Given an integer number N, find the exor of the range 1 to N
exor of 1 ^ 2 ^ 3 ^4 ^.....N;
Brute force approach:
Tc:O(n)
Sc:O(1)
public int findExor(int N){
//naive/brute force approach:
int val = 0;
for(int i=1;i<5;i++){
val = val^ i;
}
return val;
}
Optimal approach:
Tc:O(1)
Sc:O(1)
public int getExor(int N){
//better approach
/**
* one thing to observe is
* 1 = 001 = 1
* 1 ^2 = 001 ^ 010 = 011= 3
* 1^2^3 = 011 ^ 011 = 0= 0
* 1^2^3^4 = 000^100 = 100= 4
* 1^2^3^4^5 = 100^101 = 001= 1
* 1^2^3^4^5^6 = 001^110 =111= 7
* 1^2^3^4^5^6^7 = 111^111=000= 0
*
* what we can observer is :
*
* N%4==0 then result is: N
* N%4 ==1 then result is: 1
* N%4 ==2 then result is: N+1
* N%4==3 then result is: 0
*
* */
if(N%4==0) return N;
else if(N%4 ==1) return 1;
else if(N%4==2) return N+1;
else return 0;
}
What if we have to find the exor between ranges like L and R
example find an exor between numbers 4 and 7 i.e. 4^5^6^7.
For solving this we can leverage the same optimal solution above getExor()
first we will get exor till L-1 i.e getExor(L-1) = 1 ^ 2 ^ 3 (since L-1 = 3)......equation(1)
then we will find getExor(R) = 1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6 ^ 7 ----equation(2)
the finally,
Result = equation(1) ^ equation(2)
= (1 ^ 2 ^ 3) ^ (1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6 ^ 7)
= (4^5^6^7)
public int findExorOfRange(int L, int R){
return getExor(L-1) ^ getExor(R);
}
public int getExor(int N){
//better approach
/**
* one thing to observe is
* 1 = 001 = 1
* 1 ^2 = 001 ^ 010 = 011= 3
* 1^2^3 = 011 ^ 011 = 0= 0
* 1^2^3^4 = 000^100 = 100= 4
* 1^2^3^4^5 = 100^101 = 001= 1
* 1^2^3^4^5^6 = 001^110 =111= 7
* 1^2^3^4^5^6^7 = 111^111=000= 0
*
* what we can observer is :
*
* N%4==0 then result is: N
* N%4 ==1 then result is: 1
* N%4 ==2 then result is: N+1
* N%4==3 then result is: 0
*
* */
if(N%4==0) return N;
else if(N%4 ==1) return 1;
else if(N%4==2) return N+1;
else return 0;
}
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