Time: 2 ms (100.00%), Space: 58.5 MB (81.61%) - LeetHub

py3-coder · py3-coder · commit 17c890aaf692 · 2024-11-08T22:55:24.000+05:30
diff --git a/1829-maximum-xor-for-each-query/1829-maximum-xor-for-each-query.java b/1829-maximum-xor-for-each-query/1829-maximum-xor-for-each-query.java
@@ -0,0 +1,15 @@
+class Solution {
+    public int[] getMaximumXor(int[] nums, int maximumBit) {
+         int xor = 0, n = nums.length;
+        int max = (1 << maximumBit) - 1;
+        int res[] = new int[n];
+        
+        for(int e : nums) xor ^= e;//xor all elements
+        
+        for(int i = 0; i < n; i++){
+            res[i] = xor ^ max;// k = xor ^ max
+            xor ^= nums[n - i - 1];//remove last element from xor
+        }
+        return res;
+    }
+}
