | Java | C++ | [Video Solution] | flip-different-bits-only-java-c-video-so-a9lr | Intuition, approach, and complexity discussed in video solution in detail.\nhttps://youtu.be/lqSsY0EXnb0\n# Code\njava []\nclass Solution {\n public int minB | Lazy_Potato_ | NORMAL | 2024-09-11T04:36:46.484551+00:00 | 2024-09-11T04:36:46.484572+00:00 | 1,195 | false | # Intuition, approach, and complexity discussed in video solution in detail.\nhttps://youtu.be/lqSsY0EXnb0\n# Code\n``` java []\nclass Solution {\n public int minBitFlips(int start, int goal) {\n int diffNum = (start ^ goal);\n return setBitCnt(diffNum);\n }\n private int setBitCnt(int num){\n ... | 6 | 0 | ['Bit Manipulation', 'C++', 'Java'] | 4 |
minimum-bit-flips-to-convert-number | simple and easy Javascript solution 😍❤️🔥 | simple-and-easy-javascript-solution-by-s-19gw | \n\n# Complexity\n- Time complexity: O(1)\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 | shishirRsiam | NORMAL | 2024-09-11T00:36:01.324947+00:00 | 2024-09-11T00:36:01.324980+00:00 | 348 | false | \n\n# Complexity\n- Time complexity: O(1)\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```javascript []\nvar minBitFlips = function(start, goal) \n{\n // Perform bitwise XOR between \'start\' and \'goal\'.\n /... | 6 | 0 | ['Bit Manipulation', 'JavaScript'] | 8 |
minimum-bit-flips-to-convert-number | one line solution | one-line-solution-by-shishirrsiam-i4el | \n# Complexity\n- Time complexity: O(1)\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 | shishirRsiam | NORMAL | 2024-09-11T00:26:09.678792+00:00 | 2024-09-11T00:26:09.678811+00:00 | 1,359 | false | \n# Complexity\n- Time complexity: O(1)\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```python []\nclass Solution(object):\n def minBitFlips(self, start, goal):\n return bin(start ^ goal).count(\'1\') \n... | 6 | 0 | ['Bit Manipulation', 'Python', 'Python3'] | 10 |
minimum-bit-flips-to-convert-number | C 0ms simple solution by using & and ^ | c-0ms-simple-solution-by-using-and-by-ro-4rb6 | Intuition\nI thought that if I find the number of differing bits between \'start\' and \'goal\' and assign them to a variable, then I can count them.\n\n# Appro | camogluserkan | NORMAL | 2023-11-02T10:07:31.352033+00:00 | 2023-11-02T10:07:31.352062+00:00 | 293 | false | # Intuition\nI thought that if I find the number of differing bits between \'start\' and \'goal\' and assign them to a variable, then I can count them.\n\n# Approach\nFirstly, I determined how many differing bits exist between \'start\' and \'goal\'.\nThe XOR operation returns 1 if the bits of \'goal\' and \'start\' ar... | 6 | 0 | ['C'] | 1 |
minimum-bit-flips-to-convert-number | ✅[cpp || Beats 100% || O(1) TC || O(1) SC || Thorough Explanation]✅ | cpp-beats-100-o1-tc-o1-sc-thorough-expla-yf2q | Intuition\n1 ^ 1 = 0\n0 ^ 0 = 0\n1 ^ 0 = 1\n0 ^ 1 = 1\n\nIf we xor a and b we can find the bits that are distinct in a and b\n\nA ---> 1 1 1 0 0 1\nB ---> 1 0 1 | mayanksinghchouhan | NORMAL | 2023-10-19T04:27:32.620912+00:00 | 2023-10-21T01:24:40.585074+00:00 | 841 | false | # Intuition\n1 ^ 1 = 0\n0 ^ 0 = 0\n1 ^ 0 = 1\n0 ^ 1 = 1\n\nIf we xor a and b we can find the bits that are distinct in a and b\n\nA ---> 1 1 1 0 0 1\nB ---> 1 0 1 0 1 0\nA^B->0 1 0 0 1 1 \n\n# Approach\n- All you need to do is count the number of set bits in A ^ B and you will have your answer.\n- You can do this usi... | 6 | 0 | ['Bit Manipulation', 'C++'] | 3 |
minimum-bit-flips-to-convert-number | Rust solution | rust-solution-by-bigmih-u3if | \nimpl Solution {\n pub fn min_bit_flips(start: i32, goal: i32) -> i32 {\n (start ^ goal).count_ones() as _\n }\n}\n | BigMih | NORMAL | 2022-04-03T08:46:11.616165+00:00 | 2022-04-03T08:46:11.616193+00:00 | 189 | false | ```\nimpl Solution {\n pub fn min_bit_flips(start: i32, goal: i32) -> i32 {\n (start ^ goal).count_ones() as _\n }\n}\n``` | 6 | 0 | ['Rust'] | 1 |
minimum-bit-flips-to-convert-number | ✅ C++ | Easy | 2 different ways | c-easy-2-different-ways-by-chandanagrawa-cwe2 | We have to count all positions where the i\'th bit of number A and B is different , so just iterate from bit 32 to 0 and check.\n\n\nclass Solution\n{\n publ | chandanagrawal23 | NORMAL | 2022-04-02T16:11:12.953189+00:00 | 2022-04-02T16:41:10.490701+00:00 | 1,277 | false | We have to count all positions where the i\'th bit of number **A** and **B** is different , so just iterate from bit 32 to 0 and check.\n\n```\nclass Solution\n{\n public:\n int minBitFlips(int start, int goal)\n {\n int cnt = 0;\n for (int i = 32; i >= 0; i--)\n {\n\t\... | 6 | 0 | ['C'] | 6 |
minimum-bit-flips-to-convert-number | ⚡ Easy approach, O(N) time | No Bit Manipulation | easy-approach-on-time-no-bit-manipulatio-vx3h | \nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n int ans = 0;\n while(start && goal) { \n if(start%2 != goal%2 | abanoub7asaad | NORMAL | 2022-04-02T16:01:10.003140+00:00 | 2022-04-03T21:34:39.872690+00:00 | 655 | false | ```\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n int ans = 0;\n while(start && goal) { \n if(start%2 != goal%2)\n ans++;\n start /= 2;\n goal /= 2;\n } \n while(start) { \n if(start%2)\n ans... | 6 | 0 | ['C'] | 3 |
minimum-bit-flips-to-convert-number | Easy Soln✅✅||Bit Manipulation Mastery 🔥🔥||Beats 100% ✅✅ | easy-solnbit-manipulation-mastery-beats-6z7cb | Here the problem is to convert the bit value start to bit value of goal.\n\nEx. \nstart = 4 -> 0100\ngoal = 10 -> 1010\n\nwe just have to check the difference | 21eca01 | NORMAL | 2024-09-11T09:22:04.066969+00:00 | 2024-09-11T09:22:04.067006+00:00 | 222 | false | Here the problem is to convert the bit value start to bit value of goal.\n\nEx. \nstart = 4 -> 0100\ngoal = 10 -> 1010\n\nwe just have to check the differences between these two number\'s bits.\nWe can just check each bit similarity using XOR(^) method and shift every digit with the help of right shift\n \n**Walkthro... | 5 | 0 | ['Bit Manipulation', 'C++', 'Java', 'Python3'] | 1 |
minimum-bit-flips-to-convert-number | 1 Line Solution | 100 % Beats | Java | 1-line-solution-100-beats-java-by-eshwar-d319 | Code\njava []\nclass Solution {\n public int minBitFlips(int start, int goal) {\n return Integer.bitCount(start ^ goal);\n }\n}\n | eshwaraprasad | NORMAL | 2024-09-11T05:52:17.220076+00:00 | 2024-09-11T05:52:17.220115+00:00 | 196 | false | # Code\n```java []\nclass Solution {\n public int minBitFlips(int start, int goal) {\n return Integer.bitCount(start ^ goal);\n }\n}\n``` | 5 | 0 | ['Java'] | 1 |
minimum-bit-flips-to-convert-number | Java | Brute Force | 6 lines | Clean code | java-brute-force-6-lines-clean-code-by-j-ru8d | Complexity\n- Time complexity: O(max.bits)\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) \ | judgementdey | NORMAL | 2024-09-11T00:06:26.027471+00:00 | 2024-09-11T00:06:26.027502+00:00 | 561 | false | # Complexity\n- Time complexity: $$O(max.bits)$$\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```java []\nclass Solution {\n public int minBitFlips(int start, int goal) {\n var res = 0;\n\n while (start >... | 5 | 0 | ['Bit Manipulation', 'Java'] | 1 |
minimum-bit-flips-to-convert-number | Very Intuitive approach | very-intuitive-approach-by-dixon_n-dsqx | take a xor why? \n\nsee what you take a xor the 1 will remaing zero meaning we dont want to filp that bit, and if its zero it will remain as zero so we dont wan | Dixon_N | NORMAL | 2024-09-04T10:42:49.482454+00:00 | 2024-09-04T10:42:49.482486+00:00 | 423 | false | take a xor why? \n\nsee what you take a xor the 1 will remaing zero meaning we dont want to filp that bit, and if its zero it will remain as zero so we dont want to flip it \nso wha are the reamingin one\'s ? we need to count the ones\' in the bits after doing xor ithats is the problem is reduced simple\n\n\n# Code\n``... | 5 | 0 | ['Bit Manipulation', 'Java'] | 5 |
minimum-bit-flips-to-convert-number | Curious Logic Python3 | curious-logic-python3-by-ganjinaveen-1ec4 | 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 | GANJINAVEEN | NORMAL | 2023-02-26T11:21:08.016021+00:00 | 2023-02-26T11:21:08.016065+00:00 | 900 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 5 | 0 | ['Python3'] | 2 |
minimum-bit-flips-to-convert-number | 👋 CPP EZ Amazon + Google😬 Interview O(set bits) | cpp-ez-amazon-google-interview-oset-bits-dpza | UPVOTE PLZ \u2763\n# Approach-1 Always takes O(N)\nclass Solution { //O(N) soln.\npublic:\n int minBitFlips(int start, int goal) {\n | AdityaBhate | NORMAL | 2022-07-18T07:48:18.918473+00:00 | 2022-11-16T05:55:52.630071+00:00 | 1,206 | false | # ***UPVOTE PLZ \u2763***\n# **Approach-1** Always takes O(N)\nclass Solution { //O(N) soln.\npublic:\n int minBitFlips(int start, int goal) {\n int c=0, i=0;\n while(start != goal){\n int mask=1<<i;\n if((start & mask) == (goal & mask))\n i+... | 5 | 2 | ['Bit Manipulation', 'C', 'C++', 'Java', 'Python3'] | 2 |
minimum-bit-flips-to-convert-number | Python Solution | Hamming Distance Based | One Liner | python-solution-hamming-distance-based-o-ljxd | Hamming Distance\nHamming Distance between two integers is the number of bits that are different at the same position in both numbers. \n\nAlgorithm\n- XOR the | Gautam_ProMax | NORMAL | 2022-04-26T15:17:42.700131+00:00 | 2022-04-26T15:17:42.700165+00:00 | 428 | false | ## Hamming Distance\nHamming Distance between two integers is the number of bits that are different at the same position in both numbers. \n\nAlgorithm\n- XOR the numbers\n- Count set bits (1)\n\n```\nclass Solution:\n def minBitFlips(self, start: int, goal: int) -> int:\n return bin(start ^ goal).count("1")\... | 5 | 0 | ['Bit Manipulation', 'Python', 'Python3'] | 0 |
minimum-bit-flips-to-convert-number | C++ 0ms solution without bit manipulation | c-0ms-solution-without-bit-manipulation-oijvf | \nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n int count=0;\n while(start && goal){\n if(start%2!=goal%2) co | zdy012 | NORMAL | 2022-04-02T17:06:07.442494+00:00 | 2022-04-02T17:06:07.442539+00:00 | 750 | false | ```\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n int count=0;\n while(start && goal){\n if(start%2!=goal%2) count++;\n start/=2;\n goal/=2;\n }\n while(start){\n if(start%2)count++;\n start/=2;\n }\n ... | 5 | 0 | ['C++'] | 2 |
minimum-bit-flips-to-convert-number | C++ | Easy | And Operation | c-easy-and-operation-by-kamisamaaaa-6lcn | \nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n \n int res(0);\n for (int i=32; ~i; i--) {\n if ((star | kamisamaaaa | NORMAL | 2022-04-02T16:54:11.783841+00:00 | 2022-04-02T16:54:21.143616+00:00 | 327 | false | ```\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n \n int res(0);\n for (int i=32; ~i; i--) {\n if ((start & 1) != (goal & 1)) res++;\n start >>= 1; goal >>= 1;\n }\n return res;\n } \n};\n``` | 5 | 1 | ['C'] | 0 |
minimum-bit-flips-to-convert-number | Easy Solution✅✅||using XOR 🔥🔥||Beats 100% ✅✅ | easy-solutionusing-xor-beats-100-by-mr_a-9bhx | Intuition\n\nThe goal of the problem is to find the minimum number of bit flips required to convert start to goal. A bit flip means changing a bit from 0 to 1 o | mr_ajagiya | NORMAL | 2024-09-11T19:39:38.090936+00:00 | 2024-09-11T19:39:38.090957+00:00 | 127 | false | # Intuition\n\nThe goal of the problem is to find the minimum number of bit flips required to convert `start` to `goal`. A bit flip means changing a bit from 0 to 1 or from 1 to 0.\n\n1. **XOR Operation (`^`)**:\n - The XOR of two numbers, `start ^ goal`, gives a number where each bit is `1` if the corresponding bits... | 4 | 0 | ['Bit Manipulation', 'C++'] | 0 |
minimum-bit-flips-to-convert-number | Simple and Easy CPP Code!💯✅☮️ | simple-and-easy-cpp-code-by-siddharth_si-d0md | \n# Code\ncpp []\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n bitset<32> bs(start);\n bitset<32> bg(goal);\n in | siddharth_sid_k | NORMAL | 2024-09-11T16:45:21.292129+00:00 | 2024-09-11T16:45:21.292164+00:00 | 137 | false | \n# Code\n```cpp []\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n bitset<32> bs(start);\n bitset<32> bg(goal);\n int c=0;\n for(int i=bs.size();i>=0;i--)\n {\n if(bs[i]!=bg[i])\n {\n c++;\n }\n }\n ... | 4 | 0 | ['C++'] | 0 |
minimum-bit-flips-to-convert-number | ✅BEATS 100.00% ⌚2 Approaches ✔using Bit Manipulation and Second one Simplest BRUTE FORCE⛔ | beats-10000-2-approaches-using-bit-manip-ahju | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nApproach 1:\nWe use XOR to check where the two numbers (start and goal) a | ArcuLus | NORMAL | 2024-09-11T07:12:35.436835+00:00 | 2024-09-11T07:12:35.436866+00:00 | 13 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nApproach 1:\nWe use XOR to check where the two numbers (start and goal) are different. XOR gives us a new number where every 1 means a difference between the bits, and every 0 means they\u2019re the same. Then, we just count... | 4 | 0 | ['Bit Manipulation', 'Java'] | 1 |
minimum-bit-flips-to-convert-number | EASY SOLUTION | easy-solution-by-viratkohli-ts6v | USE XOR THEN COUNT EVERY SET BIT\n# Complexity\n- Time complexity:O(log n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(1)\n Add your spa | viratkohli_ | NORMAL | 2024-09-11T03:57:59.521740+00:00 | 2024-09-11T03:57:59.521772+00:00 | 264 | false | - USE XOR THEN COUNT EVERY SET BIT\n# Complexity\n- Time complexity:O(log 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. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n int ... | 4 | 0 | ['Bit Manipulation', 'C++'] | 2 |
minimum-bit-flips-to-convert-number | simple and easy C++ solution 😍❤️🔥 | simple-and-easy-c-solution-by-shishirrsi-3aqg | \n# Complexity\n- Time complexity: O(1)\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 | shishirRsiam | NORMAL | 2024-09-11T00:21:07.862880+00:00 | 2024-09-11T00:21:07.862900+00:00 | 964 | false | \n# Complexity\n- Time complexity: O(1)\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```cpp []\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) \n {\n // Convert the \'start\' and \'goa... | 4 | 1 | ['Bit Manipulation', 'C++'] | 9 |
minimum-bit-flips-to-convert-number | BEATS 100% || O(1) || EASY BEGINERS | beats-100-o1-easy-beginers-by-abhishekka-lvk1 | The provided code defines a function minBitFlips(start, goal) that calculates the minimum number of bit flips required to convert one integer (start) to another | Abhishekkant135 | NORMAL | 2024-06-24T15:23:39.192808+00:00 | 2024-06-24T15:23:39.192837+00:00 | 912 | false | The provided code defines a function `minBitFlips(start, goal)` that calculates the minimum number of bit flips required to convert one integer (`start`) to another (`goal`). Here\'s a detailed explanation:\n\n**Concept:**\n\nThe code leverages the XOR (^) operator and bit manipulation to efficiently determine the numb... | 4 | 0 | ['Bit Manipulation', 'Java'] | 1 |
minimum-bit-flips-to-convert-number | Java Simplest! Code (Beats 100%) | java-simplest-code-beats-100-by-ravinder-xzx7 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n- Compare least significant bits of both start and goal and keep right sh | Ravinder_1834 | NORMAL | 2023-10-22T18:14:26.146111+00:00 | 2023-10-22T18:14:26.146134+00:00 | 915 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n- Compare least significant bits of both start and goal and keep right shifting by 1 until it is greater than 0. \n- Simultaneously count the no. of steps(which increases when the bits are not equal).\n\n# Complexity\n- Time... | 4 | 0 | ['Java'] | 1 |
minimum-bit-flips-to-convert-number | ✅Beats 100% || Easiest Code using inbuilt function | beats-100-easiest-code-using-inbuilt-fun-fxsu | 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 | vishu_0123 | NORMAL | 2023-05-15T08:30:03.101596+00:00 | 2023-05-15T08:30:03.101636+00:00 | 1,056 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 4 | 0 | ['C++'] | 2 |
minimum-bit-flips-to-convert-number | 💡 C++ | Simple Logic | With Full Explanation ✏️ | c-simple-logic-with-full-explanation-by-m2gez | Intuition\nThe goal is to find how many bit flips (changing a bit from 0 to 1 or vice versa) are required to convert one number (start) to another number (goal) | Tusharr2004 | NORMAL | 2024-09-11T17:54:31.588632+00:00 | 2024-09-11T17:54:31.588665+00:00 | 12 | false | # Intuition\nThe goal is to find how many bit flips (changing a bit from `0` to `1` or vice versa) are required to convert one number (`start`) to another number (`goal`).\n\nThe idea is based on the following key points:\n\n1. **Binary Representation:**\n\n- Every integer can be represented as a sequence of bits (bina... | 3 | 0 | ['C++'] | 0 |
minimum-bit-flips-to-convert-number | ✅ One Line Solution | one-line-solution-by-mikposp-0fhs | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)\n\n# Code #1\nTime complexity: O(digits). Space com | MikPosp | NORMAL | 2024-09-11T09:04:55.753107+00:00 | 2024-09-11T09:04:55.753127+00:00 | 355 | 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\nTime complexity: $$O(digits)$$. Space complexity: $$O(1)$$.\n```python3\nclass Solution:\n def minBitFlips(self, v: int, u: int) -> int:\n return (v^u).bit_count()\n```\n\n# Code #2\nTime... | 3 | 0 | ['String', 'Bit Manipulation', 'Recursion', 'Python', 'Python3'] | 0 |
minimum-bit-flips-to-convert-number | Simple and Easy C++ code ||☠️💯✅ | simple-and-easy-c-code-by-shodhan_ak-g43i | Code\ncpp []\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n bitset<32> binary1(start);\n bitset<32> binary2(goal);\n | Shodhan_ak | NORMAL | 2024-09-11T07:13:03.449448+00:00 | 2024-09-11T07:13:03.449473+00:00 | 94 | false | # Code\n```cpp []\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n bitset<32> binary1(start);\n bitset<32> binary2(goal);\n int count = 0;\n for (int i = 0; i < binary1.size(); i++) {\n if (binary1[i] != binary2[i]) { \n count++;\n ... | 3 | 0 | ['C++'] | 0 |
minimum-bit-flips-to-convert-number | Easy to understand || beats 100% | easy-to-understand-beats-100-by-yash9325-3kgt | \n\n# Approach\n Describe your approach to solving the problem. \nThe goal is to calculate the minimum number of bit flips required to convert one integer (star | yash9325 | NORMAL | 2024-09-11T05:31:39.076545+00:00 | 2024-09-11T05:31:39.076581+00:00 | 4 | false | \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe goal is to calculate the minimum number of bit flips required to convert one integer (`start`) to another integer (`goal`). The basic idea is to compare the binary representation of both integers and identify the positions where the bits diffe... | 3 | 0 | ['C++', 'Java'] | 0 |
minimum-bit-flips-to-convert-number | 0ms Solution in Java👀 | 0ms-solution-in-java-by-starboy609-z5jk | Intuition\n Describe your first thoughts on how to solve this problem. \n1.XOR Operation: By using start ^ goal, you get a number that shows which bits are diff | starboy609 | NORMAL | 2024-09-11T05:14:40.461634+00:00 | 2024-09-11T05:14:40.461670+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1.XOR Operation: By using start ^ goal, you get a number that shows which bits are different between start and goal.\n\n2.Count Differences: Count how many 1s are in the result of this XOR operation. Each 1 represents a bit that is differ... | 3 | 0 | ['Java'] | 0 |
minimum-bit-flips-to-convert-number | Kotlin simple solution || Detailed explanation | kotlin-simple-solution-detailed-explanat-7ahr | Intuition\n Describe your first thoughts on how to solve this problem. \nIn order to count the number of bit flips required to convert start to goal. We need to | sushanth_grandhi | NORMAL | 2024-09-11T02:49:55.853012+00:00 | 2024-09-11T21:40:17.064038+00:00 | 43 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIn order to count the number of bit flips required to convert start to goal. We need to find the bit positions which differ in both the numbers.\n\nSo, we can go bit by bit to check if it varies or same.\n\nIf the bits in both the numbers... | 3 | 0 | ['Bit Manipulation', 'Bitmask', 'Python3', 'Kotlin'] | 1 |
minimum-bit-flips-to-convert-number | Golang solution with explanation | golang-solution-with-explanation-by-alek-w6pp | Intuition\nTo solve the problem of finding the minimum number of bit flips required to convert one integer (start) to another integer (goal), we need to identif | alekseiapa | NORMAL | 2024-09-11T01:53:17.234681+00:00 | 2024-09-11T01:53:17.234710+00:00 | 93 | false | # Intuition\nTo solve the problem of finding the minimum number of bit flips required to convert one integer (`start`) to another integer (`goal`), we need to identify the bits that differ between the two numbers. Each differing bit represents a necessary bit flip. The most efficient way to find these differing bits is... | 3 | 0 | ['Go'] | 0 |
minimum-bit-flips-to-convert-number | C++ || Easy Video Solution || Faster than 100% | c-easy-video-solution-faster-than-100-by-jboc | The video solution for the below code is\nhttps://youtu.be/kDSXbCOT6tg\n\n# Code\ncpp []\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\ | __SAI__NIVAS__ | NORMAL | 2024-09-11T01:47:45.962795+00:00 | 2024-09-11T01:47:45.962824+00:00 | 173 | false | The video solution for the below code is\nhttps://youtu.be/kDSXbCOT6tg\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n // Note:\n // Bit flip is nothing but considering... | 3 | 0 | ['C++'] | 1 |
minimum-bit-flips-to-convert-number | Cpp || Bit Manipulation | cpp-bit-manipulation-by-imsej_al-q0p1 | Intuition\n Describe your first thoughts on how to solve this problem. \n1. Each bit flip operation changes a 0 bit to a 1 bit or a 1 bit to a 0 bit.\n2. The XO | imsej_al | NORMAL | 2024-06-07T11:02:57.440588+00:00 | 2024-06-07T11:02:57.440621+00:00 | 506 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1. Each bit flip operation changes a 0 bit to a 1 bit or a 1 bit to a 0 bit.\n2. The XOR (exclusive OR) operation between two bits will be 1 if the bits are different and 0 if they are the same. Thus, start ^ goal will produce a number wh... | 3 | 0 | ['Bit Manipulation', 'C++'] | 1 |
minimum-bit-flips-to-convert-number | Most easy solution with all bit manipulation concepts covered | most-easy-solution-with-all-bit-manipula-9cia | Intuition\nBit Manipultaion\n\n\n\n\n# Approach\n- In this solution, I have taken the XOR of start and goal because, with the XOR operation, if both bits are th | Pulkit28 | NORMAL | 2024-05-19T02:58:19.489160+00:00 | 2024-05-19T20:17:30.886109+00:00 | 173 | false | # Intuition\nBit Manipultaion\n\n\n\n\n# Approach\n- In this solution, I have taken the XOR of start and goal because, with the XOR operation, if both bits are the same (b... | 3 | 0 | ['C++'] | 1 |
minimum-bit-flips-to-convert-number | One line solution 🔥🔥 | Beats 100% | Java, Python, C++ | one-line-solution-beats-100-java-python-8rij7 | \n\n# \uD83D\uDCD2Complexity\n- \u23F0Time complexity:O(1)\n Add your time complexity here, e.g. O(n) \n\n- \uD83E\uDDFASpace complexity:O(1)\n Add your space c | saikrishnanaidu | NORMAL | 2024-05-06T05:37:28.907900+00:00 | 2024-05-06T05:37:28.907936+00:00 | 457 | false | \n\n# \uD83D\uDCD2Complexity\n- \u23F0Time complexity:O(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- \uD83E\uDDFASpace complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass Solution {\n public int minBitFlips(int start, int goal) {\n return Inte... | 3 | 0 | ['Python', 'C++', 'Java', 'Python3'] | 2 |
minimum-bit-flips-to-convert-number | ✅💯Solution With Actual CONCEPT!! || UNDERSTAND THE INTUITION || NO USE OF PREDEFINED FUNCTIONS || | solution-with-actual-concept-understand-x0f0c | Intuition\n Describe your first thoughts on how to solve this problem. \nBefore starting to asses this problem lets understand the concept of Bit Shifting.\n\nB | piyuzh | NORMAL | 2024-04-29T04:40:03.611859+00:00 | 2024-04-29T16:55:55.683988+00:00 | 204 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nBefore starting to asses this problem lets understand the concept of Bit Shifting.\n\nBit shifting involves ***```moving the bits of a binary number to the left or right by a specified number of positions.```***\n\nLeft shift (<<) multipl... | 3 | 0 | ['Math', 'Bit Manipulation', 'Brainteaser', 'Bitmask', 'Java'] | 2 |
minimum-bit-flips-to-convert-number | 2220. 🔥 Java | Bit Manipulation | 100% Efficient 🔥 | 2220-java-bit-manipulation-100-efficient-6ohi | 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(n) \n\n# Co | thecompetitivedev | NORMAL | 2023-07-26T21:32:16.031584+00:00 | 2023-07-26T21:33:18.931967+00:00 | 376 | false | # 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. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public int minBitFlips(int start, int goal) {\n int count=0;\n for(int i=0;i<32;i++)\n ... | 3 | 0 | ['Bit Manipulation', 'Java'] | 3 |
minimum-bit-flips-to-convert-number | ✅ One of the most easiest solutions 💯 | one-of-the-most-easiest-solutions-by-riy-4er5 | Intuition\nWe need to just count to total number of different bits of start and end.\n\n# Approach\nFind XOR of start and end and then count 1. Because after XO | Riyad-Hossain | NORMAL | 2023-07-06T03:00:59.695207+00:00 | 2023-07-06T03:00:59.695226+00:00 | 360 | false | # Intuition\nWe need to just count to total number of different bits of `start` and `end`.\n\n# Approach\nFind `XOR` of `start` and `end` and then count 1. Because after `XOR` any two value, we get 1 only when where the bit is different.\n\n# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(1)$$\n\n... | 3 | 0 | ['Bit Manipulation', 'C++'] | 1 |
minimum-bit-flips-to-convert-number | JAVA && BIT MANIPULATION && BEATS 100% (Kernighan's Algorithm approach) | java-bit-manipulation-beats-100-kernigha-dw7m | Intuition\n Describe your first thoughts on how to solve this problem. \nUsing XOR operator the bits that are different will become 1 and same bits will become | Pratham_Upadhyay | NORMAL | 2023-06-08T13:46:59.726075+00:00 | 2023-06-08T13:46:59.726144+00:00 | 702 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUsing XOR operator the bits that are different will become 1 and same bits will become 0 . After that we just need to count the number of 1\'s using the Kernighan\'s Algorithm.\n\n# Approach\n\nFirst we store the XOR operator result in n ... | 3 | 0 | ['Bit Manipulation', 'Java'] | 1 |
minimum-bit-flips-to-convert-number | 0ms Solution || Bit Manipulation | 0ms-solution-bit-manipulation-by-vpavank-aq9k | Intuition\nJust try to make the $start$ equals to $goal$ using $Bit Manipulations$.\n\n# Approach\nIf the $goal$ bit is set, check if the $start$ bit is set, if | vpavankalyan | NORMAL | 2023-06-07T07:54:59.152628+00:00 | 2023-06-07T07:54:59.152665+00:00 | 1,712 | false | # Intuition\nJust try to make the $start$ equals to $goal$ using $Bit Manipulations$.\n\n# Approach\nIf the $goal$ bit is **set**, check if the $start$ bit is **set**, if not increase the count, as we will use one $flip$ operation.\n\nIf the $goal$ bit is **unset**, check if the $start$ bit is **unset**, if not increas... | 3 | 0 | ['Bit Manipulation', 'C++'] | 1 |
minimum-bit-flips-to-convert-number | Kernighan's algorithm || Easy to understand || With Explaination | kernighans-algorithm-easy-to-understand-41h6n | \n\n\n class Solution {\n public:\n int minBitFlips(int start, int goal) {\n int cnt=0;\n \n // # Here we get he number if set bit | harsh_patell21 | NORMAL | 2023-05-01T03:34:35.465514+00:00 | 2023-05-01T03:34:35.465559+00:00 | 234 | false | \n\n\n class Solution {\n public:\n int minBitFlips(int start, int goal) {\n int cnt=0;\n \n // # Here we get he number if set bits which are required for getting our goal number\n int set=start^goal;\n \n // # kernighan\'s algo for finding no of set bits\n whil... | 3 | 0 | ['C'] | 1 |
minimum-bit-flips-to-convert-number | Easy And Understandable Solution | easy-and-understandable-solution-by-kada-gtxb | Intuition\n Describe your first thoughts on how to solve this problem. \nFirstly On Reading The Problem, We have to find How Many Bits \nAre Different in start | kadavakallulokesh | NORMAL | 2023-02-23T18:52:49.408652+00:00 | 2023-02-23T18:52:49.408693+00:00 | 561 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirstly On Reading The Problem, We have to find How Many Bits \nAre Different in **start** And **goal**.\n\nfor finding How many different bits,we use X-OR Gate .(0^1=1 and 1^0=1)\n\n# Approach\n<!-- Describe your approach to solving the ... | 3 | 0 | ['Bit Manipulation', 'Java'] | 3 |
minimum-bit-flips-to-convert-number | C# Simple Solution With Explanation | c-simple-solution-with-explanation-by-gr-qumu | Please upvote if my solution was helpful ;)\n# Explanation\nExample: start = 10, goal = 7, in binary representation are 1010 and 0111 respectively\n\n\n\n\n1101 | gregsklyanny | NORMAL | 2022-12-10T08:04:20.437857+00:00 | 2022-12-12T15:29:34.918885+00:00 | 211 | false | **Please upvote if my solution was helpful ;)**\n# Explanation\nExample: start = 10, goal = 7, in binary representation are 1010 and 0111 respectively\n\n\n\n\n1101 is a result of XOR operation - resul... | 3 | 0 | ['Bit Manipulation', 'C#'] | 0 |
minimum-bit-flips-to-convert-number | JAVA easy solution using Kernighan's Algorithm | java-easy-solution-using-kernighans-algo-wafg | First we find xor of the two input numbers and count the number of set bits in the result using kernighan\'s algorithm\n\nhere result=start^goal\n\nkernighan\'s | 21Arka2002 | NORMAL | 2022-09-04T03:31:27.034028+00:00 | 2022-09-04T03:31:27.034055+00:00 | 370 | false | First we find xor of the two input numbers and count the number of set bits in the result using kernighan\'s algorithm\n\nhere result=start^goal\n\nkernighan\'s algorithm-\n\nIn a loop till the result > 0\nStep 1 increase the count by 1\nStep 2 result=result xor (result - 1)\n\n\n```\nclass Solution {\n public int m... | 3 | 0 | ['Bit Manipulation', 'Java'] | 1 |
minimum-bit-flips-to-convert-number | 🏌️ One Liner - Python | one-liner-python-by-robbiebusinessacc-gaxn | \n\nreturn(start^goal).bit_count()\n\nPlease upvote if you want to see more one liners and simple solutions | robbiebusinessacc | NORMAL | 2022-08-04T17:39:35.948463+00:00 | 2022-08-04T17:39:35.948502+00:00 | 310 | false | \n```\nreturn(start^goal).bit_count()\n```\n**Please upvote if you want to see more one liners and simple solutions** | 3 | 0 | ['Python'] | 2 |
minimum-bit-flips-to-convert-number | C++ solution well explained, faster than 100% | c-solution-well-explained-faster-than-10-cvmw | \nclass Solution {\npublic:\n // Function to count the set bits\n int countSetBits(int n)\n {\n int count = 0;\n while(n > 0){\n | Striver27 | NORMAL | 2022-06-15T06:11:27.168147+00:00 | 2022-06-15T06:11:27.168184+00:00 | 175 | false | ```\nclass Solution {\npublic:\n // Function to count the set bits\n int countSetBits(int n)\n {\n int count = 0;\n while(n > 0){\n n &= (n-1);\n count++;\n }\n return count;\n }\n \n int minBitFlips(int start, int goal) {\n \n return co... | 3 | 0 | ['Bit Manipulation', 'C'] | 1 |
minimum-bit-flips-to-convert-number | C++ Simple Xor Operation 100% Faster | c-simple-xor-operation-100-faster-by-kis-m2y2 | ```\n//let start = 1101\n//goal = 1000\n// Xor = 0101\n//NUmber of set bits in Xor\nclass Solution {\npublic:\n int minBitFlips(int start, int goal | kissingtheworld | NORMAL | 2022-04-11T08:17:57.589277+00:00 | 2022-04-11T08:17:57.589317+00:00 | 192 | false | ```\n//let start = 1101\n//goal = 1000\n// Xor = 0101\n//NUmber of set bits in Xor\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) \n {\n int xorr = start xor goal;\n unsigned int ans=0;\n //Count number of bits to be flipped\n while(xorr)\n {\n ... | 3 | 0 | ['Counting'] | 0 |
minimum-bit-flips-to-convert-number | Kotlin Solution: XOR then count set bit | kotlin-solution-xor-then-count-set-bit-b-vif0 | \nclass Solution {\n fun minBitFlips(start: Int, goal: Int): Int {\n var result = start xor goal\n var answer = 0\n while (result > 0) {\ | ogbe | NORMAL | 2022-04-05T00:51:24.480890+00:00 | 2022-04-05T00:51:41.479517+00:00 | 56 | false | ```\nclass Solution {\n fun minBitFlips(start: Int, goal: Int): Int {\n var result = start xor goal\n var answer = 0\n while (result > 0) {\n if (result % 2 == 1) {\n answer++\n }\n result /= 2\n }\n return answer\n }\n}\n``` | 3 | 0 | ['Kotlin'] | 1 |
minimum-bit-flips-to-convert-number | Bits Manipulation + XOR(^) | bits-manipulation-xor-by-aditya_jain_258-7nkw | \nclass Solution {\n\tpublic static int minBitFlips(int a1, int a2) {\n\t\tint n = (a1 ^ a2);\n\t\tint res = 0;\n\t\twhile (n != 0) {\n\t\t\tres++;\n\t\t\tn &= | Aditya_jain_2584550188 | NORMAL | 2022-04-03T06:21:18.397805+00:00 | 2022-04-03T06:21:18.397876+00:00 | 225 | false | ```\nclass Solution {\n\tpublic static int minBitFlips(int a1, int a2) {\n\t\tint n = (a1 ^ a2);\n\t\tint res = 0;\n\t\twhile (n != 0) {\n\t\t\tres++;\n\t\t\tn &= (n - 1);\n\t\t}\n\t\treturn res;\n\t}\n}\n``` | 3 | 0 | ['Bit Manipulation', 'Java'] | 1 |
minimum-bit-flips-to-convert-number | C++ XOR + __builtinpop_count | c-xor-__builtinpop_count-by-lzl124631x-bosw | See my latest update in repo LeetCode\n## Solution 1.\n\nXOR sets the bits that are different between start and goal, and unsets bits that are the same.\n\n__bu | lzl124631x | NORMAL | 2022-04-02T16:44:28.062210+00:00 | 2022-04-02T16:44:28.062240+00:00 | 267 | false | See my latest update in repo [LeetCode](https://github.com/lzl124631x/LeetCode)\n## Solution 1.\n\nXOR sets the bits that are different between `start` and `goal`, and unsets bits that are the same.\n\n`__builtin_popcount(mask)` counts the `1`s in `mask`.\n\nExample:\n\nExpression | Value\n---|--\nstart| `0011010`... | 3 | 0 | [] | 1 |
minimum-bit-flips-to-convert-number | ✅ || BITWISE || Easiest || LOGIC || Complexity Analysis || C++ | bitwise-easiest-logic-complexity-analysi-4z42 | Solution\n\n\n#### LOGIC\n This is simpliy asking about Hamming distance\n Take XOR, because it will set bit to 1 only when both bits are different\n Then simpl | siddp6 | NORMAL | 2022-04-02T16:05:55.058269+00:00 | 2022-04-02T16:35:19.485186+00:00 | 342 | false | ## **Solution**\n\n\n#### **LOGIC**\n* This is simpliy asking about [Hamming distance](https://en.wikipedia.org/wiki/Hamming_distance)\n* Take [XOR](https://en.wikipedia.org/wiki/Exclusive_or), because it will set bit to 1 only when both bits are different\n* Then simply count the bits\n\n\n#### **Code** \n```cpp\ncla... | 3 | 0 | ['Bit Manipulation', 'C'] | 3 |
minimum-bit-flips-to-convert-number | ✅ 🌟 JAVA SOLUTION ||🔥 BEATS 100% PROOF🔥|| 💡 CONCISE CODE ✅ || 🧑💻 BEGINNER FRIENDLY | java-solution-beats-100-proof-concise-co-2z7h | Complexity
Time complexity:O(No.OfSetBit(startgoal))
Space complexity:O(1)
Code | Shyam_jee_ | NORMAL | 2025-03-09T18:17:38.801367+00:00 | 2025-03-09T18:17:38.801367+00:00 | 64 | false | # Complexity
- Time complexity:$$O(No.OfSetBit(start^goal))$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:$$O(1)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution {
public int minBitFlips(int start, int goal) {
int num=start^goal;
... | 2 | 0 | ['Bit Manipulation', 'Java'] | 0 |
minimum-bit-flips-to-convert-number | ✅Effortless and Efficient Solution by Dhandapani✅|| 🔥🔥Beats 100% in Java🚀🚀 | effortless-and-efficient-solution-by-dha-qbjw | 🧠IntuitionTo find the minimum number of bit flips required to convert one integer to another, we can use a simple XOR operation. The XOR of two numbers will hav | Dhandapanimaruthasalam | NORMAL | 2025-01-27T08:05:14.456744+00:00 | 2025-01-27T08:05:14.456744+00:00 | 82 | false | # 🧠Intuition
To find the minimum number of bit flips required to convert one integer to another, we can use a simple XOR operation. The XOR of two numbers will have set bits (1s) in positions where the corresponding bits of the two numbers are different. Therefore, counting the number of set bits in the XOR result giv... | 2 | 0 | ['Java'] | 0 |
minimum-bit-flips-to-convert-number | Easy solution || Beats 100% | easy-solution-beats-100-by-hriii11-ypdm | Intuition\n- XOR (^) gives us 0 if same digits and 1 if different \n- we can XOR the two no to know how many have to be switched\n- different positons will have | Hriii11 | NORMAL | 2024-11-07T04:24:23.941679+00:00 | 2024-11-07T04:24:23.941717+00:00 | 8 | false | # Intuition\n- XOR (^) gives us 0 if same digits and 1 if different \n- we can XOR the two no to know how many have to be switched\n- different positons will have 1 so we will just count the no of 1s and return them\n\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n i... | 2 | 0 | ['C++'] | 1 |
minimum-bit-flips-to-convert-number | ❤️🔥 [EASY] APPROACH & EXPLANATION | easy-approach-explanation-by-ramitgangwa-j9yg | Intuition\nTo solve this problem, I thought about the nature of binary representation and how XOR can help us determine the differences between two integers. Wh | ramitgangwar | NORMAL | 2024-10-05T10:31:25.537893+00:00 | 2024-10-05T10:31:25.537932+00:00 | 53 | false | # Intuition\nTo solve this problem, I thought about the nature of binary representation and how XOR can help us determine the differences between two integers. When two bits are the same, the result of XOR is 0; when they differ, the result is 1. Therefore, by performing an XOR operation between the `start` and `goal`,... | 2 | 0 | ['Bit Manipulation', 'Java'] | 0 |
minimum-bit-flips-to-convert-number | Flipping the Minimum Bits for Transformation! | flipping-the-minimum-bits-for-transforma-q52p | Intuition\n Describe your first thoughts on how to solve this problem. \n- To find the number of bit flips required to convert start to goal, we can compare the | ROHAN_SHETTY | NORMAL | 2024-09-12T08:58:58.069666+00:00 | 2024-09-12T08:58:58.069706+00:00 | 6 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- To find the number of bit flips required to convert start to goal, we can compare the binary representation of the two numbers. The bit positions where the two numbers differ correspond to the bits that need to be flipped.\n# Approach\n... | 2 | 0 | ['C++'] | 0 |
minimum-bit-flips-to-convert-number | Beats 97.89%✅🔥| Python, C++💻| SUPER EASY, CLEAR Explanation📕 | beats-9789-python-c-super-easy-clear-exp-nzpg | Beats 97.89%\u2705\uD83D\uDD25| Python, C++\uD83D\uDCBB| SUPER EASY, CLEAR Explanation\uD83D\uDCD5\n\n## 1. Proof (Python3)\n\n\n## 2. Algorithms\n* XOR\n\n## 3 | kcp_1410 | NORMAL | 2024-09-12T02:07:41.532724+00:00 | 2024-09-12T02:07:41.532755+00:00 | 15 | false | # Beats 97.89%\u2705\uD83D\uDD25| Python, C++\uD83D\uDCBB| SUPER EASY, CLEAR Explanation\uD83D\uDCD5\n\n## 1. Proof (Python3)\n\n\n## 2. Algorithms\n* XOR\n\n## 3. Code (with inline explanation)\n```python3... | 2 | 0 | ['Bit Manipulation', 'Python', 'Python3'] | 0 |
minimum-bit-flips-to-convert-number | 1 line fast c++ 100% | 1-line-fast-c-100-by-amanraox-s4od | Approach\nPerform a bitwise XOR between start and goal. The result will have 1 in positions where the bits are different between the two numbers.\nCount the num | amanraox | NORMAL | 2024-09-11T18:17:23.097728+00:00 | 2024-09-11T18:17:23.097782+00:00 | 4 | false | # Approach\nPerform a bitwise XOR between start and goal. The result will have 1 in positions where the bits are different between the two numbers.\nCount the number of 1s in the XOR result using builtin popcount and return it.\n\n# Complexity\n- Time complexity: O(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ ... | 2 | 0 | ['C++'] | 0 |
minimum-bit-flips-to-convert-number | Beats 100%😎😎 || Easy To Understand || Explained Solution | beats-100-easy-to-understand-explained-s-e0qe | \n### Intuition\nThe problem asks us to find the minimum number of bit flips needed to convert one integer (start) into another (goal). The most intuitive appro | Yash_RajSingh | NORMAL | 2024-09-11T17:42:02.378090+00:00 | 2024-09-11T17:42:02.378131+00:00 | 13 | false | \n### Intuition\nThe problem asks us to find the minimum number of bit flips needed to convert one integer (`start`) into another (`goal`). The most intuitive approach is to compare the binary representations of both numbers bit by bit and count the differences. A difference between corresponding bits indicates a neces... | 2 | 0 | ['Bit Manipulation', 'C++'] | 1 |
minimum-bit-flips-to-convert-number | ✅🔥BEATS 100.00% || ALL LANGUAGES || BEGINNER FRIENDLY 🔥✅ | beats-10000-all-languages-beginner-frien-cwxt | Intuition\n Describe your first thoughts on how to solve this problem. \n- To convert one number into another, we need to focus on the positions where the two n | surajdivekarsd27 | NORMAL | 2024-09-11T15:26:48.096692+00:00 | 2024-09-11T15:26:48.096716+00:00 | 15 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- To convert one number into another, we need to focus on the positions where the two numbers have different bits. Each bit represents either a 0 or a 1, and if the bits at the same position in both numbers are different, we need to "flip... | 2 | 0 | ['Bit Manipulation', 'C++', 'Java', 'Python3', 'JavaScript'] | 0 |
minimum-bit-flips-to-convert-number | simplest || common sense || 4 languages | simplest-common-sense-4-languages-by-pra-yu7z | Intuition\n Describe your first thoughts on how to solve this problem. \nTo find the minimum number of bits needed to make two numbers equal, we can compare the | prajwal_nimbalkar | NORMAL | 2024-09-11T13:37:51.445076+00:00 | 2024-09-11T13:37:51.445100+00:00 | 73 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo find the minimum number of bits needed to make two numbers equal, we can compare their corresponding bits. For example, if the goal is 111011 and the start is 001011, we can see that the last four bits are already identical. Therefore,... | 2 | 0 | ['Math', 'Bit Manipulation', 'C', 'Python', 'C++', 'Java'] | 0 |
minimum-bit-flips-to-convert-number | 🔥Beats 100%🔥 || ✅2 SOLUTION✅ || WITH BITCOUNT METHOD & WITHOUT✅ | beats-100-2-solution-with-bitcount-metho-wp2l | \n# Code\njava []\nclass Solution {\n public int minBitFlips(int start, int goal) {\n return Integer.bitCount(start^goal);\n }\n}\n\n\n# Code\n```j | 717822f143 | NORMAL | 2024-09-11T09:12:49.555746+00:00 | 2024-09-11T09:12:49.555766+00:00 | 37 | false | \n# Code\n```java []\nclass Solution {\n public int minBitFlips(int start, int goal) {\n return Integer.bitCount(start^goal);\n }\n}\n\n```\n# Code\n```java []\nclass Solution {\n public int minBitFlips(int start, int goal) {\n int a=start^goal,c=0;\n while(a>0){\n if(a%2==1){\n ... | 2 | 0 | ['Java'] | 0 |
minimum-bit-flips-to-convert-number | Solution By Dare2Solve | Detailed Explanation | Clean Code | solution-by-dare2solve-detailed-explanat-b5jo | Explanation []\nauthorslog.com/blog/TtRI8mE2NY\n\n# Code\n\ncpp []\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n int num = sta | Dare2Solve | NORMAL | 2024-09-11T08:31:43.879821+00:00 | 2024-09-11T17:24:24.767541+00:00 | 21 | false | ```Explanation []\nauthorslog.com/blog/TtRI8mE2NY\n```\n# Code\n\n```cpp []\nclass Solution {\npublic:\n int minBitFlips(int start, int goal) {\n int num = start ^ goal; // XOR to find differing bits\n return countSetBits(num);\n }\n\nprivate:\n int countSetBits(int num) {\n int count = 0;... | 2 | 0 | ['Bit Manipulation', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 0 |
minimum-bit-flips-to-convert-number | "Efficient Bitwise Approach to Minimum Bit Flips Calculation" | efficient-bitwise-approach-to-minimum-bi-nt2f | Intuition\nTo solve the problem of finding the minimum number of bit flips required to convert one integer (start) to another integer (goal), the key observatio | muthupalani | NORMAL | 2024-09-11T08:12:49.664892+00:00 | 2024-09-11T08:12:49.664930+00:00 | 14 | false | **Intuition**\nTo solve the problem of finding the minimum number of bit flips required to convert one integer (start) to another integer (goal), the key observation is that a bit flip is needed whenever the corresponding bits in start and goal differ. By examining the binary representation of both integers, you can de... | 2 | 0 | ['Java'] | 0 |
minimum-bit-flips-to-convert-number | Easy solution to code and understand fast | easy-solution-to-code-and-understand-fas-jsab | Intuition\n Describe your first thoughts on how to solve this problem. \nusing some logical operator\n\n# Approach\n Describe your approach to solving the probl | KARTHICK2605 | NORMAL | 2024-09-11T07:51:20.029066+00:00 | 2024-09-11T07:51:20.029090+00:00 | 64 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nusing some logical operator\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nuse xor and make count buy reducing the bit by and\n# Complexity\n- Time complexity:0ms\n<!-- Add your time complexity here, e.g. $$O(n)$$... | 2 | 0 | ['Bit Manipulation', 'Java'] | 1 |
minimum-bit-flips-to-convert-number | Minimum Bit Flips to Convert Start to Goal | minimum-bit-flips-to-convert-start-to-go-e4d5 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem involves determining the number of bit flips required to convert one intege | Suryanshrajs | NORMAL | 2024-09-11T07:38:33.484329+00:00 | 2024-09-11T07:38:33.484353+00:00 | 2 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem involves determining the number of bit flips required to convert one integer into another. To find this, we can leverage XOR. The XOR operation between two numbers will give a result where the bits are 1 wherever the two numbe... | 2 | 0 | ['C++'] | 0 |
minimum-bit-flips-to-convert-number | Using right shift operator | using-right-shift-operator-by-piratehunt-olyo | Code\ncpp []\nclass Solution {\npublic:\n int minBitFlips(int A, int B) {\n int C=A^B;\n int ans=0;\n while(C)\n { \n | piratehunter | NORMAL | 2024-09-11T06:46:16.743807+00:00 | 2024-09-11T06:46:16.743836+00:00 | 57 | false | # Code\n```cpp []\nclass Solution {\npublic:\n int minBitFlips(int A, int B) {\n int C=A^B;\n int ans=0;\n while(C)\n { \n if(C&1)\n ans++;\n C = C >> 1;\n }\n return ans;\n }\n};\n``` | 2 | 0 | ['Bit Manipulation', 'C++'] | 2 |
minimum-bit-flips-to-convert-number | Simple Easy "%" Approach - Beats 95% - Time O(n) Space O(1) | simple-easy-approach-beats-95-time-on-sp-i4zv | Intuition\nWe can use the % operator to extract the bits of a number from right to left (from the least significant bit to the most significant oane). \n\nExamp | bogdanctdev | NORMAL | 2024-09-11T06:25:03.793893+00:00 | 2024-09-11T06:25:03.793939+00:00 | 141 | false | # Intuition\nWe can use the `%` operator to extract the bits of a number from right to left (from the least significant bit to the most significant oane). \n\nExample:\n- `10` in binary is `1010`. Using `% 2` iteratively on `10` gives the bits `0101` (starting from the least significant bit);\n- `7` in binary is `111` ... | 2 | 0 | ['Bit Manipulation', 'C#'] | 0 |
minimum-bit-flips-to-convert-number | XOR operation (Playing with bits) explained about XOR operation!! with Dry run 💯✅ | xor-operation-playing-with-bits-explaine-gbd4 | Understanding XOR operation\n- If both the bits are same results in 0, otherwise 1.\n\n\n# Approach\n Describe your approach to solving the problem. \n - Perfor | PavanKumarMeesala | NORMAL | 2024-09-11T06:20:19.612226+00:00 | 2024-09-11T06:20:19.612258+00:00 | 5 | false | # Understanding XOR operation\n- If both the bits are same results in 0, otherwise 1.\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n - Perform the XOR operation between `start and... | 2 | 0 | ['Math', 'Bit Manipulation', 'Java'] | 0 |
minimum-bit-flips-to-convert-number | One Line code with Explanation | one-line-code-with-explanation-by-psriji-a8q8 | Explanation \n Describe your first thoughts on how to solve this problem. \n For start = 10 and goal = 7:\n 10 in binary is 1010\n 7 in binary is 0111\ | psrijith | NORMAL | 2024-09-11T06:05:04.398742+00:00 | 2024-09-11T06:05:04.398774+00:00 | 29 | false | # Explanation \n<!-- Describe your first thoughts on how to solve this problem. -->\n For start = 10 and goal = 7:\n 10 in binary is 1010\n 7 in binary is 0111\n 10 ^ 7 gives 1101\n The binary 1101 contains three 1s, \n so the number of bit flips required is 3.\n\n# Code\n```python3 []\nclass Solution... | 2 | 0 | ['Python3'] | 0 |
minimum-bit-flips-to-convert-number | Minimum Bit Flips to Convert Integer - Efficient Bitwise Solution | minimum-bit-flips-to-convert-integer-eff-gkm1 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem asks us to determine how many bit flips are needed to convert one integer ( | namratha2604 | NORMAL | 2024-09-11T05:51:16.314040+00:00 | 2024-09-11T05:51:16.314073+00:00 | 15 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem asks us to determine how many bit flips are needed to convert one integer (start) into another (goal). My intuition is that this problem can be solved by comparing the binary representations of both numbers and counting the po... | 2 | 0 | ['C++'] | 0 |
minimum-bit-flips-to-convert-number | Straight forward C++ solution | Beats 100%🚀 | straight-forward-c-solution-beats-100-by-ojko | \n# Approach\nLet\'s consider start to be 10 and end to be 7\nbinary representation of these two \n1 0 1 0\n0 1 1 1\nwe have to check number of bits not matchin | ChinmayaRao | NORMAL | 2024-09-11T05:30:45.380805+00:00 | 2024-09-11T05:30:45.380843+00:00 | 31 | false | \n# Approach\nLet\'s consider start to be 10 and end to be 7\nbinary representation of these two \n1 0 1 0\n0 1 1 1\nwe have to check number of bits not matching.\n(XOR approach) if bits are different result is 1. Else result is 0.\nwe have to compare the leftmost bits.\nif the bits are different then we have to flip i... | 2 | 0 | ['Bit Manipulation', 'C++'] | 0 |
minimum-bit-flips-to-convert-number | Best way of approach in java using XOR operation ~cksolutions.. (O(1)) | best-way-of-approach-in-java-using-xor-o-r1a8 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nOne of yhe best approa | kandiah_01 | NORMAL | 2024-09-11T05:00:05.000479+00:00 | 2024-09-11T05:00:05.000507+00:00 | 76 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nOne of yhe best approach is to use *XOR OPERATION* \n#Remember XOR operation#\nRule-1: 1 ^ 0 = 1 or 0 ^ 1 = 1\nRule-2: 0 ^ 0 = 0 or 1 ^ 1 = 0\n\nFirst step is to do x... | 2 | 0 | ['Bit Manipulation', 'Java'] | 1 |
minimum-bit-flips-to-convert-number | 🔥✅💯1 Line Most Easy Solution With Detailed Explanation💯✅🔥 | 1-line-most-easy-solution-with-detailed-tgwv7 | Intuition\nThe problem is asking for the minimum number of bit flips required to convert one number (start) into another number (goal). A bit flip means changin | Ajay_Kartheek | NORMAL | 2024-09-11T04:22:22.839062+00:00 | 2024-09-11T04:22:22.839097+00:00 | 23 | false | # Intuition\nThe problem is asking for the minimum number of bit flips required to convert one number (start) into another number (goal). A bit flip means changing a 0 to 1 or a 1 to 0.\n\nTo solve this, we can use the XOR operation. XOR (^ operator in Python) will give a 1 wherever the bits of start and goal differ, a... | 2 | 0 | ['Bit Manipulation', 'C++', 'Java', 'Python3'] | 0 |
minimum-bit-flips-to-convert-number | Beginner-to-Advanced Solution with explanation CPP | beginner-to-advanced-solution-with-expla-nzth | Intuition\nThe problem asks for the minimum number of bit flips required to convert one integer (start) to another (goal). To solve this, we can leverage the bi | jaydeep-pro | NORMAL | 2024-09-11T04:11:33.072548+00:00 | 2024-09-11T04:11:33.072578+00:00 | 21 | false | # Intuition\nThe problem asks for the minimum number of bit flips required to convert one integer (start) to another (goal). To solve this, we can leverage the binary representations of both numbers. By comparing the binary digits of start and goal, we can determine how many positions differ, and each difference corres... | 2 | 0 | ['C++'] | 0 |
minimum-bit-flips-to-convert-number | Easy ✅, VERY Easy❎ || Beats 100%🚀 | easy-very-easy-beats-100-by-ayushiivarsh-1s3s | \n# Code\njava []\nclass Solution {\n public int minBitFlips(int start, int goal) {\n //checking how many bits has been flipped\n int xor = sta | AyushiiVarshney | NORMAL | 2024-09-11T04:08:15.164387+00:00 | 2024-09-11T04:08:15.164417+00:00 | 29 | false | \n# Code\n```java []\nclass Solution {\n public int minBitFlips(int start, int goal) {\n //checking how many bits has been flipped\n int xor = start ^ goal;\n \n // to count the no. of changed bits\n int count = 0;\n \n while (xor != 0) {\n // Increment count f... | 2 | 0 | ['Bit Manipulation', 'Java'] | 1 |
minimum-bit-flips-to-convert-number | 💡🔥💥 Easy Python Solution Without Bit Manipulation 💥🔥💡 | easy-python-solution-without-bit-manipul-i5fj | Approach\n Describe your approach to solving the problem. \n1. Convert to Binary Representation:\n\nConvert both start and goal to their binary representations | eknath_mali_002 | NORMAL | 2024-09-11T03:58:43.463397+00:00 | 2024-09-11T03:58:43.463427+00:00 | 28 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Convert to Binary Representation:**\n\nConvert both start and goal to their binary representations using the bin() function.\nRemove the `\'0b\'` prefix that Python includes in the output of bin() by slicing the string [2:].\n2. **Pad the Shorter... | 2 | 0 | ['Bit Manipulation', 'Python3'] | 0 |
minimum-bit-flips-to-convert-number | 𝑂(log 𝑛) | 0ms Beats 100.00% | Easy Solution | olog-n-0ms-beats-10000-easy-solution-by-a5a98 | \n\n---\n# Intuition\nTo convert the binary representation of start to goal, we need to count the number of bit positions where the two numbers differ. This is | user4612MW | NORMAL | 2024-09-11T03:08:15.322095+00:00 | 2024-09-11T03:16:19.179325+00:00 | 17 | false | #\n\n---\n# Intuition\nTo convert the binary representation of start to goal, we need to count the number of bit positions where the two numbers differ. This is equivalent to counting the number of 1\'s in the XOR of start and goal, as XOR highlights the differing bits (1 for different, 0 for the same).\n\n# Approach\n... | 2 | 0 | ['C++', 'Java', 'Python3'] | 0 |
apply-operations-to-make-all-array-elements-equal-to-zero | [Java/C++/Python] Greedy + Sliding Window | javacpython-greedy-sliding-window-by-lee-jshg | Intuition\nFor the first A[i] > 0,\nwe need to select the array A[i],A[i+1]..A[i+k-1]\nand decrease all these elements by A[i].\n\n\n# Explanation\nThe subarray | lee215 | NORMAL | 2023-07-09T04:12:16.758961+00:00 | 2023-07-09T04:12:16.758979+00:00 | 13,788 | false | # **Intuition**\nFor the first `A[i] > 0`,\nwe need to select the array `A[i],A[i+1]..A[i+k-1]`\nand decrease all these elements by `A[i]`.\n<br>\n\n# **Explanation**\nThe subarray deleted starting at `A[i]`,\nwill affect the `A[i+1], A[i+2], ...A[i+k-1]`.\n\nSo we can use `cur` to record the sum of previous `k - 1` el... | 131 | 1 | ['C', 'Python', 'Java'] | 20 |
apply-operations-to-make-all-array-elements-equal-to-zero | +1/-1 trick | c++, python, javascript and extra segment tree/BIT solution | 1-1-trick-c-python-javascript-and-extra-w7tv2 | Approach\nLet\'s look for the first non-zero value from left to right, let\'s call the index $i$, since it is the first one if we use a window starting at a low | BetoSCL | NORMAL | 2023-07-09T04:03:40.232586+00:00 | 2023-07-12T04:00:36.695988+00:00 | 5,662 | false | # Approach\nLet\'s look for the first non-zero value from left to right, let\'s call the index $i$, since it is the first one if we use a window starting at a lower index than $i$ we will affect some zero values and transform them into negative which is not valid, and a window starting greater than $i$ cannot help eith... | 29 | 2 | ['Binary Indexed Tree', 'Segment Tree', 'C++', 'Python3', 'JavaScript'] | 9 |
apply-operations-to-make-all-array-elements-equal-to-zero | | [JAVA] | Simple Solution | Sliding Window | | java-simple-solution-sliding-window-by-k-1g7u | Intuition | [JAVA] | Simple Solution | Sliding Window\n Describe your first thoughts on how to solve this problem. \n\n# Approach : Sliding Window\n Describe y | kartikeylapy | NORMAL | 2023-07-09T07:30:59.781653+00:00 | 2023-07-09T07:30:59.781683+00:00 | 954 | false | # Intuition | [JAVA] | Simple Solution | Sliding Window\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach : Sliding Window\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n*k)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Spa... | 18 | 1 | ['Java'] | 1 |
apply-operations-to-make-all-array-elements-equal-to-zero | ✅ [C++] | Simple Solution | Range Prefix Sum | c-simple-solution-range-prefix-sum-by-00-o9xm | Approach : Range Prefix Sum\n\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(n)\n\n# Code\n\nclass Solution {\npublic:\n bool checkArray(vec | 007_abhishek | NORMAL | 2023-07-09T04:03:04.979715+00:00 | 2023-07-09T16:07:24.356618+00:00 | 4,498 | false | # Approach : Range Prefix Sum\n\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(n)\n\n# Code\n```\nclass Solution {\npublic:\n bool checkArray(vector<int>& nums, int k) {\n int n=nums.size();\n if(n==1) return true;\n \n vector<int> temp(n,0);\n int i=0;\n te... | 17 | 1 | ['Prefix Sum', 'C++'] | 2 |
apply-operations-to-make-all-array-elements-equal-to-zero | C++ || Sliding window technique || Beginner Friendly | c-sliding-window-technique-beginner-frie-xtyw | Intuition\n Describe your first thoughts on how to solve this problem. \n When I am at the start position of current window, my goal will be to make this first | ChelsiGarg | NORMAL | 2023-07-10T08:53:57.862444+00:00 | 2023-07-10T08:53:57.862469+00:00 | 1,377 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n When I am at the start position of current window, my goal will be to make this first element zero so that i can go for next window (window size will be of length k). Now, let first element of this current window is ***sub*** then I will... | 13 | 0 | ['C++'] | 3 |
apply-operations-to-make-all-array-elements-equal-to-zero | Line by Line explained Short Simple C++ solution | line-by-line-explained-short-simple-c-so-qh2k | Intuition\nWindow size is k .\nWe don\'t care how many times we have to apply operations.\n \nWe just care about whether it is possible to convert all elements | Bijay293 | NORMAL | 2023-07-10T06:29:24.755074+00:00 | 2023-07-12T13:29:38.193435+00:00 | 1,286 | false | # Intuition\nWindow size is k .\nWe don\'t care how many times we have to apply operations.\n \nWe just care about whether it is possible to convert all elements to 0 or not.\n\nSo for example :\n2 2 3 1 1 taking k = 3 and window starting at index 0 we can get \n0 0 1 1 1 by applying some number of operations (we dont... | 10 | 0 | ['Sliding Window', 'C++'] | 2 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.