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:
C++
// 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 Programint main(){ int q = 3; int l[] = { 2, 2, 5 }; int r[] = { 4, 8, 9 }; allQueries(q, l, r); return 0;} |
Java
// 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
# 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 Codeq = 3l = [ 2, 2, 5 ] r = [ 4, 8, 9 ] allQueries(q, l, r) # This code is contributed # by sahishelangia |
C#
// 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// 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?> |
Javascript
<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> |
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)
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