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Maximize the number of subarrays with XOR as zero in C++
We are given an array Arr[] containing integer values. The goal is to find the maximum number of subarrays with XOR as 0. The bits of any subarray can be swapped any number of times.
Note:- 1<=Arr[i]<=1018
In order to make any subarray’s XOR as 0 by swapping bits, two conditions have to be met:-
If the number of set bits in range left to right is even.
For any given range sum of bits <= 2 (largest number in set bits)
Let us see various input output scenarios for this -
In −Arr[] = { 1,2,5,4 }
Out −
Subarrays satisfying only 1st condition : 4
Subarrays satisfying both condition : 3
In − Arr[] = { 3,7,2,9 }
Out −
Subarrays satisfying only 1st condition : 6
Subarrays satisfying both condition : 3
Approach used in the below program is as follows −
In this approach we observed that in order to make any subarray’s XOR as 0 by swapping bits, two conditions have to be met:- If the number of set bits in range left to right is even or for any given range sum of bits <= 2 (largest number in set bits)
Take the input array Arr[] and calculate its length.
Function removeSubarr(int arr[], int len) returns the count of subarrays not satisfying condition 2.
Take the initial count as 0.
Iterate array using for loop and take variables sum and maxVal.
Take another for loop to iterate in range of 60 subarrays as beyond 60 the condition 2 can never be false.
Add element to sum and take maximum in maxVal.
If sum is even and 2 * maxVal > sum then increment count as condition 2 is not met.
At the end of both loops return count.
Function findSubarrays(int arr1[], int len1) takes an input array and its length and returns the count of subarrays satisfying both conditions mentioned above.
Take a prefix array to calculate the count of subarrays that follow condition 1 only.
Traverse array using for loop and set each element with __builtin_popcountll(arr1[i]) which is the number of set bits in it.
Populate prefix array using for loop and set prefix[i] = prefix[i] + prefix[i - 1] where except first element.
Count odd and even values in the prefix array.
Set tmp1= ( oddcount * (oddcount-1) )/2 and tmp2= ( evencount * (evencount-1) )/2 and result as sum of both.
Result will be the sum of subarrays satisfying condition 1 only.
Print result.
Now update result with result=result - removeSubarr(arr1, len1).
Now the result contains subarrays satisfying both conditions.
Print result again.
Example
#include <bits/stdc++.h>
using namespace std;
// Function to count subarrays not satisfying condition 2
int removeSubarr(int arr[], int len){
int count = 0;
for (int i = 0; i < len; i++){
int sum = 0;
int maxVal = 0;
for (int j = i; j < min(len, i + 60); j++){
sum = sum + arr[j];
maxVal = arr[j] > maxVal ? arr[j]: maxVal;
if (sum % 2 == 0){
if( 2 * maxVal > sum)
{ count++; }
}
}
}
return count;
}
int findSubarrays(int arr1[], int len1){
int prefix[len1];
int oddcount, evencount;
int result;
for (int i = 0; i < len1; i++)
{ arr1[i] = __builtin_popcountll(arr1[i]); }
for (int i = 0; i < len1; i++){
prefix[i] = arr1[i];
if (i != 0)
{ prefix[i] = prefix[i] + prefix[i - 1]; }
}
oddcount = evencount = 0;
for (int i = 0; i < len1; i++){
if (prefix[i] % 2 == 0)
{ evencount = evencount +1; }
else
{ oddcount = oddcount +1; }
}
evencount++;
int tmp1= ( oddcount * (oddcount-1) )/2;
int tmp2= ( evencount * (evencount-1) )/2;
result = tmp1+tmp2;
cout << "Subarrays satisfying only 1st condition : "<<result << endl;
cout << "Subarrays satisfying both condition : ";
result = result - removeSubarr(arr1, len1);
return result;
}
int main()
{ int Arr[] = { 1,2,5,4 };
int length = sizeof(Arr) / sizeof(Arr[0]);
cout << findSubarrays(Arr, length);
return 0;
}Output
If we run the above code it will generate the following Out
Subarrays satisfying only 1st condition : 4 Subarrays satisfying both condition : 3
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