id int64 1 3.58k | problem_description stringlengths 516 21.8k | instruction int64 0 3 | solution_c dict |
|---|---|---|---|
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "const int maxn = 2002;\nint ans[maxn][maxn], best[maxn][maxn];\nclass Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& a, vector<vector<int>>& queries) {\n int n = a.size();\n for(int i = 0; i < n; i++){\n ans[i][i] = a[i];\n best[i][i] = a[i];\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "const int inf = 1e9; \nclass SegTree {\n public: \n #define left (p<<1) \n #define right (left|1)\n #define mid ((l+r)>>1)\n #define toleft left,l,mid\n #define toright right,mid+1,r\n int n; \n vector<int> t; \n SegTree(int n):n(n),t(4*n, 0){}\n void update(int p,int l,int r,... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": " int dp[2001][2001];\n int xorScore[2001][2001];\n vector<int> a;\nclass Solution {\npublic:\n int maket(int i, int j){\n if(i==j){\n xorScore[i][j] = a[i];\n return a[i];\n }\n if(xorScore[i][j] == -1){\n xorScore[i][j] = (maket(i, j-1)^mak... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& a, vector<vector<int>>& queries) {\n int n=a.size();\n int val[n][n];\n for(int len=0;len<n;len++){\n for(int i=0;i<n;i++){\n if(i+len>=n)continue;\n if(len==0){\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n=nums.size();\n int dp[n][n];\n int mi[n][n];\n for(int i=0;i<n;i++){\n for(int j=0;i+j<n;j++){\n if(!i){\n dp[j][j]=... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\nint x[2001][2001];\nint dp[2001][2001];\nint n;\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n memset(dp, 0, sizeof(dp));\n memset(x, 0, sizeof(x));\n n=nums.size();\n for(int i=0;i<n;i++)x[i][i]=dp[i][i]=nums[... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n int dp[2001][2001][2]={0};\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& q) {\n int n=nums.size();\n for(int i=0;i<n;i++){\n dp[0][i][0]=nums[i];\n dp[0][i][1]=nums[i];\n }\n for(int i=1;i<n;i++){... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n \n int B[2002][2002];\n \n int dp[2002][2002];\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n=nums.size();\n \n for(int i=0;i<=n;i++) {\n for(int j=0;j<=i;j++) {\n if (j==0... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n int dp[n][n];\n int dp2[n][n] , dpp[n][n];\n for(int i = 0 ; i < n ;i++){\n dp[i][i] = nums[i];\n dp2[i][i] = nums[i];\... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<pair<int, int>>> heap(n);\n heap[0].resize(n);\n for (int i = 0; i < n; ++i)\n heap[0][i] = {nums[i], nums[i]};\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<pair<int, int>>> heap(n);\n heap[0].resize(n);\n for (int i = 0; i < n; ++i)\n heap[0][i] = {nums[i], nums[i]};\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<pair<int, int>>> heap(n);\n heap[0].resize(n);\n for (int i = 0; i < n; ++i)\n heap[0][i] = {nums[i], nums[i]};\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<pair<int, int>>> heap(n);\n heap[0].resize(n);\n for (int i = 0; i < n; ++i)\n heap[0][i] = {nums[i], nums[i]};\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<pair<int, int>>> heap(n);\n heap[0].resize(n);\n for (int i = 0; i < n; ++i)\n heap[0][i] = {nums[i], nums[i]};\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n int f[2002][2002];\n int a[2002][2002];\n int res[2002][2002];\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n vector<int> ans;\n int n = nums.size();\n //init\n for(int len = 0; len <n; len++){\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n /*\n Sol01.Get all subarryas and getther max xor for eah query TC = O(k^2 *q)\n Sol02.Use Sliding WIndow to calculate XOR of each subarray\n Sol03.Maybe use prefix sum similar and getthe xor\n\n Sol:-\n 0 1 2 3 4 ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& a, vector<vector<int>>& que) {\n vector<vector<int>> b;\n int n = a.size();\n b.push_back(a);\n for (int i = 1; i < n; i ++) {\n b.push_back(vector<int>(n - i));\n for (int j = 0; j ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& a, vector<vector<int>>& que) {\n vector<vector<int>> b;\n int n = a.size();\n b.push_back(a);\n for (int i = 1; i < n; i ++) {\n b.push_back(vector<int>(n - i));\n for (int j = 0; j ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n vector<int> res;\n \n struct Tree {\n int val=0;\n vector<Tree*> c;\n \n Tree(){\n c=vector<Tree*>(2, nullptr);\n }\n };\n \n void Insert(Tree* root, int val) {\n auto curr=root;\n for (int i=31; i>=0;... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "#include <iostream>\n#define ll long long\n\nusing namespace std;\n\nclass Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n ll n = nums.size(), q = queries.size();\n ll X[n][n], dp[n][n];\n for (int i=0; i<n; ++i) X[i][i] = dp... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 0 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n=nums.size();\n vector<vector<int>> dp(n,vector<int>(n));\n for(int i=0;i<n;i++){\n dp[i][i]=nums[i];\n }\n for(int len=2;len<=n;len++){\n for(... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "template <class T>\nbool umax(T &a, const T b) { return a < b ? a = b, 1 : 0; }\n\nclass Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& a, vector<vector<int>>& queries) {\n // Key Observation: dp[i][j] = dp[i + 1][j] ^ dp[i][j - 1]\n int n = a.size(), m = queries.size();... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> dp(n, vector<int>{});\n for (int i = 0; i < n; i++) {\n dp[i].reserve(n-i);\n }\n dp[0] = nums;\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& A, vector<vector<int>>& Q) {\n unordered_map<int, unordered_map<int, vector<int>>> mp;\n for(int i = 0; i < Q.size(); i++) mp[Q[i][1]-Q[i][0]+1][Q[i][0]].push_back(i);\n vector<int> res(Q.size()), B(A.size());\n... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& A, vector<vector<int>>& Q) {\n unordered_map<int, unordered_map<int, vector<int>>> mp;\n for(int i = 0; i < Q.size(); i++) mp[Q[i][1]-Q[i][0]+1][Q[i][0]].push_back(i);\n vector<int> res(Q.size()), B(A.size());\n... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n vector<int> l,r; \n for(int i = 0; i < queries.size(); i++){ \n \tl.push_back(queries[i][0]); \n r.push_back(queries[i][1]); \n }\n\t\tint n = nums.size(... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& a, vector<vector<int>>& q) {\n int n = a.size(), m = q.size();\n int val[n][n];\n vector<vector<int>> dp(n, vector<int>(n));\n for(int i = 0; i < n; i ++) {\n val[i][i] = a[i];\n dp[... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> xors(n, vector<int>(n, 0));\n for(int i = 0; i < n; i++) xors[i][i] = nums[i];\n for(int l = 1; l < n; l++) {\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "\nclass Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> t(n, vector<int>(n, 0));\n\n // Initialize the base cases\n for (int j = 0; j < n; ++j) {\n t[j][j] = nums[j... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\n\n vector<int> t;\n\n int getMax(int v, int tl, int tr, int l, int r) {\n if (l > r) \n return 0;\n if (l == tl && r == tr) {\n return t[v];\n }\n int tm = (tl + tr) / 2;\n return max(getMax(v*2, tl, tm, l, min(r, tm)),\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> dp(n, vector<int>(n));\n for(int i = 0; i < n; i++)\n dp[i][i] = nums[i];\n \n for(int len = 2; len <= ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n\n vector<vector<int>> xors(n, vector<int>(n));\n\n for (int i = 0; i < n; i++) xors[0][i] = nums[i];\n for (int i = 1; i < n; i++)\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> dp(n, vector<int>(n, -1));\n\n for (int i = n - 1; i >= 0; i--) {\n for (int j = i; j < n; j++) {\n if... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n=nums.size(), i, j, x, y;\n vector < int > res(queries.size());\n vector < vector < int > > dp(n, vector < int > (n, 0));\n for(i=0;i<n;i++) dp[0][i]=nums[i];\... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n void debug(vector<vector<int>> grid) {\n for (auto i : grid) {\n for (int j : i) {\n cout<<j<<\" \";\n }\n cout<<endl;\n }\n }\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& Q) {\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> arr(n, vector<int>(n));\n vector<int> times, pos;\n for(int size = 1; size <= n; size++) {\n if(size == 1) {\n... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\n\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<int> v[n];\n v[0]=nums;\n // BRUTE FORCE\n for(int depth=1;depth<n;depth++) {\n for(int i=0;i+1<v[depth-1].size();... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n \n int n=nums.size();\n vector<vector<int>> dp(n,vector<int>(n));\n\n for(int i=0;i<n;i++) dp[i][i]=nums[i];\n\n for(int len=2;len<=n;len++){\n fo... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 1 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums,\n vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> dp(n, vector<int>(n));\n for (int i = 0; i < n; i++)\n dp[i][i] = nums[i];\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n\n void singleSubarray(int l, int r, vector<vector<int>>& maxSubscore, vector<vector<int>>& subscore) {\n if (subscore[r][l] == -1) {\n singleSubarray(l + 1, r, maxSubscore, subscore);\n singleSubarray(l, r - 1, maxSubscore, subscore);\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int> &nums, vector<vector<int>> &queries) {\n int n = nums.size();\n vector<int> xors(n * n, -1);\n for(int i = 0; i < n; i++) {\n for(int j = i; j < n; j++){\n xors[i * n + j] = i == 0 ? num... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n const int n = nums.size();\n vector<vector<int>> levels(n);\n levels[n - 1] = nums;\n for (int l = n - 1; l --> 0; ) {\n levels[l].resize(l + 1);\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "#ifndef ATCODER_SEGTREE_HPP\n#define ATCODER_SEGTREE_HPP 1\n\n#include <algorithm>\n#include <cassert>\n#include <functional>\n#include <vector>\n\nnamespace atcoder {\n\nnamespace internal {\n\n#if __cplusplus >= 202002L\n\nusing std::bit_ceil;\n\n#else\n\n// @return same with std::bit::bit_ceil\nunsigned... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n \n vector<vector<int>> p;\n \n int curr = 0;\n int dp[2001][2001];\n\n void create(int x){\n p.push_back({});\n curr = max(curr,x);\n p[0].push_back(x);\n for(int i = 1;i<p.size();i++){\n int sz = p[i].size();\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> dp(n, vector<int> (n));\n vector<vector<int>> vv = {{0}};\n for(int i=0; i<n-1; i++){\n vector<int> u = vv.bac... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "int dp[2001][2001];\nclass Solution {\npublic:\n vector<int>v;\n vector<vector<int>>pre;\n int calc(int l,int r){\n return pre[l][r];\n }\n int rec(int l,int r){\n if(l>r)return 0;\n int &ret=dp[l][r];\n if(~ret)return ret;\n ret=calc(l,r);\n ret=max... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n // dp[i][j]: the largest XOR score of the subarray a[i:j]\n int n = nums.size();\n vector<vector<int>> dp(n, vector(n, 0));\n vector<int> result(queries.size(), 0);... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "#define ll long long\nclass Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n vector<int>v;\n int n = nums.size();\n vector<vector<ll>>dp(n,vector<ll>(n+1,0));\n for(int i=0;i<n;i++){\n dp[i][1]=nums[i];\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n // dp[i][j]: the largest XOR score of the subarray a[i:j]\n int n = nums.size();\n vector<vector<int>> dp(n, vector(n, 0));\n vector<int> result(queries.size(), 0);... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int len = nums.size();\n vector<vector<int>> xorDP(len, vector<int>(len, 0));\n // vector<vector<int>> xorScore(len, vector<int>(len, 0));\n for (int i = 0; i < len... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n #define pb push_back\n vector<int> maximumSubarrayXor(vector<int>& v, vector<vector<int>>& query) {\n int n=v.size();\n int p[n][n];memset(p,0,sizeof(p));\n vector<vector<int>>d(n);\n for(int i=0;i<n;i++){\n if(i==0){\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n vector <vector <int>> grid;\n grid.push_back(nums);\n while(grid.back().size()>1){\n vector <int> arr = grid.back();\n for(int i = 0;i<arr.size()-1;i... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n vector <vector <int>> grid;\n grid.push_back(nums);\n while(grid.back().size()>1){\n vector <int> arr = grid.back();\n for(int i = 0;i<arr.size()-1;i... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n=nums.size();\n vector<vector<int>> dp;\n dp.push_back(nums);\n for(int i=1;i<n;i++){\n vector <int> temp;\n for(int j=1;j<dp[i-1].size()... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& A, vector<vector<int>>& queries) {\n vector<vector<int>> B = { A };\n \n while (B.back().size() != 1) {\n vector<int> nextLevel;\n for (size_t i = 0; i < B.back().size() - 1; ++i) {\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "\nclass Solution {\npublic:\n std::vector<int> maximumSubarrayXor(std::vector<int>& A, std::vector<std::vector<int>>& queries) {\n std::vector<std::vector<int>> B = {A};\n \n while (B.back().size() != 1) {\n std::vector<int> newRow;\n for (size_t i = 0; i < B.b... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n std::vector<int> maximumSubarrayXor(std::vector<int>& nums, std::vector<std::vector<int>>& queries) {\n int n = nums.size();\n std::vector<std::vector<int>> dp(n);\n\n dp[0] = nums;\n \n for (int i = 1; i < n; ++i) {\n std::vector... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> dp(n);\n\n dp[0] = nums;\n \n for (int i = 1; i < n; ++i) {\n vector<int> xorLevel;\n for (i... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n pair<int,int> find(int i, int j, vector<int>&nums, vector<vector<pair<int, int>>>&dp){\n if(i == j)return {nums[i], nums[i]};\n\n if(dp[i][j].first!=-1)return dp[i][j];\n\n pair<int, int> left = find(i, j-1, nums, dp);\n pair<int, int> right = find... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n=nums.size();\n vector<vector<int>> adj(n);\n adj[0] = nums;\n for(int i=1;i<n;i++){\n for(int j=0;j+1<adj[i-1].size();j++){\n adj[i]... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> xorVal( n + 1, vector<int> ( n + 1, 0 ) );\n for( int len = 1; len <= n; len++ )\n {\n for( int i = 0; i < n; ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> dp(n, vector<int>(n));\n vector<vector<int>> max_dp(n, vector<int>(n));\n for (int i = 0; i < n; i++) {\n dp[i... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n=nums.size();\n vector<vector<pair<int,int>>>xor1(n);\n for(int i=0;i<n;i++){\n for(int j=0;j+i<n;j++){\n if(i==0){\n xor... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n struct Return_Value {\n int score;\n int max_score;\n Return_Value() {\n score = -1;\n max_score = -1;\n }\n Return_Value(int score, int max_score) : score(score), max_score(max_score) {}\n bool empty() {\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n #define ll long long\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<ll>>dp(n,vector<ll>(n,0));\n for(int i=0;i<n;i++){\n dp[i][i]=nums[i];\n }\n int sz... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& q)\n {\n int n=nums.size();\n vector<vector<pair<int,int>>> w(n,vector<pair<int,int>>(n));\n for(int i=0;i<n;i++)\n {\n w[i][i]={nums[i],nums[i]};\n }\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "\nclass Solution {\npublic:\n pair<int,int> solve(int i,int j, vector<vector<pair<int,int>>>&dp,vector<int>&nums)\n {\n if(i==j)return {nums[i],nums[i]};\n if(dp[i][j]!=make_pair(-1,-1))return dp[i][j];\n pair<int,int> left=solve(i,j-1,dp,nums);\n pair<int,int> down=solve... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& q) {\n int n = nums.size();\n vector<vector<int>> v(n);\n for(int i = 0; i < nums.size(); i++) {\n v[0].push_back(nums[i]);\n }\n for(int i = 1; i < n; i++) {\... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n vector<vector<int>> pyramid(nums.size(), vector<int>());\n vector<vector<int>> maxor(nums.size(), vector<int>());\n pyramid[0] = nums;\n maxor[0] = nums;\n f... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n int dp[2000][2000];\n int find(int i,int n,vector<vector<int>>&v){\n if(i>n)return 0;\n if(dp[i][n]!=-1)return dp[i][n];\n return dp[i][n]=max({v[n-i][i],find(i+1,n,v),find(i,n-1,v)});\n }\n vector<int> maximumSubarrayXor(vector<int>& nums, vecto... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<long>> dp(n,vector<long>(n));\n\n for(int i = 0 ; i < n ; i++) dp[i][i] = nums[i];\n\n for(int i = 2 ; i <= n ; i++){\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n\n const static int N = 2e3 + 1;\n\n pair<int, int> solve(int l, int r, vector<int>& a, vector<vector<pair<int, int>>>& dp){\n if (l == r){\n return dp[l][r] = {a[l], a[l]};\n }\n if (l > r) return {0, 0};\n\n if (dp[l][r].first != -1)... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<pair<int,int>> ans[2001],ans2[2001];\n int dp[2001][2001];\n \n #define dpp dp[l][r]\n int solve(int l, int r) {\n if(l>r) return 0;\n if(dpp!=-1) return dpp;\n int where=lower_bound(ans[l].begin(), ans[l].end(), make_pair(r+1,0))-ans[l... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "#define faster ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)\nclass Solution {\npublic:\n // unordered_map<int, unordered_map<int, pair<int, int>>> dp;\n pair<int, int> helper(vector<int> &nums, int i, int j, vector<vector<pair<int, int>>> &dp) {\n if(i > j) return {0, 0};\n if(i == ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n=nums.size();\n vector<vector<int>>xr(n, vector<int>(n)), mx(n, vector<int>(n));\n for(int i=0; i<n; i++){\n xr[i][i] = nums[i];\n mx[i][i] = nu... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 2 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n=nums.size();\n vector<vector<int>> xs(n,vector<int>(n)),mxs(n,vector<int>(n));\n for(int i=0;i<n;i++){ \n xs[i][i]=nums[i];\n mxs[i][i]=nums[i]... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\n vector<int> v;\n vector<vector<int>> dp,dp2;\n int calc(int s,int e){\n if(s==e) return v[s];\n if(dp2[s][e]>=0) return dp2[s][e];\n dp2[s][e]=calc(s,e-1)^calc(s+1,e);\n return dp2[s][e];\n }\n int rec(int s,int e){\n if(s==e) return v[s]... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\n vector<int> v;\n vector<vector<int>> dp,dp2;\n int rec(int s,int e){\n if(s==e) return v[s];\n if(dp[s][e]>=0) return dp[s][e];\n else{\n dp[s][e]=max(dp2[s][e],max(rec(s,e-1),rec(s+1,e)));;\n return dp[s][e];\n }\n }\npublic:... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "// segtree from atcoder\n// under CC0 lincense\n// https://atcoder.github.io/ac-library/document_en/segtree.html\n\nunsigned int bit_ceil(unsigned int n) {\n unsigned int x = 1;\n while (x < (unsigned int)(n)) x *= 2;\n return x;\n}\n\nint countr_zero(unsigned int n) {\n return __builtin_ctz(n)... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\npublic:\ntypedef long long int ll;\n ll doit(vector<int>&nums, vector<vector<ll> >&dp, int l, int r){\n if(l == r){\n return nums[l];\n }\n if(dp[l][r]!=-1)return dp[l][r];\n ll val = doit(nums,dp,l,r-1) ^ doit(nums,dp,l+1,r);\n dp[l][r] = ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\npublic:\n vector<int> solve(vector<int>& v){\n vector<int> v1;\n for(int i=0;i<v.size()-1;i++){\n v1.push_back(v[i]^v[i+1]);\n }\n return v1;\n }\n\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& q) {\n int n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& A, vector<vector<int>>& Q) {\n unordered_map<int, unordered_map<int, unordered_set<int>>> mp;\n for(int i = 0; i < Q.size(); i++) mp[Q[i][1]-Q[i][0]+1][Q[i][0]].insert(i);\n vector<int> res(Q.size()), B(A.size()... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& A, vector<vector<int>>& Q) {\n unordered_map<int, unordered_map<int, unordered_set<int>>> mp;\n for(int i = 0; i < Q.size(); i++) mp[Q[i][1]-Q[i][0]+1][Q[i][0]].insert(i);\n vector<int> res(Q.size()), B(A.size()... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\npublic:\n \n vector<vector<int>> build(vector<int> nums) {\n vector<vector<int>> f(nums.size(), vector<int>(nums.size()));\n for(int i=0;i<nums.size();i++) {\n for(int j=0;j+i<nums.size();j++) {\n f[j][j+i] = nums[j];\n }\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "\nint t[10000][10000];\nint m;\nint n;\n\nvector<vector<int>> a;\n\nvoid build_y(int vx, int lx, int rx, int vy, int ly, int ry) {\n if (ly == ry) {\n if (lx == rx) {\n t[vx][vy] = a[lx][ly];\n } else {\n t[vx][vy] = max(t[vx * 2][vy], t[vx * 2 + 1][vy]);\n }\n } else {\n int my = (... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> xor_score(n,vector<int>(n+1,0));\n for(int len=1;len<=n;len++){\n for(int i=0;i<n-len+1;i++){\n if(len... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "auto speedup = []() {\n ios::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n return 'a';\n}();\n\nclass Solution\n{\npublic:\n vector<int> maximumSubarrayXor(vector<int> &nums, vector<vector<int>> &queries)\n {\n vector<int> ans(queries.size(), 0);\n vector<vect... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "static const int init = [](){\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n cout.tie(nullptr);\n return 0;\n}();\n\n// #define dbg(x) cout << #x << \":\" << (x) << endl\n#define dbg(x)\n\nclass Solution {\n vector<vector<int>> DP;\n vector<vector<int>> MP;\n\n int F(int i, int j... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums,vector<vector<int>>& queries) {\n vector<vector<int>>tbl1(nums.size()+1,vector<int>(nums.size()+1,-1)),tbl2(nums.size()+1,vector<int>(nums.size()+1,-1));\n vector<int> ans;\n for(int i = 0;i < queries.size(... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\npublic:\n int f(int i,int j,vector<vector<int>> &prescore,vector<vector<int>> &dp){\n if(i>j) return INT_MIN;\n if(dp[i][j]!=-1) return dp[i][j];\n return dp[i][j]=max({f(i+1,j,prescore,dp),f(i,j-1,prescore,dp),prescore[i][j]});\n }\n vector<int> maximumSubarray... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n short i, j, n = nums.size();\n if(n == 1) return {nums[0]};\n\n int initialOp, nestedOp;\n vector<vector<int>> dp(n, vector<int>(n));\n\n for(i = 0; i < n; i... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "#define DEBUG(x...) { cout << \"(\" << #x << \")\" << \" = (\"; Print(x); }\ntemplate <typename T1> void Print(T1 t1) { cout << t1 << \" )\" << endl; }\ntemplate <typename T1, typename... T2>\nvoid Print(T1 t1, T2... t2) { cout << t1 << \",\"; Print(t2...); }\n#define ll long long\n#define sz(x) (int)x.siz... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n=nums.size();\n vector<vector<int>>xor_vals(n+1,vector<int>(n+1));\n \n for(int len=1;len<=n;len++){\n for(int i=0;i<n;i++){\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& arr, vector<vector<int>>& queries) {\n int n=arr.size();\n // [i][j] denotes subarray of length i+1, starting at index j\n vector<vector<int>> valueOfFunction(n, vector<int>(n, 0));\n for (int j = 0; j < ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> temp(n, vector<int>(n));\n vector<vector<int>> final(n, vector<int>(n));\n for (int i = n - 1; i >= 0; --i) {\n ... |
3,551 | <p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>
<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">su... | 3 | {
"code": "class Solution {\npublic:\n vector<int> maximumSubarrayXor(vector<int>& nums, vector<vector<int>>& queries) {\n int n = nums.size();\n vector<vector<int>> dp(n, vector<int>(n));\n for(int i = 0; i < n; i++) dp[i][i] = nums[i];\n for(int L = 2; L <= n; L++){\n for(i... |
3,567 | <p>You are given a string <code>date</code> representing a Gregorian calendar date in the <code>yyyy-mm-dd</code> format.</p>
<p><code>date</code> can be written in its binary representation obtained by converting year, month, and day to their binary representations without any leading zeroes and writing them down in ... | 0 | {
"code": "#include <bits/stdc++.h>\nusing namespace std;\ntypedef signed long long ll;\n\n#undef _P\n#define _P(...) (void)printf(__VA_ARGS__)\n#define FOR(x,to) for(x=0;x<(to);x++)\n#define FORR(x,arr) for(auto& x:arr)\n#define FORR2(x,y,arr) for(auto& [x,y]:arr)\n#define ALL(a) (a.begin()),(a.end())\n#define ZERO(... |
3,567 | <p>You are given a string <code>date</code> representing a Gregorian calendar date in the <code>yyyy-mm-dd</code> format.</p>
<p><code>date</code> can be written in its binary representation obtained by converting year, month, and day to their binary representations without any leading zeroes and writing them down in ... | 0 | {
"code": "class Solution {\npublic:\n string convertDateToBinary(string date) {\n int y = (date[0] - '0') * 1000 + (date[1] - '0') * 100 + (date[2] - '0') * 10 + (date[3] - '0');\n int m = (date[5] - '0') * 10 + (date[6] - '0');\n int d = (date[8] - '0') * 10 + (date[9] - '0');\n strin... |
3,567 | <p>You are given a string <code>date</code> representing a Gregorian calendar date in the <code>yyyy-mm-dd</code> format.</p>
<p><code>date</code> can be written in its binary representation obtained by converting year, month, and day to their binary representations without any leading zeroes and writing them down in ... | 0 | {
"code": "class Solution {\npublic:\n string convertDateToBinary(string date) {\n string ans;\n int year = 1000 * (date[0] - '0') + 100 * (date[1] - '0') + 10 * (date[2] - '0') + (date[3] - '0');\n int mon = 10 * (date[5] - '0') + (date[6] - '0');\n int day = 10 * (date[8] - '0') + (da... |
3,567 | <p>You are given a string <code>date</code> representing a Gregorian calendar date in the <code>yyyy-mm-dd</code> format.</p>
<p><code>date</code> can be written in its binary representation obtained by converting year, month, and day to their binary representations without any leading zeroes and writing them down in ... | 0 | {
"code": "class Solution {\npublic:\n string inttoBinary(int val){\n string ans = \"\";\n while(val > 0){\n if(val % 2 == 1){\n ans += '1';\n }\n else{\n ans += '0';\n }\n val = val / 2;\n }\n reverse(... |
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