Runtime: 603 ms (Top 88.8%) | Memory: 180.29 MB (Top 72.2%)

Author: AnasImloul

scripts/algorithms/M/Maximum XOR of Two Numbers in an Array/Maximum XOR of Two Numbers in an Array.cpp

@@ -1,20 +1,70 @@
+// Runtime: 603 ms (Top 88.8%) | Memory: 180.29 MB (Top 72.2%)
+
 class Solution {
 public:
-    int findMaximumXOR(vector<int>& nums) {
-        int ans=0,mask=0;
-        unordered_set<int>st;
-        for(int i=31;i>=0;i--){
-            mask|=(1<<i);
-            int temp=ans|(1<<i);
-            for (int j = 0; j < nums.size(); j++) {
-                int num = nums[j] & mask;
-                if (st.find(temp ^ num)!=st.end()) {
-                    ans = temp;
-                    break;
+    struct TrieNode {
+        //trie with max 2 child, not taking any bool or 26 size value because no need
+        TrieNode* one;
+        TrieNode* zero;
+    };
+    void insert(TrieNode* root, int n) {
+        TrieNode* curr = root;
+        for (int i = 31; i >= 0; i--) {
+            int bit = (n >> i) & 1;  //it will find 31st bit and check it is 1 or 0
+            if (bit == 0) {
+                if (curr->zero == nullptr) {   //if 0 then we will continue filling on zero side
+                    TrieNode* newNode = new TrieNode();    
+                    curr->zero = newNode;  
+                }
+                curr = curr->zero;   //increase cur to next zero position
+            }
+            else {
+                //similarly if we get 1 
+                if (curr->one == nullptr) {
+                    TrieNode* newNode = new TrieNode();
+                    curr->one = newNode;
+                }
+                curr = curr->one;
+            }
+        }
+    }
+     int findmax(TrieNode* root, int n) {
+        TrieNode* curr = root;
+        int ans = 0;
+        for (int i = 31; i >= 0; i--) {
+            int bit = (n >> i) & 1;
+            if (bit == 1) {
+                if (curr->zero != nullptr) {  //finding complement , if find 1 then we will check on zero side
+                    ans += (1 << i); //push values in ans
+                    curr = curr->zero;
+                }
+                else {
+                    curr = curr->one;  //if we don't get then go to one's side
                 }
-                st.insert(num);
             }
-            st.clear();
+            else {
+                //similarly on zero side if we get 0 then we will check on 1 s side
+                if (curr->one != nullptr) {   
+                    ans += (1 << i);
+                    curr = curr->one;
+                }
+                else {
+                    curr = curr->zero;
+                }
+            }
+        }
+        return ans;
+    }
+
+    int findMaximumXOR(vector<int>& nums) {
+        int n = nums.size();
+        TrieNode* root = new TrieNode();
+        int ans = 0;
+        for (int i = 0; i < n; i++) {
+            insert(root, nums[i]);    //it will make trie by inserting values
+        }
+        for (int i = 1; i < n; i++) {
+            ans = max(ans, findmax(root, nums[i]));  //find the necessary complementory values and maximum store
         }
         return ans;
     }