C/C++ Program to Find remainder of array multiplication divided by n
Write a C/C++ program for a given multiple numbers and a number n, the task is to print the remainder after multiplying all the numbers divided by n.
Examples:
Input: arr[] = {100, 10, 5, 25, 35, 14},
n = 11
Output: 9
Explanation: 100 x 10 x 5 x 25 x 35 x 14 = 61250000 % 11 = 9
Input : arr[] = {100, 10},
n = 5
Output : 0
Explanation: 100 x 10 = 1000 % 5 = 0
C/C++ Program to Find remainder of array multiplication divided by n using Naive approach:
First multiple all the number then take % by n then find the remainder, But in this approach, if the number is maximum of 2^64 then it give the wrong answer.
Below is the implementation of the above approach:
CPP
#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;
}
|
Time Complexity: O(n)
Auxiliary Space: 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
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
int findremainder( int arr[], int len, int n)
{
int mul = 1;
for ( int i = 0; i < len; i++)
mul = (mul * (arr[i] % n)) % n;
return mul % 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);
}
|
Time Complexity: O(len), where len is the size of the given array
Auxiliary Space: O(1)
Please refer complete article on Find remainder of array multiplication divided by n for more details!
Last Updated :
18 Oct, 2023
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