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C++ program to find maximum possible value of XORed sum
Suppose we have an array A with N elements and another value K. For an integer X in range 0 to K, let f(X) = (X xor A[1]) + (X xor A[2]) + ... + (X xor A[N]). We have to find the maximum possible value of f.
So, if the input is like K = 7; A = [1, 6, 3], then the output will be 14, because f(4) = (4 XOR 1) + (4 XOR 6) + (4 XOR 3) = 5 + 2 + 7 = 14.
Steps
To solve this, we will follow these steps −
n := size of A for initialize i := 45, when i >= 0, update (decrease i by 1), do: p := 2^i m := 0 for initialize j := 0, when j < n, update (increase j by 1), do: if A[j] AND p is non-zero, then: (increase m by 1) if o + p <= k, then: if m < n - m, then: m := n - m o := o + p d := d + p * m return d
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
long solve(int k, vector<int> A){
long n = A.size(), d = 0, m, p, o = 0;
for (long i = 45; i >= 0; i--){
p = pow(2, i);
m = 0;
for (int j = 0; j < n; j++){
if (A[j] & p)
m++;
}
if (o + p <= k){
if (m < n - m){
m = n - m;
o += p;
}
}
d += p * m;
}
return d;
}
int main(){
int K = 7;
vector<int> A = { 1, 6, 3 };
cout << solve(K, A) << endl;
}Input
7, { 1, 6, 3 }Output
14
Updated on: 03-Mar-2022
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