Find a partition point in array to maximize its xor sum
Last Updated :
13 Oct, 2022
Given an array a of size N. The task is to find an index 'i' (1 <= i <= N) such that (a[1] ^ ... ^ a[i]) + (a[i+1] ^ ... ^ a[N]) (x^y represents the xor value of x and y) is maximum possible.
Examples:
Input : arr[] = {1, 4, 6, 3, 8, 13, 34, 2, 21, 10}
Output : 2
Explanation : The maximum value is 68 at index 2
Input : arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Output : 4
Naive Approach: A naive approach is to use nested loops. Traverse the array and find the xor of the array till the i’th index and find the xor of elements from index i+1 to and calculate the maximum sum possible.
Below is the implementation of the above approach:
C++
// CPP program to find partition point in
// array to maximize xor sum
#include <bits/stdc++.h>
using namespace std;
// Function to find partition point in
// array to maximize xor sum
int Xor_Sum(int arr[], int n)
{
int sum = 0, index, left_xor = 0, right_xor = 0;
// Traverse through the array
for (int i = 0; i < n; i++)
{
// Calculate xor of elements left of index i
// including ith element
left_xor = left_xor ^ arr[i];
right_xor = 0;
for (int j = i + 1; j < n; j++)
{
// Calculate xor of the elements right of
// index i
right_xor = right_xor ^ arr[j];
}
// Keep the maximum possible xor sum
if (left_xor + right_xor > sum)
{
sum = left_xor + right_xor;
index = i;
}
}
// Return the 1 based index of the array
return index+1;
}
// Driver code
int main()
{
int arr[] = { 1, 4, 6, 3, 8, 13, 34, 2, 21, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
cout << Xor_Sum(arr, n);
return 0;
}
// Java program to find partition point in
// array to maximize xor sum
class GFG
{
// Function to find partition point in
// array to maximize xor sum
public static int Xor_Sum(int[] arr, int n)
{
int sum = 0, index = -1;
int left_xor = 0, right_xor = 0;
// Traverse through the array
for (int i = 0; i < n; i++)
{
// Calculate xor of elements left of index i
// including ith element
left_xor = left_xor ^ arr[i];
right_xor = 0;
for (int j = i + 1; j < n; j++)
{
// Calculate xor of the elements right of
// index i
right_xor = right_xor ^ arr[j];
}
// Keep the maximum possible xor sum
if (left_xor + right_xor > sum)
{
sum = left_xor + right_xor;
index = i;
}
}
// Return the 1 based index of the array
return index + 1;
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 1, 4, 6, 3, 8,
13, 34, 2, 21, 10 };
int n = arr.length;
// Function call
System.out.println(Xor_Sum(arr, n));
}
}
// This code is contributed by sanjeev2552
# Python3 program to find partition point in
# array to maximize xor sum
# Function to find partition point in
# array to maximize xor sum
def Xor_Sum(arr, n):
sum = 0
index, left_xor = 0, 0
right_xor = 0
# Traverse through the array
for i in range(n):
# Calculate xor of elements left of index i
# including ith element
left_xor = left_xor ^ arr[i]
right_xor = 0
for j in range(i + 1, n):
# Calculate xor of the elements
# right of index i
right_xor = right_xor ^ arr[j]
# Keep the maximum possible xor sum
if (left_xor + right_xor > sum):
sum = left_xor + right_xor
index = i
# Return the 1 based index of the array
return index + 1
# Driver code
arr = [ 1, 4, 6, 3, 8,
13, 34, 2, 21, 10]
n = len(arr)
# Function call
print(Xor_Sum(arr, n))
# This code is contributed by Mohit Kumar
// C# program to find partition point in
// array to maximize xor sum
using System;
class GFG
{
// Function to find partition point in
// array to maximize xor sum
public static int Xor_Sum(int[] arr,
int n)
{
int sum = 0, index = -1;
int left_xor = 0, right_xor = 0;
// Traverse through the array
for (int i = 0; i < n; i++)
{
// Calculate xor of elements left of index i
// including ith element
left_xor = left_xor ^ arr[i];
right_xor = 0;
for (int j = i + 1; j < n; j++)
{
// Calculate xor of the elements
// right of index i
right_xor = right_xor ^ arr[j];
}
// Keep the maximum possible xor sum
if (left_xor + right_xor > sum)
{
sum = left_xor + right_xor;
index = i;
}
}
// Return the 1 based index of the array
return index + 1;
}
// Driver code
public static void Main(String[] args)
{
int[] arr = { 1, 4, 6, 3, 8,
13, 34, 2, 21, 10 };
int n = arr.Length;
// Function call
Console.WriteLine (Xor_Sum(arr, n));
}
}
// This code is contributed by PrinciRaj1992
<script>
// Javascript program to
// find partition point in
// array to maximize xor sum
// Function to find partition point in
// array to maximize xor sum
function Xor_Sum(arr, n)
{
let sum = 0, index, left_xor = 0,
right_xor = 0;
// Traverse through the array
for (let i = 0; i < n; i++)
{
// Calculate xor of elements
// left of index i
// including ith element
left_xor = left_xor ^ arr[i];
right_xor = 0;
for (let j = i + 1; j < n; j++)
{
// Calculate xor of the
// elements right of
// index i
right_xor = right_xor ^ arr[j];
}
// Keep the maximum possible xor sum
if (left_xor + right_xor > sum)
{
sum = left_xor + right_xor;
index = i;
}
}
// Return the 1 based index of the array
return index+1;
}
// Driver code
let arr = [ 1, 4, 6, 3, 8, 13, 34, 2, 21, 10 ];
let n = arr.length;
// Function call
document.write(Xor_Sum(arr, n));
</script>
Time complexity: O( N^2 )
Auxiliary space: O(1)
Efficient Approach: An efficient approach is to use a prefix xor array. At any index ‘i’ PrefixXor[i] gives us arr[1] ^ arr[1] ^....^ arr[i] and to get arr[i+1] ^ arr[i+2] ^ . . ^ arr[n-1], find PrefixXor[i] ^ PrefixXor[n] .
Below is the implementation of the above approach:
C++
// CPP program to find partition point in
// array to maximize xor sum
#include <bits/stdc++.h>
using namespace std;
// Function to calculate Prefix Xor array
void ComputePrefixXor(int arr[], int PrefixXor[], int n)
{
PrefixXor[0] = arr[0];
// Calculating prefix xor
for (int i = 1; i < n; i++)
PrefixXor[i] = PrefixXor[i - 1] ^ arr[i];
}
// Function to find partition point in
// array to maximize xor sum
int Xor_Sum(int arr[], int n)
{
// To store prefix xor
int PrefixXor[n];
// Compute the prefix xor
ComputePrefixXor(arr, PrefixXor, n);
// To store sum and index
int sum = 0, index;
// Calculate the maximum sum that can be obtained
// splitting the array at some index i
for (int i = 0; i < n; i++)
{
// PrefixXor[i] = Xor of all arr
// elements till i'th index PrefixXor[n-1]
// ^ PrefixXor[i] = Xor of all elements
// from i+1' th index to n-1'th index
if (PrefixXor[i] + (PrefixXor[n - 1] ^
PrefixXor[i]) > sum)
{
sum = PrefixXor[i] +
(PrefixXor[n - 1] ^ PrefixXor[i]);
index = i;
}
}
// Return the index
return index+1;
}
// Driver code
int main()
{
int arr[] = { 1, 4, 6, 3, 8, 13, 34, 2, 21, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
cout << Xor_Sum(arr, n);
return 0;
}
// Java program to find partition point in
// array to maximize xor sum
import java.util.*;
class GFG
{
// Function to calculate Prefix Xor array
static void ComputePrefixXor(int arr[],
int PrefixXor[],
int n)
{
PrefixXor[0] = arr[0];
// Calculating prefix xor
for (int i = 1; i < n; i++)
PrefixXor[i] = PrefixXor[i - 1] ^ arr[i];
}
// Function to find partition point in
// array to maximize xor sum
static int Xor_Sum(int arr[], int n)
{
// To store prefix xor
int []PrefixXor = new int[n];
// Compute the prefix xor
ComputePrefixXor(arr, PrefixXor, n);
// To store sum and index
int sum = 0, index = 0;
// Calculate the maximum sum that can be obtained
// splitting the array at some index i
for (int i = 0; i < n; i++)
{
// PrefixXor[i] = Xor of all arr
// elements till i'th index PrefixXor[n-1]
// ^ PrefixXor[i] = Xor of all elements
// from i+1' th index to n-1'th index
if (PrefixXor[i] + (PrefixXor[n - 1] ^
PrefixXor[i]) > sum)
{
sum = PrefixXor[i] +
(PrefixXor[n - 1] ^ PrefixXor[i]);
index = i;
}
}
// Return the index
return index+1;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 4, 6, 3, 8,
13, 34, 2, 21, 10 };
int n = arr.length;
// Function call
System.out.println(Xor_Sum(arr, n));
}
}
// This code is contributed by 29AjayKumar
# Python3 program to find partition point in
# array to maximize xor sum
# Function to calculate Prefix Xor array
def ComputePrefixXor(arr, PrefixXor, n):
PrefixXor[0] = arr[0];
# Calculating prefix xor
for i in range(1, n):
PrefixXor[i] = PrefixXor[i - 1] ^ arr[i];
# Function to find partition point in
# array to maximize xor sum
def Xor_Sum(arr, n):
# To store prefix xor
PrefixXor = [0] * n;
# Compute the prefix xor
ComputePrefixXor(arr, PrefixXor, n);
# To store sum and index
sum, index = 0, 0;
# Calculate the maximum sum that can be obtained
# splitting the array at some index i
for i in range(n):
# PrefixXor[i] = Xor of all arr
# elements till i'th index PrefixXor[n-1]
# ^ PrefixXor[i] = Xor of all elements
# from i+1' th index to n-1'th index
if (PrefixXor[i] + (PrefixXor[n - 1] ^
PrefixXor[i]) > sum):
sum = PrefixXor[i] +\
(PrefixXor[n - 1] ^ PrefixXor[i]);
index = i;
# Return the index
return index + 1;
# Driver code
arr = [ 1, 4, 6, 3, 8, 13, 34, 2, 21, 10 ];
n = len(arr);
# Function call
print(Xor_Sum(arr, n));
# This code is contributed by Rajput-Ji
// C# program to find partition point in
// array to maximize xor sum
using System;
class GFG
{
// Function to calculate Prefix Xor array
static void ComputePrefixXor(int[] arr,
int[] PrefixXor,
int n)
{
PrefixXor[0] = arr[0];
// Calculating prefix xor
for (int i = 1; i < n; i++)
PrefixXor[i] = PrefixXor[i - 1] ^ arr[i];
}
// Function to find partition point in
// array to maximize xor sum
static int Xor_Sum(int[] arr, int n)
{
// To store prefix xor
int []PrefixXor = new int[n];
// Compute the prefix xor
ComputePrefixXor(arr, PrefixXor, n);
// To store sum and index
int sum = 0, index = 0;
// Calculate the maximum sum that can be obtained
// splitting the array at some index i
for (int i = 0; i < n; i++)
{
// PrefixXor[i] = Xor of all arr
// elements till i'th index PrefixXor[n-1]
// ^ PrefixXor[i] = Xor of all elements
// from i+1' th index to n-1'th index
if (PrefixXor[i] + (PrefixXor[n - 1] ^
PrefixXor[i]) > sum)
{
sum = PrefixXor[i] + (PrefixXor[n - 1] ^
PrefixXor[i]);
index = i;
}
}
// Return the index
return index + 1;
}
// Driver code
public static void Main()
{
int[] arr = { 1, 4, 6, 3, 8,
13, 34, 2, 21, 10 };
int n = arr.Length;
// Function call
Console.WriteLine(Xor_Sum(arr, n));
}
}
// This code is contributed by Code_Mech
<script>
// Javascript program to find partition point in
// array to maximize xor sum
// Function to calculate Prefix Xor array
function ComputePrefixXor(arr, PrefixXor, n)
{
PrefixXor[0] = arr[0];
// Calculating prefix xor
for (let i = 1; i < n; i++)
PrefixXor[i] = PrefixXor[i - 1] ^ arr[i];
}
// Function to find partition point in
// array to maximize xor sum
function Xor_Sum(arr, n)
{
// To store prefix xor
let PrefixXor = new Array(n);
// Compute the prefix xor
ComputePrefixXor(arr, PrefixXor, n);
// To store sum and index
let sum = 0, index;
// Calculate the maximum sum that can be obtained
// splitting the array at some index i
for (let i = 0; i < n; i++)
{
// PrefixXor[i] = Xor of all arr
// elements till i'th index PrefixXor[n-1]
// ^ PrefixXor[i] = Xor of all elements
// from i+1' th index to n-1'th index
if (PrefixXor[i] + (PrefixXor[n - 1] ^
PrefixXor[i]) > sum)
{
sum = PrefixXor[i] +
(PrefixXor[n - 1] ^ PrefixXor[i]);
index = i;
}
}
// Return the index
return index+1;
}
// Driver code
let arr = [ 1, 4, 6, 3, 8, 13, 34, 2, 21, 10 ];
let n = arr.length;
// Function call
document.write(Xor_Sum(arr, n));
</script>
Time complexity: O(N) where N is the size of the given array
Auxiliary space: O(N)
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