Range Query on array whose each element is XOR of index value and previous element
Consider an arr[] which can be defined as:
You are given Q queries of the form [l, r]. The task is to output the value of arr[l] ? arr[l+1] ? ..... ? arr[r-1] ? arr[r] for each query.
Examples :
Input : q = 3
q1 = { 2, 4 }
q2 = { 2, 8 }
q3 = { 5, 9 }
Output : 7
9
15
The beginning of the array with constraint look like:
arr[] = { 0, 1, 3, 0, 4, 1, 7, 0, 8, 1, 11, .... }
For q1, 3 ? 0 ? 4 = 7
For q2, 3 ? 0 ? 4 ? 1 ? 7 ? 0 ? 8 = 9
For q3, 1 ? 7 ? 0 ? 8 ? 1 = 15
Let's observe arr[]
arr[0] = 0 arr[1] = 1 arr[2] = 1 ? 2 arr[3] = 1 ? 2 ? 3 arr[4] = 1 ? 2 ? 3 ? 4 arr[5] = 1 ? 2 ? 3 ? 4 ? 5 ....
Let's make another array, say brr[], where brr[i] = arr[0] ? arr[1] ? arr[2] ? ..... ? arr[i].
brr[i] = arr[0] ? arr[1] ? arr[2] ? ... ? arr[i-1] ? arr[i] = brr[j] ? arr[j+1] ? arr[j+2] ? ..... ? arr[i], for any 0 <= j <= i.
So, arr[l] ? arr[l+1] ? ..... ? arr[r] = brr[l-1] ? brr[r].
Now, let's observe brr[]:
brr[1] = 1 brr[2] = 2 brr[3] = 1 ? 3 brr[4] = 2 ? 4 brr[5] = 1 ? 3 ? 5 brr[6] = 2 ? 4 ? 6 brr[7] = 1 ? 3 ? 5 ? 7 brr[8] = 2 ? 4 ? 6 ? 8
It's easy to observe that in odd indexes brr[i] = 1 ? 3 ? 5 ? .... ? i and for even indexes brr[i] = 2 ? 4 ? 6 ? .... ? i.
For even indexes there are numbers from 1 to i/2 multipliedby 2, that means bits are moved to left by 1, so, brr[i] = 2 ? 4 ? 6 ? .... ? i = (1 ? 2 ? 3 ? ..... ? i/2) * 2.
And for odd indexes there are numbers from 0 to (i - 1)/2 multiplied by 2 and plus 1. That means bits are moved to left by 1, and last bit is made 1. So, brr[i] = 1 ? 3 ? 5 ? .... ? i = (0 ? 1 ? 2 ? .... ? (i - 1)/2) * 2 + x.
x is 1 ? 1 ? 1 ? ..... ? 1 "ones" are repeated (i - 1)/2 + 1 times. So, if (i-1)/2 + 1 is odd then x = 1 else x = 0.
Now, calculation of 1 ? 2 ? 3 ? .... ? x.
Let's prove that (4K) ? (4K + 1) ? (4K + 2) ? (4K + 3) = 0 for 0 <= k.
bitmask(K)00=4K
xorsum bitmask(K)01=4K+1
bitmask(K)10=4K+2
bitmask(K)11=4k+3
---------------------
000000000000=0
So as 0 ? Y = Y then 1 ? 2 ? 3 ? ... ? x = (floor(x/4) x 4) ? ... ? x here are maximum 3 numbers so we can calculate in O(1).
Below is the implementation of this approach:
// CPP Program to solve range query on array
// whose each element is XOR of index value
// and previous element.
#include <bits/stdc++.h>
using namespace std;
// function return derived formula value.
int fun(int x)
{
int y = (x / 4) * 4;
// finding xor value of range [y...x]
int ans = 0;
for (int i = y; i <= x; i++)
ans ^= i;
return ans;
}
// function to solve query for l and r.
int query(int x)
{
// if l or r is 0.
if (x == 0)
return 0;
int k = (x + 1) / 2;
// finding x is divisible by 2 or not.
return (x %= 2) ? 2 * fun(k) : ((fun(k - 1) * 2) ^ (k & 1));
}
void allQueries(int q, int l[], int r[])
{
for (int i = 0; i < q; i++)
cout << (query(r[i]) ^ query(l[i] - 1)) << endl;
}
// Driven Program
int main()
{
int q = 3;
int l[] = { 2, 2, 5 };
int r[] = { 4, 8, 9 };
allQueries(q, l, r);
return 0;
}
// Java Program to solve range query on array
// whose each element is XOR of index value
// and previous element.
import java.io.*;
class GFG {
// function return derived formula value.
static int fun(int x)
{
int y = (x / 4) * 4;
// finding xor value of range [y...x]
int ans = 0;
for (int i = y; i <= x; i++)
ans ^= i;
return ans;
}
// function to solve query for l and r.
static int query(int x)
{
// if l or r is 0.
if (x == 0)
return 0;
int k = (x + 1) / 2;
// finding x is divisible by 2 or not.
return ((x %= 2) != 0) ? 2 * fun(k) :
((fun(k - 1) * 2) ^ (k & 1));
}
static void allQueries(int q, int l[], int r[])
{
for (int i = 0; i < q; i++)
System.out.println((query(r[i]) ^
query(l[i] - 1))) ;
}
// Driven Program
public static void main (String[] args) {
int q = 3;
int []l = { 2, 2, 5 };
int []r = { 4, 8, 9 };
allQueries(q, l, r);
}
}
// This code is contributed by vt_m.
# Python3 Program to solve range query
# on array whose each element is XOR of
# index value and previous element.
# function return derived formula value.
def fun(x):
y = (x // 4) * 4
# finding xor value of range [y...x]
ans = 0
for i in range(y, x + 1):
ans ^= i
return ans
# function to solve query for l and r.
def query(x):
# if l or r is 0.
if (x == 0):
return 0
k = (x + 1) // 2
# finding x is divisible by 2 or not.
if x % 2 == 0:
return((fun(k - 1) * 2) ^ (k & 1))
else:
return(2 * fun(k))
def allQueries(q, l, r):
for i in range(q):
print(query(r[i]) ^ query(l[i] - 1))
# Driver Code
q = 3
l = [ 2, 2, 5 ]
r = [ 4, 8, 9 ]
allQueries(q, l, r)
# This code is contributed
# by sahishelangia
// C# Program to solve range query on array
// whose each element is XOR of index value
// and previous element.
using System;
class GFG {
// function return derived formula value.
static int fun(int x)
{
int y = (x / 4) * 4;
// finding xor value of range [y...x]
int ans = 0;
for (int i = y; i <= x; i++)
ans ^= i;
return ans;
}
// function to solve query for l and r.
static int query(int x)
{
// if l or r is 0.
if (x == 0)
return 0;
int k = (x + 1) / 2;
// finding x is divisible by 2 or not.
return ((x %= 2)!=0) ? 2 * fun(k) :
((fun(k - 1) * 2) ^ (k & 1));
}
static void allQueries(int q, int []l, int []r)
{
for (int i = 0; i < q; i++)
Console.WriteLine((query(r[i])
^ query(l[i] - 1))) ;
}
// Driven Program
public static void Main ()
{
int q = 3;
int []l = { 2, 2, 5 };
int []r = { 4, 8, 9 };
allQueries(q, l, r);
}
}
// This code is contributed by vt_m.
<?php
// PHP Program to solve range
// query on array whose each
// element is XOR of index
// value and previous element.
// function return derived
// formula value.
function fun($x)
{
$y = ((int)($x / 4) * 4);
// finding xor value
// of range [y...x]
$ans = 0;
for ($i = $y; $i <= $x; $i++)
$ans ^= $i;
return $ans;
}
// function to solve
// query for l and r.
function query($x)
{
// if l or r is 0.
if ($x == 0)
return 0;
$k = (int)(($x + 1) / 2);
// finding x is divisible
// by 2 or not.
return ($x %= 2) ? 2 * fun($k) :
((fun($k - 1) * 2) ^ ($k & 1));
}
function allQueries($q, $l, $r)
{
for ($i = 0; $i < $q; $i++)
echo (query($r[$i]) ^
query($l[$i] - 1)) , "\n";
}
// Driver Code
$q = 3;
$l = array( 2, 2, 5 );
$r = array ( 4, 8, 9 );
allQueries($q, $l, $r);
// This code is contributed by ajit
?>
<script>
// Javascript Program to solve range query on array
// whose each element is XOR of index value
// function return derived formula value.
function fun(x)
{
let y = parseInt(x / 4) * 4;
// finding xor value of range [y...x]
let ans = 0;
for (let i = y; i <= x; i++)
ans ^= i;
return ans;
}
// function to solve query for l and r.
function query(x)
{
// if l or r is 0.
if (x == 0)
return 0;
let k = parseInt((x + 1) / 2);
// finding x is divisible by 2 or not.
return (x %= 2) ? 2 * fun(k) : ((fun(k - 1) * 2) ^ (k & 1));
}
function allQueries(q, l, r)
{
for (let i = 0; i < q; i++)
document.write((query(r[i]) ^ query(l[i] - 1)) + "<br>");
}
// Driven Program
let q = 3;
let l = [ 2, 2, 5 ];
let r = [ 4, 8, 9 ];
allQueries(q, l, r);
</script>
Output
0 2 0
Time Complexity: O(q* log(n)) where q is the number of queries and n is the largest value of r in the queries.
Auxiliary Space: O(1)
