Easy Code for Solution 2 in Question 1829.Maximum XOR for Each Query

Author: piyushlasane

solution/1800-1899/1829.Maximum XOR for Each Query/README_EN.md

@@ -303,20 +303,16 @@ class Solution:
 ```java
 class Solution {
     public int[] getMaximumXor(int[] nums, int maximumBit) {
-        int xs = 0;
-        for (int x : nums) {
-            xs ^= x;
+        int xor = (int)(Math.pow(2, maximumBit))-1;
+        int[] answer = new int[nums.length];
+        for(int i = 1; i < nums.length; i++){
+            nums[i] ^= nums[i-1];
         }
-        int mask = (1 << maximumBit) - 1;
-        int n = nums.length;
-        int[] ans = new int[n];
-        for (int i = 0; i < n; ++i) {
-            int x = nums[n - i - 1];
-            int k = xs ^ mask;
-            ans[i] = k;
-            xs ^= x;
+        int k=0;
+        for(int i=nums.length-1; i>=0; i--){
+            answer[k++] ^= nums[i]^xor; 
         }
-        return ans;
+        return answer;
     }
 }
 ```