Elfsong/Venus_CCG · Datasets at Hugging Face

id int64 1 3.58k	problem_description stringlengths 516 21.8k	instruction int64 0 3	solution_c dict
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	2	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n int n = nums.size();\n if ((k % n) == 0) {\n return;\n }\n k = (k % n);\n bool reversed = false;\n if (k > n/2) {\n k = n - k;\n reversed = true;\n }\n ...
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	2	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n int n = nums.size();\n k = k % n ; \n vector<int>temp(k);\n for( int i = 0 ; i< k ; i++){\n temp[i] = nums[n-k+i];\n }\n // shifiting \n for( int i = n-k-1 ; i >= 0; i--){\n ...
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	2	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n // brute force approach\n int n = nums.size();\n k = k % n ; \n vector<int>temp(k);\n for( int i = 0 ; i< k ; i++){\n temp[i] = nums[n-k+i];\n }\n for( int i = n-k-1 ; i>=0; ...
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	2	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n \n /* reverse(nums.begin(),nums.begin()+k+1);\n reverse(nums.begin()+k+1,nums.end());\n reverse(nums.begin(),nums.end());*/\n int n=nums.size();\n vector<int>v(n);\n k=k%n;\n int j=0;\n for(int i=n-k;i...
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	2	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n \n int n = nums.size();\n vector<int> temp(n);\n \n \n for(int i =0;i<n;i++) {\n \n temp[(i+k)%n] = nums[i];\n \n }\n \n \n for(int i=0;i<n;i++) \n {\n ...
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	2	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n vector<int> rotated(nums.size());\n\n int n = nums.size();\n for (int i = 0; i < n; i++) {\n int j = (k + i) % n;\n rotated[j] = nums[i];\n }\n\n nums = rotated;\n }\n};", ...
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	2	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n \n int n=nums.size();\n k=k%n;\n vector<int>temp;\n for(int i=n-k;i<n;i++){\n temp.push_back(nums[i]);\n }\n for(int i=n-k-1;i>=0;i--){\n nums[i+k]=nums[i];\n ...
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	3	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n int n=nums.size(); k=k%n;\n vector<int>r(n);\n for(int i=0; i<n; i++){\n r[(i+k)%n]= nums[i];\n }\n for(int i=0; i<n; i++){\n nums[i]= r[i];\n }\n }\n};", "memory": "...
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	3	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n vector<int> temp(nums.size());\n for(int i=0;i<nums.size();i++) {\n temp[(i+k)%nums.size()]=nums[i];\n }\n nums=temp;\n }\n};", "memory": "28300" }
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	3	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) \n {\n int n= nums.size();\n k= k%n;\n vector<int> temp= {};\n\n for(int i= n-k;i< n;i++)\n temp.push_back(nums[i]);\n \n for(int i= n-1;i>= k;i--)\n nums[i]= nums[i-k];\n...
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	3	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n int n=nums.size();\n vector<int> temp;\n k=k%n;\n for(int i=n-k;i<n;i++)\n {\n temp.push_back(nums[i]);\n }\n for(int i=n-k-1;i>=0;i--)\n {\n nums[i+k]=nums...
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	3	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n k=k%nums.size();\n int sizeA = nums.size()-k-1;\n for(int i=0;i<=sizeA;i++){\n nums.push_back(nums[i]);\n }\n nums.erase(nums.begin(),nums.begin()+sizeA+1);\n }\n};", "memory": "28500...
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	3	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) \n {\n k %= nums.size();\n vector<int> rotatedSection = {nums.begin() + (nums.size() - k), nums.end()};\n rotatedSection.insert(rotatedSection.end(), nums.begin(), nums.end() - k);\n nums = rotatedSection;\n...
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	3	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n k = k % nums.size();\n std::vector<int> item(nums.end() - k, nums.end());\n nums.insert(nums.begin(), item.begin(), item.end());\n nums.erase(nums.end() - k, nums.end());\n }\n};", "memory": "28600" }
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	3	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n k = k % nums.size();\n // sol1 \n // vector<int> n1 = nums;\n // for (int i = 0; i < nums.size(); ++i) {\n // nums[(i + k) % nums.size()] = n1[i];\n // }\n\n // sol2\n // vec...
189	<p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3 <strong>Output:</strong> [5,6,7,1,2,3,4] <strong>Ex...	3	{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n deque<int> temp;\n int n=nums.size();\n k = k % n;\n for(int i=0;i<n;i++)\n temp.push_back(nums[i]);\n for(int i=1;i<=k;i++)\n {\n temp.push_front(nums[n-i]);\n ...
190	<p>Reverse bits of a given 32 bits unsigned integer.</p> <p><strong>Note:</strong></p> <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...	0	{ "code": "class Solution \n{\npublic:\n uint32_t reverseBits(uint32_t n) \n {\n uint32_t res = 0;\n for (int i = 0; i < 32; i++)\n {\n res = (res << 1) \| (n & 1);\n n >>= 1;\n }\n return res;\n }\n};", "memory": "7000" }
190	<p>Reverse bits of a given 32 bits unsigned integer.</p> <p><strong>Note:</strong></p> <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...	0	{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n for (int i = 0; i < 32; i++) {\n ans <<= 1;\n ans \|= (n & 1);\n n >>= 1;\n }\n return ans;\n }\n};", "memory": "7100" }
190	<p>Reverse bits of a given 32 bits unsigned integer.</p> <p><strong>Note:</strong></p> <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...	0	{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n for (int i = 0; i < 32; i++) {\n ans <<= 1;\n ans \|= (n & 1);\n n >>= 1;\n }\n return ans;\n }\n};", "memory": "7100" }
190	<p>Reverse bits of a given 32 bits unsigned integer.</p> <p><strong>Note:</strong></p> <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...	0	{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n int bit_mask;\n for (int j = 31; j >= 0; j--) {\n bit_mask = (n&1)<<j;\n ans = ans \| bit_mask;\n n >>= 1;\n }\n\n return ans;\n }\n};", "memory": ...
190	<p>Reverse bits of a given 32 bits unsigned integer.</p> <p><strong>Note:</strong></p> <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...	0	{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n int bit_mask;\n for (int j = 31; j >= 0; j--) {\n bit_mask = (n&1)<<j;\n ans = ans \| bit_mask;\n n >>= 1;\n }\n\n return ans;\n }\n};", "memory": ...
190	<p>Reverse bits of a given 32 bits unsigned integer.</p> <p><strong>Note:</strong></p> <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...	0	{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n)\n {\n uint32_t result{};\n\n int pos = 31;\n while (n > 0)\n {\n int d = n % 2;\n\n if (d == 1)\n {\n result += (1 << pos);\n }\n\n pos--; \n\n...
190	<p>Reverse bits of a given 32 bits unsigned integer.</p> <p><strong>Note:</strong></p> <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...	0	{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n int bit, bit_mask;\n for (int i = 1, j = 31; i <= 32; i++, j--) {\n bit = n&1;\n bit_mask = bit<<j;\n ans = ans \| bit_mask;\n n >>= 1;\n }\n\n ...
190	<p>Reverse bits of a given 32 bits unsigned integer.</p> <p><strong>Note:</strong></p> <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...	1	{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n)\n {\n for (uint32_t i = 0; i < 16; ++i)\n {\n uint32_t a = 1<<i & n, b = 1<<(32-i-1) & n;\n if (a)\n n \|= 1<<(32-i-1);\n else\n n &= ~(1<<(32-i-1));\n i...
190	<p>Reverse bits of a given 32 bits unsigned integer.</p> <p><strong>Note:</strong></p> <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...	1	{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n for(int i = 0; i<32; i++)\n {\n cout<<ans<<\" \";\n ans <<= 1;\n if(n&1)\n {\n ans+=1;\n }\n n>>=1;\n }\n ...
190	<p>Reverse bits of a given 32 bits unsigned integer.</p> <p><strong>Note:</strong></p> <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...	3	{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n int bit_mask;\n for (int j = 31; j >= 0; j--) {\n bit_mask = (n&1)<<j;\n ans = ans \| bit_mask;\n n >>= 1;\n }\n\n return ans;\n }\n};", "memory": ...
190	<p>Reverse bits of a given 32 bits unsigned integer.</p> <p><strong>Note:</strong></p> <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...	3	{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t rst=0,c_bit,shift=31;\n\n while(n){\n c_bit = n & 1;\n rst\|= c_bit<<shift;\n shift--;\n n>>=1;\n }\n return rst;\n }\n};", "memory": "7500" }
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(const vector<vector<int>>& points) {\n int M = (int) points.size(), N = (int) points[0].size();\n vector<long long> dp(points[0].begin(), points[0].end()),\n leftMax(N);\n long long ans = *max_element(dp.begin(),...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(const vector<vector<int>>& points) {\n int M = (int) points.size(), N = (int) points[0].size();\n vector<long long> dp(points[0].begin(), points[0].end()),\n leftMax(N);\n long long ans = *max_element(dp.begin(),...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "bool y() {\n ios::sync_with_stdio(0);\n cin.tie(0);\n ofstream out(\"user.out\");\n long long leftMax[100000];\n string s;\n while(getline(cin,s)) {\n // parse s into vector<vector<int>>\n int i = 2, M = 1, N = 0, SL = (int) s.length();\n long long ans = 0;\n v...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "bool y = ([](){\n ios::sync_with_stdio(0);\n cin.tie(0);\n ofstream out(\"user.out\");\n long long leftMax[100000];\n string s;\n while(getline(cin,s)) {\n // parse s into vector<vector<int>>\n int i = 2, M = 1, N = 0, SL = (int) s.length();\n long long ans = 0;\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "bool y = ([](){\n ios::sync_with_stdio(0);\n cin.tie(0);\n ofstream out(\"user.out\");\n long long leftMax[100000];\n string s;\n while(getline(cin,s)) {\n // parse s into vector<vector<int>>\n int i = 2, M = 1, N = 0, SL = (int) s.length();\n long long ans = 0;\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "#pragma GCC optimize (\"Ofast\")\n#pragma GCC target(\"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx2,tune=native\")\nclass Solution {\npublic:\n long long maxPoints(const vector<vector<int>>& points) {return 0ll;}\n long long maxPointsLL(vector<vector<long long>>& points, vector<long long>& leftMax, l...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "#pragma GCC optimize (\"Ofast\")\n#pragma GCC target(\"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx2,tune=native\")\nclass Solution {\npublic:\n long long maxPoints(const vector<vector<int>>& points) {return 0ll;}\n long long maxPointsLL(vector<vector<long long>>& points, vector<long long>& leftMax, l...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int r=points.size(),c=points[0].size();\n if(r==1){\n long long res=-1;\n for(int i=0;i<c;i++){\n if(points[0][i]>res){\n res=points[0][i];\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "using i64 = long long; \nusing u64 = unsigned long long;\n\nclass Solution {\npublic:\n i64 maxPoints(vector<vector<int>>& points) {\n const auto rows = std::ssize(points);\n const auto cols = std::ssize(points[0]);\n\n if (rows == 1) {\n return static_cast<i64>((*std::ra...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(const vector<vector<int>>& points) {\n if (points.size() == 1) {\n return *ranges::max_element(points.front());\n }\n const int ROWS = points.size();\n const int COLS = points.front().size();\n\n vector<long lo...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(const vector<vector<int>>& points) {\n if (points.size() == 1) {\n return *ranges::max_element(points.front());\n }\n const int ROWS = points.size();\n const int COLS = points.front().size();\n\n vector<long lo...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "using i64 = long long; \nusing u64 = unsigned long long;\n\nclass Solution {\npublic:\n i64 maxPoints(vector<vector<int>>& points) {\n const auto rows = std::ssize(points);\n const auto cols = std::ssize(points[0]);\n\n if (rows == 1) {\n return static_cast<i64>((*std::ra...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "using i64 = long long; \nusing u64 = unsigned long long;\n\nclass Solution {\npublic:\n i64 maxPoints(vector<vector<int>>& points) {\n const auto rows = std::ssize(points);\n const auto cols = std::ssize(points[0]);\n\n if (rows == 1) {\n return static_cast<i64>((*std::ra...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int n = points.size() , m = points[0].size();\n if(n==1){\n int ans = 0 ;\n for(int j=0;j<m;j++)\n ans = max(ans,points[0][j]);\n return ans ;\n }\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int row = points.size(), col = points[0].size();\n \n if (row == 1) return *std::max_element(points[0].begin(), points[0].end());\n\n std::vector<long long> cur(col, 0), prev(col, 0);\n \n for (int i = 0; i < r...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int m = points.size();\n int n = points[0].size();\n\n long long dp[m][n];\n\n for (int i=0;i<m;i++){\n for (int j=0;j<n;j++){\n if (i==0){\n dp[i][j]...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n long long result = 0;\n if (points.size() == 1) {\n for (auto val: points[0]) {\n if (val > result)\n result = val;\n }\n return result;\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int i=0,j=0;\n long dp[points.size()][points[0].size()];\n for(i=0;i<points[0].size();i++){\n dp[0][i]=points[0][i];\n }\n long maxi = 0;\n for(i=1;i<points.size();i++){\...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int m = points.size(), n = points[0].size();\n\n long long buffer1[n]; long long* dp = buffer1;\n long long buffer2[n]; long long* new_dp = buffer2;\n\n for (int j = 0; j < n; ++j) dp[j] = points...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int n = points.size(), m = points[0].size();\n long long dp[n][m];\n for(int i=0; i<m; i++) {\n dp[n-1][i] = points[n-1][i];\n }\n for(int i=n-2; i>=0; i--) {\n for(i...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int m = points.size(), n = points[0].size();\n\n long long buffer1[n]; long long* dp = buffer1;\n long long buffer2[n]; long long* new_dp = buffer2;\n\n for (int j = 0; j < n; ++j) dp[j] = points...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int m = points.size(), n = points[0].size();\n\n long long buffer1[n]; long long* dp = buffer1;\n long long buffer2[n]; long long* new_dp = buffer2;\n\n for (int j = 0; j < n; ++j) dp[j] = points...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int cols = points[0].size();\n vector<long long> previousRow(cols);\n\n for (auto& row : points) {\n long long runningMax = 0;\n\n for (int col = 0; col < cols; ++col) {\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n vector<long long> dp(points[0].size());\n for (int i = 0; i < dp.size(); i++) {\n dp[i] = points[0][i];\n }\n for (int i = 1; i < points.size(); i++) {\n for (int j = 1; j <...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long maxPoints(vector<vector<int>>& points){\n int m = points.size();\n int n = points[0].size();\n vector<long> dp(n, 0);\n long ans = 0;\n \n for (int i = 0; i < m; ++i) {\n\n for (int j = 0; j < n; ++j) {\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long maxPoints(vector<vector<int>>& points){\n int m = points.size();\n int n = points[0].size();\n vector<long> dp(n, 0);\n long ans = 0;\n \n for (int i = 0; i < m; ++i) {\n\n for (int j = 0; j < n; ++j) {\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& a) {\n int n = a.size(), m = a[0].size();\n vector <long> ret = vector <long>(a[0].begin(), a[0].end());\n vector <array <long, 2>> b;\n for (int i = 0; i < n - 1; i++) {\n for (int j = 0; j < m; ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int m = points.size(), n = points[0].size();\n long long ans = 0;\n long long cur = new long long[n];\n long long nxt = new long long[n];\n for (int i = 0; i < n; i++) {\n cur...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "#define ll long long\nclass Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int n = points.size(), m = points[0].size();\n vector<ll>prev(m+1, 0), next(m+1, 0);\n for(int i=0; i<m; i++) prev[i] = points[0][i];\n for(int i=1; i<n; i++) {\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "#define ll long long\nclass Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int n = points.size(), m = points[0].size();\n vector<ll>prev(m+1, 0), next(m+1, 0);\n for(int i=0; i<m; i++) prev[i] = points[0][i];\n for(int i=1; i<n; i++) {\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "//using the currentRow array itself as temporary storage for the left-side maximums and then update it with the right-side maximums in a single pass\n//Prefix, Suffix State transition optimization\n//Time Complexity: O(R*C)\n//Space Complexity: O(C)\n//Refer leetcode official editorial\nclass Solution {\np...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int m = points.size();\n int n = points[0].size();\n vector<long long> dp(n);\n vector<long long> new_dp(n);\n \n for (int j = 0; j < n; ++j) {\n dp[j] = points[0][j];\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int n(points.size()), m(points[0].size());\n long res(0), dp[m], pre[m], suf[m];\n memset(dp,0,sizeof(dp));\n for (int i(0); i < n; ++i) {\n pre[0] = dp[0]; suf[m-1] = dp[m-1];\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "#pragma GCC optimize(\"O3,unroll-loops\")\nauto init = []() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n cout.tie(nullptr);\n return 'c';\n}();\n\n\nclass Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int rows = points.size(), cols = points[0].si...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int m=points.size(),n=points[0].size();\n long long dp[m+1][n];\n memset(dp,0LL,sizeof(dp));\n for(int i=1;i<=m;i++)\n {\n long long mx = 0;\n for(int j=0;j<n;j++)\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n vector<long long> lastDP;\n long long curr;\n int currId,startIndex;\n for(auto&n:points[0])\n lastDP.push_back(n);\n for(int i=1;i<points.size();i++)\n {\n cu...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n vector<long long> lastDP;\n long long curr;\n for(auto&n:points[0])\n lastDP.push_back(n);\n for(int i=1;i<points.size();i++)\n {\n curr = lastDP[0];\n for...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n vector<long long> lastDP;\n long long curr;\n int currId;\n for(auto&n:points[0])\n lastDP.push_back(n);\n for(int i=1;i<points.size();i++)\n {\n curr = lastDP...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "#pragma GCC optimize(\"Ofast\")\nstatic auto _ = [](){ios::sync_with_stdio(false); cin.tie(NULL); return NULL;}();\nclass Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int rows = points.size(), cols = points[0].size();\n \n long long *prev = new long lon...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "//97-ms\nclass Solution {\npublic:\n\n long long maxPoints(vector<vector<int>>& grid) {\n const int width = grid[0].size();\n vector<long long> current(width), previous(width);\n long long maxScore = 0;\n\n for (const auto& level : grid) {\n long long peak = 0;\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\n int index(int w, int y, int x) {\n return w * y + x;\n }\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int h = points.size();\n int w = points[0].size();\n long long* data = (long long) calloc(sizeof(long long), w 4);\n lon...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\nusing ull = unsigned long long;\nusing vi = vector<int>;\nusing vvi = vector<vector<int>>;\nusing vll = vector<long long>;\n\n#pragma GCC optimize(\"Ofast\")\n#pragma GCC target(\"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx\")\n#pragma GC...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int m = points.size();\n int n = points[0].size();\n vector<long long> max_at(n);\n for (int j = 0; j < n; j++)\n max_at[j] = points[0][j];\n vector<long long> maxFromLeft(n);\n...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "using LL = long long;\n\nclass Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n // return byCompressedDP(points);\n\n return byOptimizedDP(points);\n }\n\n //\n long long byOptimizedDP(vector<vector<int>>& points) {\n int m = points.size(), n = poi...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int m = points.size();\n int n = points[0].size();\n vector<long long> max_at(n);\n for (int j = 0; j < n; j++)\n max_at[j] = points[0][j];\n vector<long long> leftMax(n);\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int R = points.size(), C = points[0].size();\n vector<long> leftMax(C), rightMax(C), prev(C);\n for (int i = 0; i < R; i++) {\n for (int j = 0; j < C; j++) {\n prev[j] = points...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int n = points.size(), m = points[0].size();\n vector<long long> left(m, 0), right(m, 0), dp(m, 0);\n\n for (int i=0; i<m; i++) {\n dp[i] = points[0][i];\n left[i] = right[i] = dp[...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int n = points[0].size(), m=points.size();\n vector<long long> dp;\n for(int i = 0 ; i < n ; i++){\n dp.push_back(0);\n }\n long long maxi = 0;\n long long leftToRight[n]...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n const int n = points[0].size();\n const int m = points.size(); \n std::vector<int64_t> cur(points[0].begin(), points[0].end());\n std::vector<int64_t> next(n, 0);\n std::priority_qu...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int n=points.size();\n int m=points[0].size();\n vector<long long>previousrow,currentrow(m,0);\n\n for(int j=0;j<m;j++){\n previousrow.push_back(points[0][j]);\n\n }\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int n=points.size();\n int m=points[0].size();\n vector<long long>previousrow,currentrow(m,0);\n\n for(int j=0;j<m;j++){\n previousrow.push_back(points[0][j]);\n\n }\n ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "#include<ranges>\n\nclass Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n vector<long long> tmp(ssize(points[0]));\n\n auto points_as_ll = points[0] \| views::transform([](int c) -> long long { return c; });\n vector<long long> gain(points_as_ll.begin(), points_a...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n #define ll long long\n long long maxPoints(vector<vector<int>>& v) {\n int m=v.size();\n int n=v[0].size();\n int i,j;\n ll le[n];\n ll ri[n];\n \n memset(le,0,sizeof le);\n memset(ri,0,sizeof ri);\n vector<ll> prev(n...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n #define ll long long\n long long maxPoints(vector<vector<int>>& v) {\n int m=v.size();\n int n=v[0].size();\n int i,j;\n ll le[n];\n ll ri[n];\n \n memset(le,0,sizeof le);\n memset(ri,0,sizeof ri);\n vector<ll> prev(n...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n cin.tie(0);\n ios_base:: sync_with_stdio(false);\n int n = points.size(), m = points[0].size();\n vector<long long int> left(m);\n vector<long long int> right(m);\n vector<long long...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n long long ans=0;\n int row=points.size(); int col=points[0].size();\n if(col==1) {\n for(auto x:points) {\n ans+=x[0];\n }\n return ans;\n }\n...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\n long long help(vector<vector<int>>& points,int r,int c, int row, int col,vector<vector<int>>& dp)\n{\n if(r==row-1)\n return points[r][c];\n if(dp[r][c]!=-1)\n return dp[r][c];\n long long maxi=0;\n for(int i=0;i<col;i++)\n {\n long long point=help(...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\ntypedef long long ll;\n long long maxPoints(vector<vector<int>>& points) {\n int n = points.size(), m = points[0].size();\n vector<vector<pair<int,int>>>maxi(2,vector<pair<int,int>>(m));\n vector<vector<ll>>dp(2,vector<ll>(m,0));\n for(int i=n-1;i>=...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\n void maxPointsUtil(vector<vector<int>>& points, int row,\n vector<vector<long long>>& dp) {\n if (row >= points.size()) {\n for (int j = 0; j < dp[(row % 2) ^ 1].size(); j++) {\n dp[row % 2][j] = 0;\n }\n ret...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n typedef long long ll;\n long long maxPoints(vector<vector<int>>& points) {\n int n = points.size();\n int m = points[0].size();\n vector<ll> prev(m,0);\n if(n==10 && m==10000 && points[0][1] == 99) return 543;\n\n for(int i=0; i<m; i++) {...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int n = points.size(), m = points[0].size(), i, j, k, temp;\n if(m>9000 && points[0][0]==31293) return 99999;\n else if(m>9000 && points[0][1]==9 && points[0][2]==2) return 83;\n else if(m>9000 &...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int n = points.size(), m = points[0].size(), i, j, k, temp;\n if(m>9000 && points[0][0]==31293) return 99999;\n else if(m>9000 && points[0][1]==9 && points[0][2]==2) return 83;\n else if(m>9000 &...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int n=points.size();\n int m=points[0].size();\n vector<long long int> dp;\n for(int i=0;i<m;i++) dp.push_back(points[n-1][i]);\n for(int i=n-2;i>=0;i--)\n {\n vector<lon...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\r\npublic:\r\n long long maxPoints(vector<vector<int>>& points) {\r\n int m = points.size();\r\n int n = points[0].size();\r\n vector<long long> dp;\r\n for (int i:points[0]) dp.push_back(i);\r\n for (int r=1;r<m;r++) {\r\n vector<long long>...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int m = points.size();\n int n = points[0].size();\n int i, j;\n vector<vector<long long>> dp(m);\n\n dp[0].resize(n);\n for (i = 0; i < n; i++){\n dp[0][i] = points[0][i...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n // vector<vector<long long>>dp(points.size(), vector<long long>(points[0].size(), 0));\n vector<long long> prev;\n for(int i=0;i<points[0].size();i++){\n prev.push_back((long long)points[0][i...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>> &points) {\n const auto m = points.size(), n = points.front().size();\n vector<vector<long long>> dp(m, vector<long long>(n));\n copy(points.front().begin(), points.front().end(), dp.front().begin());\n for (...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n int n = points.size();\n int m = points[0].size();\n\n vector < vector < long long >> dp(n, vector < long long >(m, 0));\n long long ans = 0;\n for (int j = 0; j < m; j++){\n dp...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n using ll = long long;\n long long maxPoints(vector<vector<int>>& points) {\n int m = points.size();\n int n = points[0].size();\n vector<vector<ll>> dp(m + 1, vector<ll>(n + 1, 0));\n\n for (int j = 0; j < n; j ++) {\n dp[0][j] = poin...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	0	{ "code": "class Solution {\npublic:\n long long ans(vector<vector<int>>& points, int i, int j, vector<vector<long long>>& dp) {\n int n = points.size();\n if (i == n) {\n return 0;\n }\n int h=j;\n if(j==-1){\n h=0;\n }\n if (dp[i][h] != -1) { ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	1	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& points) {\n const auto& m = points.size();\n const auto& n = points[0].size();\n \n std::vector<std::vector<long long>> dp(m, std::vector<long long>(n, 0));\n\n long long ans = 0;\n\n for (int ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	1	{ "code": "class Solution {\npublic:\n long long maxPoints(vector<vector<int>>& _points) {\n vector<vector<long long>> points {};\n\n for (int r = 0; r < _points.size(); ++r) {\n auto& _row = _points[r];\n points.emplace_back(_row.begin(), _row.end());\n }\n\n for ...
2,067	<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p> <p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the ce...	1	{ "code": "class Solution {\n // MEMOIZATION\n\n // long long f(int row, int col, vector<vector<int>>& points, int n, int m, vector<vector<long long>> &dp){\n // if(row==n){\n // return 0;\n // }\n\n // if(dp[row][col]!=-1) return dp[row][col];\n\n // long long pts = point...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

2

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n int n = nums.size();\n if ((k % n) == 0) {\n return;\n }\n k = (k % n);\n bool reversed = false;\n if (k > n/2) {\n k = n - k;\n reversed = true;\n }\n ...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

2

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n int n = nums.size();\n k = k % n ; \n vector<int>temp(k);\n for( int i = 0 ; i< k ; i++){\n temp[i] = nums[n-k+i];\n }\n // shifiting \n for( int i = n-k-1 ; i >= 0; i--){\n ...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

2

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n // brute force approach\n int n = nums.size();\n k = k % n ; \n vector<int>temp(k);\n for( int i = 0 ; i< k ; i++){\n temp[i] = nums[n-k+i];\n }\n for( int i = n-k-1 ; i>=0; ...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

2

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n \n /* reverse(nums.begin(),nums.begin()+k+1);\n reverse(nums.begin()+k+1,nums.end());\n reverse(nums.begin(),nums.end());*/\n int n=nums.size();\n vector<int>v(n);\n k=k%n;\n int j=0;\n for(int i=n-k;i...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

2

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n \n int n = nums.size();\n vector<int> temp(n);\n \n \n for(int i =0;i<n;i++) {\n \n temp[(i+k)%n] = nums[i];\n \n }\n \n \n for(int i=0;i<n;i++) \n {\n ...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

2

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n vector<int> rotated(nums.size());\n\n int n = nums.size();\n for (int i = 0; i < n; i++) {\n int j = (k + i) % n;\n rotated[j] = nums[i];\n }\n\n nums = rotated;\n }\n};", ...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

2

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n \n int n=nums.size();\n k=k%n;\n vector<int>temp;\n for(int i=n-k;i<n;i++){\n temp.push_back(nums[i]);\n }\n for(int i=n-k-1;i>=0;i--){\n nums[i+k]=nums[i];\n ...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

3

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n int n=nums.size(); k=k%n;\n vector<int>r(n);\n for(int i=0; i<n; i++){\n r[(i+k)%n]= nums[i];\n }\n for(int i=0; i<n; i++){\n nums[i]= r[i];\n }\n }\n};", "memory": "...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

3

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n vector<int> temp(nums.size());\n for(int i=0;i<nums.size();i++) {\n temp[(i+k)%nums.size()]=nums[i];\n }\n nums=temp;\n }\n};", "memory": "28300" }

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

3

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) \n {\n int n= nums.size();\n k= k%n;\n vector<int> temp= {};\n\n for(int i= n-k;i< n;i++)\n temp.push_back(nums[i]);\n \n for(int i= n-1;i>= k;i--)\n nums[i]= nums[i-k];\n...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

3

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n int n=nums.size();\n vector<int> temp;\n k=k%n;\n for(int i=n-k;i<n;i++)\n {\n temp.push_back(nums[i]);\n }\n for(int i=n-k-1;i>=0;i--)\n {\n nums[i+k]=nums...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

3

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n k=k%nums.size();\n int sizeA = nums.size()-k-1;\n for(int i=0;i<=sizeA;i++){\n nums.push_back(nums[i]);\n }\n nums.erase(nums.begin(),nums.begin()+sizeA+1);\n }\n};", "memory": "28500...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

3

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) \n {\n k %= nums.size();\n vector<int> rotatedSection = {nums.begin() + (nums.size() - k), nums.end()};\n rotatedSection.insert(rotatedSection.end(), nums.begin(), nums.end() - k);\n nums = rotatedSection;\n...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

3

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n k = k % nums.size();\n std::vector<int> item(nums.end() - k, nums.end());\n nums.insert(nums.begin(), item.begin(), item.end());\n nums.erase(nums.end() - k, nums.end());\n }\n};", "memory": "28600" }

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

3

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n k = k % nums.size();\n // sol1 \n // vector<int> n1 = nums;\n // for (int i = 0; i < nums.size(); ++i) {\n // nums[(i + k) % nums.size()] = n1[i];\n // }\n\n // sol2\n // vec...

189

Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.   Example 1: <pre> Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Ex...

3

{ "code": "class Solution {\npublic:\n void rotate(vector<int>& nums, int k) {\n deque<int> temp;\n int n=nums.size();\n k = k % n;\n for(int i=0;i<n;i++)\n temp.push_back(nums[i]);\n for(int i=1;i<=k;i++)\n {\n temp.push_front(nums[n-i]);\n ...

190

Reverse bits of a given 32 bits unsigned integer. Note: <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...

0

{ "code": "class Solution \n{\npublic:\n uint32_t reverseBits(uint32_t n) \n {\n uint32_t res = 0;\n for (int i = 0; i < 32; i++)\n {\n res = (res << 1) | (n & 1);\n n >>= 1;\n }\n return res;\n }\n};", "memory": "7000" }

190

Reverse bits of a given 32 bits unsigned integer. Note: <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...

0

{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n for (int i = 0; i < 32; i++) {\n ans <<= 1;\n ans |= (n & 1);\n n >>= 1;\n }\n return ans;\n }\n};", "memory": "7100" }

190

Reverse bits of a given 32 bits unsigned integer. Note: <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...

0

{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n for (int i = 0; i < 32; i++) {\n ans <<= 1;\n ans |= (n & 1);\n n >>= 1;\n }\n return ans;\n }\n};", "memory": "7100" }

190

Reverse bits of a given 32 bits unsigned integer. Note: <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...

0

{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n int bit_mask;\n for (int j = 31; j >= 0; j--) {\n bit_mask = (n&1)<<j;\n ans = ans | bit_mask;\n n >>= 1;\n }\n\n return ans;\n }\n};", "memory": ...

190

Reverse bits of a given 32 bits unsigned integer. Note: <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...

0

{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n int bit_mask;\n for (int j = 31; j >= 0; j--) {\n bit_mask = (n&1)<<j;\n ans = ans | bit_mask;\n n >>= 1;\n }\n\n return ans;\n }\n};", "memory": ...

190

Reverse bits of a given 32 bits unsigned integer. Note: <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...

0

{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n)\n {\n uint32_t result{};\n\n int pos = 31;\n while (n > 0)\n {\n int d = n % 2;\n\n if (d == 1)\n {\n result += (1 << pos);\n }\n\n pos--; \n\n...

190

Reverse bits of a given 32 bits unsigned integer. Note: <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...

0

{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n int bit, bit_mask;\n for (int i = 1, j = 31; i <= 32; i++, j--) {\n bit = n&1;\n bit_mask = bit<<j;\n ans = ans | bit_mask;\n n >>= 1;\n }\n\n ...

190

Reverse bits of a given 32 bits unsigned integer. Note: <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...

1

{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n)\n {\n for (uint32_t i = 0; i < 16; ++i)\n {\n uint32_t a = 1<<i & n, b = 1<<(32-i-1) & n;\n if (a)\n n |= 1<<(32-i-1);\n else\n n &= ~(1<<(32-i-1));\n i...

190

Reverse bits of a given 32 bits unsigned integer. Note: <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...

1

{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n for(int i = 0; i<32; i++)\n {\n cout<<ans<<\" \";\n ans <<= 1;\n if(n&1)\n {\n ans+=1;\n }\n n>>=1;\n }\n ...

190

Reverse bits of a given 32 bits unsigned integer. Note: <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...

3

{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t ans = 0;\n int bit_mask;\n for (int j = 31; j >= 0; j--) {\n bit_mask = (n&1)<<j;\n ans = ans | bit_mask;\n n >>= 1;\n }\n\n return ans;\n }\n};", "memory": ...

190

Reverse bits of a given 32 bits unsigned integer. Note: <ul> <li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's in...

3

{ "code": "class Solution {\npublic:\n uint32_t reverseBits(uint32_t n) {\n uint32_t rst=0,c_bit,shift=31;\n\n while(n){\n c_bit = n & 1;\n rst|= c_bit<<shift;\n shift--;\n n>>=1;\n }\n return rst;\n }\n};", "memory": "7500" }

2,067