Count number of trailing zeros in (1^1)*(2^2)*(3^3)*(4^4)*..
Last Updated :
23 Jun, 2022
Given an integer n, the task is to find the number of trailing zeros in the function f(n)=\prod\limits_{i = 1}^{n} i^{i} i.e. f(n) = 11 * 22 * 33 * ... * nn.
Examples:
Input: n = 5
Output: 5
f(5) = 11 * 22 * 33 * 44 * 55 = 1 * 4 * 27 * 256 * 3125 = 86400000
Input: n = 12
Output: 15
Approach: We know that 5 * 2 = 10 i.e. 1 trailing zero is the result of the multiplication of a single 5 and a single 2. So, if we have x number of 5 and y number of 2 then the number of trailing zeros will be min(x, y).
Now, for every number i in the series, we need to count the number of 2 and 5 in its factors say x and y but the number of 2s and 5s will be x * i and y * i respectively because in the series i is raised to the power itself i.e. ii. Count the number of 2s and 5s in the complete series and print the minimum of them which is the required answer.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the number of
// trailing zeros
int trailing_zeros(int N)
{
// To store the number of 2s and 5s
int count_of_two = 0, count_of_five = 0;
for (int i = 1; i <= N; i++) {
int val = i;
while (val % 2 == 0 && val > 0) {
val /= 2;
// If we get a factor 2 then we
// have i number of 2s because
// the power of the number is
// raised to i
count_of_two += i;
}
while (val % 5 == 0 && val > 0) {
val /= 5;
// If we get a factor 5 then
// we have i number of 5s
// because the power of the
// number is raised to i
count_of_five += i;
}
}
// Take the minimum of them
int ans = min(count_of_two, count_of_five);
return ans;
}
// Driver code
int main()
{
int N = 12;
cout << trailing_zeros(N);
return 0;
}
// Java implementation of the approach
class GFG
{
// Function to return the number of
// trailing zeros
static int trailing_zeros(int N)
{
// To store the number of 2s and 5s
int count_of_two = 0, count_of_five = 0;
for (int i = 1; i <= N; i++)
{
int val = i;
while (val % 2 == 0 && val > 0)
{
val /= 2;
// If we get a factor 2 then we
// have i number of 2s because
// the power of the number is
// raised to i
count_of_two += i;
}
while (val % 5 == 0 && val > 0)
{
val /= 5;
// If we get a factor 5 then
// we have i number of 5s
// because the power of the
// number is raised to i
count_of_five += i;
}
}
// Take the minimum of them
int ans = Math.min(count_of_two, count_of_five);
return ans;
}
// Driver code
public static void main (String[] args)
{
int N = 12;
System.out.println(trailing_zeros(N));
}
}
// This code is contributed by chandan_jnu
# Python 3 implementation of the approach
# Function to return the number of
# trailing zeros
def trailing_zeros(N):
# To store the number of 2s and 5s
count_of_two = 0
count_of_five = 0
for i in range(1, N + 1, 1):
val = i
while (val % 2 == 0 and val > 0):
val /= 2
# If we get a factor 2 then we
# have i number of 2s because
# the power of the number is
# raised to i
count_of_two += i
while (val % 5 == 0 and val > 0):
val /= 5
# If we get a factor 5 then we
# have i number of 5s because
# the power of the number is
# raised to i
count_of_five += i
# Take the minimum of them
ans = min(count_of_two, count_of_five)
return ans
# Driver code
if __name__ == '__main__':
N = 12
print(trailing_zeros(N))
# This code is contributed by
# Sanjit_Prasad
// C# implementation of the above approach
using System;
class GFG
{
// Function to return the number of
// trailing zeros
static int trailing_zeros(int N)
{
// To store the number of 2s and 5s
int count_of_two = 0, count_of_five = 0;
for (int i = 1; i <= N; i++)
{
int val = i;
while (val % 2 == 0 && val > 0)
{
val /= 2;
// If we get a factor 2 then we
// have i number of 2s because
// the power of the number is
// raised to i
count_of_two += i;
}
while (val % 5 == 0 && val > 0)
{
val /= 5;
// If we get a factor 5 then
// we have i number of 5s
// because the power of the
// number is raised to i
count_of_five += i;
}
}
// Take the minimum of them
int ans = Math.Min(count_of_two, count_of_five);
return ans;
}
// Driver code
public static void Main()
{
int N = 12;
Console.WriteLine(trailing_zeros(N));
}
}
// This code is contributed by Ryuga
<?php
// PHP implementation of the approach
// Function to return the number of
// trailing zeros
function trailing_zeros($N)
{
// To store the number of 2s and 5s
$count_of_two = 0;
$count_of_five = 0;
for ($i = 1; $i <= $N; $i++)
{
$val = $i;
while ($val % 2 == 0 && $val > 0)
{
$val /= 2;
// If we get a factor 2 then we
// have i number of 2s because
// the power of the number is
// raised to i
$count_of_two += $i;
}
while ($val % 5 == 0 && $val > 0)
{
$val /= 5;
// If we get a factor 5 then
// we have i number of 5s
// because the power of the
// number is raised to i
$count_of_five += $i;
}
}
// Take the minimum of them
$ans = min($count_of_two, $count_of_five);
return $ans;
}
// Driver code
$N = 12;
echo trailing_zeros($N);
// This code is contributed by ita_c
?>
<script>
// Javascript implementation of the approach
// Function to return the number of
// trailing zeros
function trailing_zeros(N)
{
// To store the number of 2s and 5s
let count_of_two = 0, count_of_five = 0;
for(let i = 1; i <= N; i++)
{
let val = i;
while (val % 2 == 0 && val > 0)
{
val = parseInt(val / 2);
// If we get a factor 2 then we
// have i number of 2s because
// the power of the number is
// raised to i
count_of_two += i;
}
while (val % 5 == 0 && val > 0)
{
val = parseInt(val / 5);
// If we get a factor 5 then
// we have i number of 5s
// because the power of the
// number is raised to i
count_of_five += i;
}
}
// Take the minimum of them
let ans = Math.min(count_of_two,
count_of_five);
return ans;
}
// Driver code
let N = 12;
document.write(trailing_zeros(N));
// This code is contributed by subhammahato348
</script>
Time Complexity: O(N * (log2N + log5N))
Auxiliary Space: O(1), since no extra space has been taken.
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