Create 1829. Maximum XOR for Each Query (#629)

Author: Chayandas07

1829. Maximum XOR for Each Query

@@ -0,0 +1,15 @@
+class Solution {
+public:
+vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {
+    int n = nums.size();
+    int mx = pow(2,maximumBit)-1;
+    vector<int> ans(n,0);
+    int xr = 0;
+    for(int i=0;i<n;i++)
+    {
+        xr ^= nums[i]; 
+        ans[n-i-1] = (mx)^(xr&mx);
+    }
+    return ans;  
+}
+};