Refine 1310_XOR_Queries_of_a_Subarray

Author: qiyuangong

README.md

@@ -228,6 +228,7 @@ I'm currently working on [Analytics-Zoo](https://github.com/intel-analytics/anal
 | 1260 | [Shift 2D Grid](https://leetcode.com/problems/shift-2d-grid/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1260_Shift_2D_Grid.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1260_Shift_2D_Grid.java) | Final position of each element can be computed according to k, m and n, e.g., k == mn, then don't move, O(mn) and O(mn) |
 | 1290 | [Convert Binary Number in a Linked List to Integer](https://leetcode.com/problems/convert-binary-number-in-a-linked-list-to-integer/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1290_Convert_Binary_Number_in_a_Linked_List_to_Integer.py) | Take 2 to the power digit position from right (starting from 0) and multiply it with the digit |
 | 1304 | [Find N Unique Integers Sum up to Zero](https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1304_Find_N_Unique_Integers_Sum_up_to_Zero.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1304_Find_N_Unique_Integers_Sum_up_to_Zero.java) | [1,n-1] and its negative sum |
+| 1310 | [XOR Queries of a Subarray](https://leetcode.com/problems/xor-queries-of-a-subarray) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1310_XOR_Queries_of_a_Subarray.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1310_XOR_Queries_of_a_Subarray.java) [Cpp](https://github.com/qiyuangong/leetcode/blob/master/cpp/1310_XOR_Queries_of_a_Subarray.cpp) | Compute accumulated xor from head, qeury result equals to xor[0, l] xor x[0, r], O(n) and O(n) |
 | 1323 | [Maximum 69 Number](https://leetcode.com/problems/maximum-69-number/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1323_Maximum_69_Number.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1323_Maximum_69_Number.java) | 9 is greater than 6, so change first 6 to 9 from left if exist, O(n) and O(1) |
 | 1337 | [The K Weakest Rows in a Matrix](https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1337_The_K_Weakest_Rows_in_a_Matrix.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1337_The_K_Weakest_Rows_in_a_Matrix.java) | Check by row, from left to right, until encount first zero, O(mn) and O(1) |
 | 1342 | [Number of Steps to Reduce a Number to Zero](https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1342_Number_of_Steps_to_Reduce_a_Number_to_Zero.py) | If number is divisible by 2, divide the number by 2, else subtract 1 from the number, and output the number of steps, O(logn) and O(1) |

cpp/1310_XOR_Queries_of_a_Subarray.cpp

@@ -0,0 +1,12 @@
+class Solution {
+public:
+    vector<int> xorQueries(vector<int>& arr, vector<vector<int>>& queries) {
+        vector<int> res;
+        // Compute accumulated xor from head
+        for (int i = 1; i < arr.size(); i++)
+            arr[i] ^= arr[i - 1];
+        for (auto &q: queries)
+            res.push_back(q[0] > 0 ? arr[q[0] - 1] ^ arr[q[1]] : arr[q[1]]);
+        return res;
+    }
+};

java/1310_XOR_Queries_of_a_Subarray.java

@@ -0,0 +1,14 @@
+class Solution {
+    public int[] xorQueries(int[] arr, int[][] queries) {
+        int[] res = new int[queries.length], q;
+        // Compute accumulated xor from head
+        for (int i = 1; i < arr.length; i++)
+            arr[i] ^= arr[i - 1];
+        // query result equals to xor[0, l] xor xor[0, r]
+        for (int i = 0; i < queries.length; i++) {
+            q = queries[i];
+            res[i] = q[0] > 0 ? arr[q[0] - 1] ^ arr[q[1]] : arr[q[1]];
+        }
+        return res;
+    }
+}

python/1310_XOR_Queries_of_a_Subarray.py

@@ -1,10 +1,16 @@
 class Solution:
     def xorQueries(self, arr: List[int], queries: List[List[int]]) -> List[int]:
         pref = [0]
+        # Compute accumulated xor from head
         for e in arr:
             pref.append(e ^ pref[-1])
         ans = []
+        # query result equal to xor[0, l] xor x[0, r]
         for [l, r] in queries:
             ans.append(pref[r+1] ^ pref[l])
         return ans
-        
+
+    # def xorQueries(self, arr: List[int], queries: List[List[int]]) -> List[int]:
+    #     for i in range(len(arr) - 1):
+    #         arr[i + 1] ^= arr[i]
+    #     return [arr[j] ^ arr[i - 1] if i else arr[j] for i, j in queries]