Given multiple numbers and a number n, the task is to print the remainder after multiply all the number divide by n.
Examples:
Input : arr[] = {100, 10, 5, 25, 35, 14},
n = 11
Output : 9
100 x 10 x 5 x 25 x 35 x 14 = 61250000 % 11 = 9
Input : arr[] = {100, 10},
n = 5
Output : 0
100 x 10 = 1000 % 5 = 0
Naive approach: First multiple all the number then take % by n then find the remainder, But in this approach if number is maximum of 2^64 then it give wrong answer.
Implementation of the above approach:
Try It Yourself
#include <bits/stdc++.h>
using namespace std;
int findremainder(int arr[], int len, int n)
{
long long int product = 1;
for (int i = 0; i < len; i++) {
product = product * arr[i];
}
return product % n;
}
int main()
{
int arr[] = { 100, 10, 5, 25, 35, 14 };
int len = sizeof(arr) / sizeof(arr[0]);
int n = 11;
cout << findremainder(arr, len, n);
return 0;
}
public class Main {
// This code finds the remainder of the product of all
// the elements in the array arr divided by 'n'.
public static int findRemainder(int[] arr, int len,
int n)
{
int product = 1;
for (int i = 0; i < len; i++) {
product = product * arr[i];
}
return product % n;
}
public static void main(String[] args)
{
int[] arr = { 100, 10, 5, 25, 35, 14 };
int len = arr.length;
int n = 11;
System.out.println(findRemainder(arr, len, n));
}
}
# This code finds the remainder of the product of all the elements in the array arr divided by 'n'.
def findremainder(arr, len, n):
product = 1
for i in range(len):
product = product * arr[i]
return product % n
arr = [100, 10, 5, 25, 35, 14]
len = len(arr)
n = 11
print(findremainder(arr, len, n))
using System;
// This code finds the remainder of the product of all
// the elements in the array arr divided by 'n'.
class Program
{
static int findRemainder(int[] arr, int len, int n)
{
long product = 1;
for (int i = 0; i < len; i++)
{
product *= arr[i];
}
return (int)(product % n);
}
static void Main()
{
int[] arr = { 100, 10, 5, 25, 35, 14 };
int len = arr.Length;
int n = 11;
Console.WriteLine(findRemainder(arr, len, n));
}
}
// Define a function to find the remainder
// of product of array elements divided by n
function findRemainder(arr, len, n) {
let product = 1;
for (let i = 0; i < len; i++) {
product = product * arr[i];
}
return product % n;
}
// Driver Code
let arr = [100, 10, 5, 25, 35, 14];
let len = arr.length;
let n = 11;
console.log(findRemainder(arr, len, n));
Output
9
Time Complexity: O(n)
Space Complexity: O(1)
Approach that avoids overflow : First take a remainder or individual number like arr[i] % n. Then multiply the remainder with current result. After multiplication, again take remainder to avoid overflow. This works because of distributive properties of modular arithmetic. ( a * b) % c = ( ( a % c ) * ( b % c ) ) % c
Implementation:
// C++ program to find
// remainder when all
// array elements are
// multiplied.
#include <iostream>
using namespace std;
// Find remainder of arr[0] * arr[1] *
// .. * arr[n-1]
int findremainder(int arr[], int len, int n)
{
int mul = 1;
// find the individual remainder
// and multiple with mul.
for (int i = 0; i < len; i++)
mul = (mul * (arr[i] % n)) % n;
return mul % n;
}
// Driver code
int main()
{
int arr[] = { 100, 10, 5, 25, 35, 14 };
int len = sizeof(arr) / sizeof(arr[0]);
int n = 11;
// print the remainder of after
// multiple all the numbers
cout << findremainder(arr, len, n);
}
// Java program to find
// remainder when all
// array elements are
// multiplied.
import java.util.*;
import java.lang.*;
public class GfG{
// Find remainder of arr[0] * arr[1] *
// .. * arr[n-1]
public static int findremainder(int arr[],
int len, int n)
{
int mul = 1;
// find the individual remainder
// and multiple with mul.
for (int i = 0; i < len; i++)
mul = (mul * (arr[i] % n)) % n;
return mul % n;
}
// Driver function
public static void main(String argc[])
{
int[] arr = new int []{ 100, 10, 5,
25, 35, 14 };
int len = 6;
int n = 11;
// print the remainder of after
// multiple all the numbers
System.out.println(findremainder(arr, len, n));
}
}
/* This code is contributed by Sagar Shukla */
# Python3 program to
# find remainder when
# all array elements
# are multiplied.
# Find remainder of arr[0] * arr[1]
# * .. * arr[n-1]
def findremainder(arr, lens, n):
mul = 1
# find the individual
# remainder and
# multiple with mul.
for i in range(lens):
mul = (mul * (arr[i] % n)) % n
return mul % n
# Driven code
arr = [ 100, 10, 5, 25, 35, 14 ]
lens = len(arr)
n = 11
# print the remainder
# of after multiple
# all the numbers
print( findremainder(arr, lens, n))
# This code is contributed by "rishabh_jain".
<script>
// Javascript program to find
// remainder when all
// array elements are
// multiplied.
// Find remainder of arr[0] * arr[1] *
// .. * arr[n-1]
function findremainder(arr, len, n)
{
let mul = 1;
// find the individual remainder
// and multiple with mul.
for (let i = 0; i < len; i++)
mul = (mul * (arr[i] % n)) % n;
return mul % n;
}
let arr = [ 100, 10, 5, 25, 35, 14 ];
let len = 6;
let n = 11;
// print the remainder of after
// multiple all the numbers
document.write(findremainder(arr, len, n));
// This code is contributed by rameshtravel07.
</script>
// C# program to find
// remainder when all
// array elements are
// multiplied.
using System;
public class GfG{
// Find remainder of arr[0] * arr[1] *
// .. * arr[n-1]
public static int findremainder(int []arr,
int len, int n)
{
int mul = 1;
// find the individual remainder
// and multiple with mul.
for (int i = 0; i < len; i++)
mul = (mul * (arr[i] % n)) % n;
return mul % n;
}
// Driver function
public static void Main()
{
int[] arr = new int []{ 100, 10, 5,
25, 35, 14 };
int len = 6;
int n = 11;
// print the remainder of after
// multiple all the numbers
Console.WriteLine(findremainder(arr, len, n));
}
}
/* This code is contributed by vt_m */
<?php
// PHP program to find
// remainder when all
// array elements are
// multiplied.
// Find remainder of arr[0] * arr[1] *
// .. * arr[n-1]
function findremainder($arr, $len, $n)
{
$mul = 1;
// find the individual remainder
// and multiple with mul.
for ($i = 0; $i < $len; $i++)
$mul = ($mul * ($arr[$i] % $n)) % $n;
return $mul % $n;
}
// Driver code
$arr = array(100, 10, 5, 25, 35, 14);
$len = sizeof($arr);
$n = 11;
// print the remainder of after
// multiple all the numbers
echo(findremainder($arr, $len, $n));
// This code is contributed by Ajit.
?>
Output
9
Time Complexity: O(len), where len is the size of the given array
Auxiliary Space: O(1)
