Maximum XOR subarray | Practice | GeeksforGeeks

Given an array arr[] of size, N. Find the subarray with maximum XOR. A subarray is a contiguous part of the array.

Example 1:
Input:
N = 4
arr[] = {1,2,3,4}
Output: 7
Explanation: 
The subarray {3,4} has maximum xor 
value equal to 7.


Your Task:  
You don't need to read input or print anything. Your task is to complete the function maxSubarrayXOR() which takes the array arr[], its size N as input parameters and returns the maximum xor of any subarray.
 
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)

 
Constraints:
1 <= N <= 105
1 <= arr[i] <= 105
0
apoorvmishra2 months ago
struct node{   node* child[2];   int value;   node(){       child[0]=NULL;       child[1]=NULL;       value=0;   }};class trie{   public:   node* root;   trie(){       root= new node;   }   void insert(int num){       node* temp=root;       for(int i=31;i>=0;i--){           bool bit=num&(1<<i);           if(temp->child[bit]==NULL){               temp->child[bit]=new node;           }           temp=temp->child[bit];       }       temp->value=num;   }   int query(int num){       node* temp=root;       for(int i=31;i>=0;i--){           bool bit=num&(1<<i);           if(temp->child[1-bit]!=NULL){               temp=temp->child[1-bit];           }else if(temp->child[bit]!=NULL){               temp=temp->child[bit];           }       }       return (temp->value)^num;   }};class Solution{   public:   int maxSubarrayXOR(int N, int arr[]){           trie t;       t.insert(0);       int res=0,prefixXor=0;       for(int i=0;i<N;i++){           prefixXor=prefixXor^arr[i];           t.insert(prefixXor);           res=max(res,t.query(prefixXor));       }       return res;   }};
0
mayank20212 months ago
@gfg please see.. backend code seems wrong in this case.for this testcase, it's showing 100001 as output. I tried manually but i can't get above 10000. what am I missing here ?3 100000 99999 99998
-1
aloksinghbais023 months ago
C++ solution using kadane's algorithm having time complexity as O(N) and space complexity as O(1) is as follows :-
 
Execution Time :- 0.1 / 1.2 sec
 
int maxSubarrayXOR(int N, int arr[]){           if(N==1) return (arr[0]);       int max_so_far = 0;       int curr_max = 1;       for(int i=0; i<N; i++){           curr_max = max(arr[i],arr[i] ^ curr_max);           max_so_far = max(max_so_far,curr_max);       }       return (max_so_far);   }
-1
devshukla07
This comment was deleted.
-9
amail4muni4 months ago
for java solution
 
class Solution{
   static int maxSubarrayXOR(int N, int arr[]){       // code here         int ans = 1; // Initialize result       int curr_xor = 1;        // Pick starting points of subarrays        for (int i=0; i<N; i++)        {               curr_xor = Math.max(arr[i],curr_xor^arr[i]);                ans = Math.max(ans, curr_xor);        }        return ans;    }}
-11
sreenathmopuri4 months ago
class Solution{   public:     int maxSubarrayXOR(int N, int arr[]){              //code here          int max_so_far = 1, maxXor = 1;          for (int i = 0; i < N; i++)          {               max_so_far = max(arr[i], max_so_far ^ arr[i]);               maxXor = max(maxXor, max_so_far);         }         return maxXor;     }};
0
dmaddalena694 months ago
This is the solution but I cannot decrease the execution time of the program. Anyone can help me?
int maxSubarrayXOR(int N, int arr[]){    
        
        int max_xor = 0, ele_subarr = 1, re_xor = 0;
        
        while(ele_subarr <= N)
        {
            for(int i = 0; i < N-(ele_subarr-1); i++)
            {
                for(int j = i; j < ele_subarr+i; j++)
                    re_xor ^= arr[j];
                
                if(re_xor > max_xor)
                    max_xor = re_xor;
                    
                re_xor = 0;
            }
            
            ele_subarr++;
        }
            
        return max_xor;
    }
0
tanmaymm585 months ago
Why isn't this solution working? I am gettig the output 2 and the XOR of 4 and 6 is 2 but it is saying the answer is 6. Can some explain me? class Solution:   def maxSubarrayXOR (self, N, arr):       # code here        for i in range(N-1):           arr[i] = arr[i] ^ arr[i+1]       for j in range(len(arr)-1):           if arr[j] > arr[j+1]:               pass       return arr[j]       
0
sudarshans35 months ago
ss
-10
lapimpale7 months ago
class Solution{

    static int maxSubarrayXOR(int N, int arr[]){
        int ans = 1; // Initialize result 
       int curr_xor = 1; 
        // Pick starting points of subarrays 
        for (int i=0; i<N; i++) 
        {
                curr_xor = Math.max(arr[i],curr_xor^arr[i]); 
                ans = Math.max(ans, curr_xor); 
        } 
        return ans; 
    }
}
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