Find last k digits in product of an array numbers

Given a array size of n, find the last k digits (1 <= k < 10) of product of array numbers

Examples:

Input  : a[] = {22, 31, 44, 27, 37, 43}
Output : 56

Input  : a[] = {24, 7, 144, 77, 29, 19}
Output : 84

A simple solution is to multiply all numbers, then find last k digits of product. This solution may lead overflow as the array product may be high.

A better solution is to multiply array elements under modulo of 10k

C++

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// CPP program to find the last k digits in
// product of array
#include <bits/stdc++.h>
using namespace std;
  
// Returns last k digits in product of a[]
int lastKDigits(int a[], int n, int k)
{
    int num = (int)pow(10, k);
  
    // Multiplying array elements under
    // modulo 10^k.
    int mul = a[0] % num;
    for (int i = 1; i < n; i++) {
        a[i] = a[i] % num;
        mul = (a[i] * mul) % num;
    }
    return mul;
}
  
// Driven program
int main()
{
    int a[] = { 22, 31, 44, 27, 37, 43 };
    int k = 2;
    int n = sizeof(a) / sizeof(a[0]);
    cout << lastKDigits(a, n, k);
    return 0;
}

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Java

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// Java program to find 
// the last k digits in 
// product of array
import java.io.*;
import java.math.*;
  
class GFG {
      
    // Returns last k digits in product of a[]
    static int lastKDigits(int a[], int n, int k)
    {
        int num = (int)(Math.pow(10, k));
      
        // Multiplying array elements 
        // under modulo 10^k.
        int mul = a[0] % num;
          
        for (int i = 1; i < n; i++) {
            a[i] = a[i] % num;
            mul = (a[i] * mul) % num;
        }
        return mul;
    }
      
// Driven program
public static void main(String args[])
{
    int a[] = { 22, 31, 44, 27, 37, 43 };
    int k = 2;
    int n = a.length;
      
    System.out.println(lastKDigits(a, n, k));
}
  
}
  
  
/*This code is contributed by Nikita Tiwari.*/

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Python 3

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# Python 3 program to find the last
# k digits inproduct of array
import math
  
# Returns last k digits 
# in product of a[]
def lastKDigits(a, n, k) :
      
    num = (int)(math.pow(10, k))
      
    # Multiplying array elements
    # under modulo 10^k.
    mul = a[0] % num
      
    for i in range(1,n) :
        a[i] = a[i] % num
        mul = (a[i] * mul) % num
      
    return mul
      
  
# Driven program
a = [ 22, 31, 44, 27, 37, 43 ]
k = 2
n = len(a)
print(lastKDigits(a, n, k))
  
  
# This code is contributed by Nikita Tiwari.

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C#

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// C# program to find 
// the last k digits in 
// product of array
using System;
  
class GFG {
      
    // Returns last k digits in product of a[]
    static int lastKDigits(int []a, int n, int k)
    {
        int num = (int)(Math.Pow(10, k));
      
        // Multiplying array elements 
        // under modulo 10^k.
        int mul = a[0] % num;
          
        for (int i = 1; i < n; i++) {
            a[i] = a[i] % num;
            mul = (a[i] * mul) % num;
        }
        return mul;
    }
      
    // Driven program
    public static void Main()
    {
        int []a = { 22, 31, 44, 27, 37, 43 };
        int k = 2;
        int n = a.Length;
          
        Console.WriteLine(lastKDigits(a, n, k));
    }
  
}
  
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP program to find
// the last k digits in
// product of array
  
// Returns last k digits
// in product of a[]
function lastKDigits($a, $n, $k)
{
      
    $num = (int)pow(10, $k);
  
    // Multiplying array elements 
    // under modulo 10^k.
    $mul = $a[0] % $num;
    for ($i = 1; $i < $n; $i++) 
    {
        $a[$i] = $a[$i] % $num;
        $mul = ($a[$i] * $mul) % $num;
    }
    return $mul;
}
  
// Driver Code
$a = array( 22, 31, 44, 27, 37, 43 );
$k = 2;
$n = sizeof($a);
echo(lastKDigits($a, $n, $k));
  
// This code is contributed by Ajit.
?>

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Output:

 56

Time Complexity : O(n)



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Improved By : jit_t