Count of Pairs such that modulo of product of one with their XOR and X is 1
Last Updated :
10 Jan, 2023
Given two arrays A[] and B[] of length N and M respectively and a prime number X, the task is to find the number of ordered pair of indices (i, j) that satisfies the following conditions:
- 1 ? i ? N and 1 ? j ? M
- ( Ai ? Bj ) < X
- (( Ai ? ( Ai ? Bj )) ? 1) % X = 0
Note: ? is XOR operator.
Examples:
Input: A[] = {7, 14} , B = {2, 13}, X = 17
Output: 1
Explanation: There are 4 ordered pairs of indices. Looking at them individually,
(1, 1) satisfies because (7 ? 2) = 5 < 17, and ((7? (7 ? 2)) ? 1) = 34 is divisible by 17.
(1, 2) satisfies because (7 (7 ? 13)) ? 1) = 69 is not divisible by 17.
(2, 1) satisfies because ((14?(14 ? 2)) ? 1) = 167 is not divisible by 17
(2, 2) satisfies because ((14?(14 ? 13)) ? 1) = 41 is not divisible by 17.
Input: A[] = {3} , B = {3}, X = 11
Output: 0
Approach: The problem can be solved based on the following mathematical observation:
Consider two values Ai and Bj which satisfies the condition.
So, ( Ai?( Ai ? Bj )?1) mod X = 0
(( Ai?( Ai ? Bj )) mod X = 1
( Ai ? Bj ) mod X = Ai?1 mod X
( Ai ? Bj )= Ai?1 mod X (according to the last condition)
Bj = ( Ai ? ( Ai-1 mod X ))
So if the value of Ai is known we can easily check if there is any value in B[] that satisfies the condition.
Follow the below steps to implement the idea:
- Store all the elements of B[] in a map.
- Traverse through the array A[]:
- Note if the values A[i] and X are not prime then the condition cannot be satisfied, as we cannot get some multiple of A[i] which will give 1 as a remainder when divided by X.
- Otherwise, calculate the value from B that will satisfy the condition using the above formula.
- Find the count of that value and add that to the answer.
- Return the final value of the answer.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
#include <iostream>
#include <map>
using namespace std;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int mI(int a, int m)
{
int m0 = m;
int y = 0, x = 1;
if (m == 1)
return 0;
while (a > 1) {
int q = a / m;
int t = m;
m = a % m;
a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0)
x += m0;
return x;
}
int findPairs(int a[], int b[], int n, int m, int p)
{
map<int, int> cnt;
for (int j = 0; j < m; j++) {
cnt[b[j]]++;
}
int ans = 0;
for (int i = 0; i < n; i++) {
if (gcd(a[i], p) != 1) {
continue;
}
int x = mI(a[i], p) ^ a[i];
ans += cnt[x];
}
return ans;
}
int main()
{
int A[] = { 7, 14 };
int B[] = { 2, 13 };
int N = sizeof(A) / sizeof(A[0]);
int M = sizeof(B) / sizeof(B[0]);
int X = 17;
cout << findPairs(A, B, N, M, X);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static int mI(int a, int m)
{
int m0 = m;
int y = 0, x = 1;
if (m == 1)
return 0;
while (a > 1) {
int q = a / m;
int t = m;
m = a % m;
a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0)
x += m0;
return x;
}
public static int findPairs(int a[],
int b[], int n,
int m, int p)
{
HashMap<Integer, Integer> map
= new HashMap<>();
for (int i = 0; i < m; i++) {
if (!map.containsKey(b[i]))
map.put(b[i], 1);
else
map.put(b[i], map.get(b[i]) + 1);
}
int ans = 0;
for (int i = 0; i < n; i++) {
if (gcd(a[i], p) != 1)
continue;
int x = mI(a[i], p) ^ a[i];
if (!map.containsKey(x))
continue;
ans = ans + map.get(x);
}
return ans;
}
public static void main(String[] args)
{
int A[] = { 7, 14 };
int B[] = { 2, 13 };
int N = A.length;
int M = B.length;
int X = 17;
System.out.println(findPairs(A, B, N, M, X));
}
}
|
Python3
def gcd(a, b) :
if b == 0 :
return a
else :
return gcd(b, a % b);
def mI(a, m) :
m0 = m;
y = 0;
x = 1;
if (m == 1) :
return 0;
while (a > 1) :
q = a // m;
t = m;
m = a % m;
a = t;
t = y;
y = x - q * y;
x = t;
if (x < 0) :
x += m0;
return x;
def findPairs(a, b, n, m, p) :
cnt = dict.fromkeys(b, 0);
for j in range(m) :
cnt[b[j]] += 1;
ans = 0;
for i in range(n) :
if (gcd(a[i], p) != 1) :
continue;
x = mI(a[i], p) ^ a[i];
if x in cnt:
ans += cnt[x];
else :
continue;
return ans;
if __name__ == "__main__" :
A = [ 7, 14 ];
B = [ 2, 13 ];
N = len(A);
M = len(B);
X = 17;
print(findPairs(A, B, N, M, X));
|
C#
using System;
using System.Collections.Generic;
using System.Collections;
public class GFG
{
public static int gcd(int a, int b)
{
if (b == 0)
{
return a;
}
return GFG.gcd(b, a % b);
}
public static int mI(int a, int m)
{
var m0 = m;
var y = 0;
var x = 1;
if (m == 1)
{
return 0;
}
while (a > 1)
{
var q = (int)(a / m);
var t = m;
m = a % m;
a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0)
{
x += m0;
}
return x;
}
public static int findPairs(int[] a, int[] b, int n, int m, int p)
{
var map = new Dictionary<int, int>();
for (int i = 0; i < m; i++)
{
if (!map.ContainsKey(b[i]))
{
map[b[i]] = 1;
}
else
{
map[b[i]] = map[b[i]] + 1;
}
}
var ans = 0;
for (int i = 0; i < n; i++)
{
if (GFG.gcd(a[i], p) != 1)
{
continue;
}
var x = GFG.mI(a[i], p) ^ a[i];
if (!map.ContainsKey(x))
{
continue;
}
ans = ans + map[x];
}
return ans;
}
public static void Main(String[] args)
{
int[] A = {7, 14};
int[] B = {2, 13};
var N = A.Length;
var M = B.Length;
var X = 17;
Console.WriteLine(GFG.findPairs(A, B, N, M, X));
}
}
|
Javascript
function gcd(a, b) {
return b == 0 ? a : gcd(b, a % b);
}
function mI(a, m) {
let m0 = m;
let y = 0,
x = 1;
if (m == 1) return 0;
while (a > 1) {
let q = Math.floor(a / m);
let t = m;
m = a % m;
a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0) x += m0;
return Math.floor(x);
}
function findPairs(a, b, n, m, p) {
let cnt = {};
for (let j = 0; j < m; j++) {
if (cnt.hasOwnProperty(b[j])) cnt[b[j]]++;
else cnt[b[j]] = 1;
}
let ans = 0;
for (let i = 0; i < n; i++) {
if (gcd(a[i], p) != 1) {
continue;
}
let x = mI(a[i], p) ^ a[i];
if (cnt.hasOwnProperty(x)) ans += cnt[x];
}
return ans;
}
let A = [7, 14];
let B = [2, 13];
let N = A.length;
let M = B.length;
let X = 17;
console.log(findPairs(A, B, N, M, X));
|
Time Complexity: O(N * log X)
Auxiliary Space: O(m), space used by map.
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