Preimage Size of Factorial Zeroes Function

Preimage Size of Factorial Zeroes Function - Problem

Let f(x) be the number of zeroes at the end of x!. Recall that x! = 1 * 2 * 3 * ... * x and by convention, 0! = 1.

For example, f(3) = 0 because 3! = 6 has no zeroes at the end, while f(11) = 2 because 11! = 39916800 has two zeroes at the end.

Given an integer k, return the number of non-negative integers x have the property that f(x) = k.

Input & Output

Example 1 — Basic Case

$ Input: k = 3

› Output: 5

💡 Note: f(15) = f(16) = f(17) = f(18) = f(19) = 3, so exactly 5 numbers give 3 trailing zeros

Example 2 — Special Case Zero

$ Input: k = 0

› Output: 5

💡 Note: f(0) = f(1) = f(2) = f(3) = f(4) = 0, so exactly 5 numbers give 0 trailing zeros

Example 3 — Impossible Value

$ Input: k = 4

› Output: 5

💡 Note: f(20) = f(21) = f(22) = f(23) = f(24) = 4, so exactly 5 numbers give 4 trailing zeros

Constraints

0 ≤ k ≤ 10⁹

Visualization

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Asked in

G Google 15 M Microsoft 8

Brief HTML summary: The key insight is that factorial trailing zeros come from factors of 10 = 2×5, and there are always more factors of 2 than 5, so we count factors of 5. The function f(x) is monotonically increasing but has gaps - some values are impossible. Best approach uses binary search to find boundaries efficiently. Time: O(log²k), Space: O(1)

Common Approaches

✓ Binary Search on Answer Space

⏱️ Time: O(log²k) Space: O(1)

Since f(x) is monotonically increasing, we can binary search to find the first and last x where f(x) = k. The key insight is that f(x) has gaps - some values of k are impossible.

Brute Force Linear Search

⏱️ Time: O(k × log k) Space: O(1)

Start from x=0 and count trailing zeros for each factorial until we find numbers that give exactly k zeros. Count how many consecutive numbers give this count.

Binary Search on Answer Space — Algorithm Steps

Step 1: Binary search to find smallest x where f(x) >= k
Step 2: Check if f(x) exactly equals k
Step 3: Return 5 if found (always 5 consecutive), 0 if impossible

Visualization

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Step-by-Step Walkthrough

Set search range

left=0, right=5*k

Binary search

Find first x where f(x) >= k

Verify exact match

Check if f(x) = k exactly

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

long long countTrailingZeros(long long n) {
    long long count = 0;
    for (long long i = 5; i <= n; i *= 5) {
        count += n / i;
        if (i > n / 5) break;  // Prevent overflow
    }
    return count;
}

int solution(int k) {
    // Binary search for first x where f(x) >= k
    long long left = 0, right = 5LL * k + 10;  // Upper bound with some buffer
    
    while (left < right) {
        long long mid = (left + right) / 2;
        if (countTrailingZeros(mid) < k) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    
    // Check if f(left) == k
    if (countTrailingZeros(left) == k) {
        return 5;  // Always exactly 5 consecutive numbers
    } else {
        return 0;  // k is impossible
    }
}

int main() {
    int k;
    scanf("%d", &k);
    printf("%d\n", solution(k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log²k)

Binary search takes O(log k) iterations, each iteration calculates trailing zeros in O(log k)

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra variables for binary search

✓ Linear Space

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Common Approaches

Binary Search on Answer Space — Algorithm Steps

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Code -

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