2549. Count Distinct Numbers on Board

EasyArrayHash TableMathSimulation

Problem Description

You start with a positive integer n placed on a board. Every day for 10^9 days, you perform this procedure:

For each number x currently on the board, find all integers i where 1 ≤ i ≤ n such that x % i == 1
Add all those integers i to the board

Numbers remain on the board once placed. Your task is to find how many distinct integers will be on the board after 10^9 days.

For example, if x = 5 is on the board and n = 10, you would check which values of i from 1 to 10 satisfy 5 % i == 1. Since 5 % 1 = 0, 5 % 2 = 1, 5 % 3 = 2, and 5 % 4 = 1, you would add 2 and 4 to the board.

The key insight is that when you have a number x on the board, any number i where x % i == 1 means that x = k*i + 1 for some integer k. This is equivalent to saying i divides (x-1).

Starting with n, you'll generate n-1 (since n % (n-1) = 1). Then n-1 will generate n-2 (since (n-1) % (n-2) = 1), and this cascade continues. Eventually, you'll have all integers from 2 to n on the board, giving you n-1 distinct integers total.

The special case is when n = 1, where no new numbers can be generated (since there's no i where 1 % i == 1 for 1 ≤ i ≤ 1), so the answer remains 1.

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Intuition

Let's think about what happens when we have a number x on the board. We're looking for all values i where x % i == 1.

The condition x % i == 1 means that when we divide x by i, we get a remainder of 1. This can be rewritten as x = k*i + 1 for some integer k ≥ 1. Rearranging, we get x - 1 = k*i, which means i must be a divisor of (x-1).

Now, let's trace through what happens starting with n:

When we have n on the board, we check: what values of i satisfy n % i == 1?
One obvious answer is i = n-1, because n % (n-1) = 1 (since n = 1*(n-1) + 1)
So n-1 gets added to the board

Next, with n-1 on the board:

We need (n-1) % i == 1, which means i divides (n-2)
One such value is i = n-2, because (n-1) % (n-2) = 1
So n-2 gets added to the board

This pattern continues: each number x will generate x-1 because x % (x-1) = 1 is always true for x > 2.

Following this chain: n → n-1 → n-2 → ... → 3 → 2

The chain stops at 2 because 2 % 1 = 0 (not 1), so 1 never gets added to the board.

Therefore, starting from n, we eventually generate all integers from 2 to n, giving us exactly n-1 distinct integers. The only exception is when n = 1, where no new numbers can be generated, leaving us with just 1 integer on the board.

Learn more about Math patterns.

Solution Approach

Based on our intuition that every integer from 2 to n will eventually appear on the board, the solution becomes remarkably simple.

The key observation is that for any number x ≥ 2 on the board, the number x-1 will be added because x % (x-1) = 1. This creates a cascading effect:

n generates n-1
n-1 generates n-2
n-2 generates n-3
... and so on until we reach 2

Since this cascade ensures all numbers from 2 to n appear on the board, the total count of distinct integers is n - 1.

However, we need to handle the edge case where n = 1. When only 1 is on the board, no new numbers can be generated (since 1 % 1 = 0, not 1), so the answer remains 1.

Therefore, the implementation is simply:

def distinctIntegers(self, n: int) -> int:
    return max(1, n - 1)

The max(1, n - 1) elegantly handles both cases:

When n = 1: max(1, 0) = 1
When n > 1: max(1, n - 1) = n - 1

This solution has O(1) time and space complexity, as we don't need to simulate the entire process or store any intermediate values. The mathematical pattern allows us to directly compute the final answer.

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Example Walkthrough

Let's walk through a small example with n = 5 to see how the solution works.

Day 0: Board contains [5]

Check which values i (where 1 ≤ i ≤ 5) satisfy 5 % i == 1
5 % 1 = 0 ❌
5 % 2 = 1 ✓ → Add 2 to board
5 % 3 = 2 ❌
5 % 4 = 1 ✓ → Add 4 to board
5 % 5 = 0 ❌

Day 1: Board contains [5, 2, 4]

For 5: Already processed
For 2: Check 2 % i == 1 for i from 1 to 5
- Only 2 % 1 = 0 ❌ (no new numbers)
For 4: Check 4 % i == 1 for i from 1 to 5
- 4 % 3 = 1 ✓ → Add 3 to board

Day 2: Board contains [5, 2, 4, 3]

For 3: Check 3 % i == 1 for i from 1 to 5
- 3 % 2 = 1 ✓ → But 2 is already on board

Day 3 onwards: Board remains [5, 2, 4, 3]

No new numbers can be generated

Final board has 4 distinct integers: {2, 3, 4, 5}

Notice how we obtained all integers from 2 to 5, which is exactly n - 1 = 4 distinct integers. The cascade happened as: 5 generated 4, then 4 generated 3, and 3 would generate 2 (but 2 was already added directly from 5). This confirms our formula: for n = 5, the answer is max(1, 5 - 1) = 4.

Solution Implementation

1class Solution:
2    def distinctIntegers(self, n: int) -> int:
3        """
4        Calculate the number of distinct integers that can be obtained.
5      
6        For n = 1: Only 1 distinct integer (itself)
7        For n > 1: Can obtain n-1 distinct integers (typically 1 through n-1)
8      
9        Args:
10            n: The starting integer value
11          
12        Returns:
13            The count of distinct integers that can be obtained
14        """
15        # When n is 1, we can only have 1 distinct integer
16        # When n > 1, we can obtain n-1 distinct integers
17        return max(1, n - 1)
18

1class Solution {
2    /**
3     * Calculates the number of distinct integers that can be generated
4     * starting from integer n using the following rule:
5     * - For each integer x on the board, you can add x-1 if x-1 doesn't equal 0
6     * 
7     * The pattern emerges that starting from n, we can generate n-1, n-2, ..., 2, 1
8     * except when n=1, where only 1 remains on the board.
9     * 
10     * @param n The starting integer (1 <= n <= 100)
11     * @return The count of distinct integers that will eventually be on the board
12     */
13    public int distinctIntegers(int n) {
14        // Special case: when n is 1, only 1 can be on the board
15        // General case: when n > 1, we can generate all integers from 1 to n-1, plus n itself
16        // This gives us n distinct integers in total, but simplified to n-1 distinct additions
17        // The formula simplifies to max(1, n-1) to handle both cases
18        return Math.max(1, n - 1);
19    }
20}
21

1class Solution {
2public:
3    /**
4     * Calculate the number of distinct integers on the board after all operations.
5     * 
6     * Starting with integer n on the board, we can repeatedly add new integers
7     * by choosing an existing integer x and adding (x - 1) to the board if x - 1 >= 1.
8     * 
9     * The pattern is: starting from n, we can generate n-1, n-2, ..., 2, 1
10     * Therefore, the total distinct integers will be n-1 (all integers from 2 to n)
11     * except when n = 1, where only 1 remains on the board.
12     * 
13     * @param n The initial integer on the board (1 <= n <= 100)
14     * @return The number of distinct integers on the board after all operations
15     */
16    int distinctIntegers(int n) {
17        // Special case: when n is 1, only 1 remains on the board
18        // General case: we can generate all integers from 1 to n except 1 is generated from 2
19        // So we have n-1 distinct integers (2, 3, ..., n)
20        return std::max(1, n - 1);
21    }
22};
23

1/**
2 * Calculates the number of distinct integers that can be present on the board
3 * after performing operations starting with integer n.
4 * 
5 * The operation: For each integer x on the board, place another integer x-1 on the board.
6 * After infinite operations, all integers from 1 to n-1 will be present (except when n=1).
7 * 
8 * @param n - The starting integer on the board (1 <= n <= 100)
9 * @returns The number of distinct integers on the board after all operations
10 */
11function distinctIntegers(n: number): number {
12    // When n = 1, only 1 remains on the board (cannot generate 0)
13    // When n > 1, we can generate all integers from n down to 1
14    // Total distinct integers: n, n-1, n-2, ..., 2, 1 = n integers
15    // But since we cannot have 0, the count is n - 1 + 1 = n - 1 distinct positive integers
16    // Actually, starting from n, we get: n, n-1, n-2, ..., 2, 1
17    // That's n distinct integers when n > 1, but the formula n - 1 works because:
18    // We exclude n itself from counting and count all integers from 1 to n-1
19    return Math.max(1, n - 1);
20}
21

Time and Space Complexity

The time complexity is O(1) because the function performs a single comparison operation (max) between two values (1 and n-1) and returns the result. Both the subtraction n - 1 and the max comparison are constant-time operations that don't depend on the size of the input.

The space complexity is O(1) because the function only uses a fixed amount of additional memory regardless of the input value n. No extra data structures are created, and only primitive values are used for the computation and return value.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Attempting to Simulate the Entire Process

The Pitfall: Many developers initially try to simulate the entire 10^9 days process using sets or arrays to track all numbers on the board:

def distinctIntegers(self, n: int) -> int:
    board = {n}
    for day in range(10**9):  # This will timeout!
        new_numbers = set()
        for x in board:
            for i in range(1, n + 1):
                if x % i == 1:
                    new_numbers.add(i)
        if not new_numbers:
            break
        board.update(new_numbers)
    return len(board)

Why It's Wrong:

Time Limit Exceeded: Even with early termination, simulating potentially millions of iterations is unnecessary
Space inefficiency: Storing all intermediate states wastes memory
Misses the mathematical pattern that solves the problem instantly

The Solution: Recognize that the process stabilizes quickly. After analyzing a few examples, you'll notice that all numbers from 2 to n always appear, making simulation unnecessary.

2. Forgetting the Special Case n = 1

The Pitfall: Writing a simple return statement without considering edge cases:

def distinctIntegers(self, n: int) -> int:
    return n - 1  # Wrong for n = 1!

Why It's Wrong: When n = 1, this returns 0, but there's actually 1 number on the board (the initial 1 itself).

The Solution: Use max(1, n - 1) or add explicit condition checking:

def distinctIntegers(self, n: int) -> int:
    if n == 1:
        return 1
    return n - 1

3. Misunderstanding the Modulo Condition

The Pitfall: Some might think they need to find numbers where i % x == 1 instead of x % i == 1:

def distinctIntegers(self, n: int) -> int:
    # Incorrectly looking for i where i % x == 1
    # This would give completely different results!

Why It's Wrong: The problem specifically states we're looking for divisors i where x % i == 1, meaning x leaves remainder 1 when divided by i. This is fundamentally different from i % x == 1.

The Solution: Carefully read the problem statement and understand that for a number x on the board, we add all i where x % i == 1, which means i divides (x-1).

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