Problem Description

You need to find a set of positive integers where each integer ends with the digit k (units digit is k), and the sum of all integers in the set equals num.

The goal is to return the minimum possible size of such a set. If it's impossible to create such a set, return -1.

Key points to understand:

• Every number in your set must have k as its last digit (e.g., if k = 3, valid numbers could be 3, 13, 23, 33, etc.)
• The same number can appear multiple times in the set
• The sum of all numbers in the set must equal exactly num
• You want the smallest possible set size
• An empty set has a sum of 0

For example:

• If num = 58 and k = 9, you could use the set {9, 49} which has size 2
• If num = 37 and k = 2, you could use the set {2, 12, 12, 12} or other combinations, but you want the minimum size

The solution works by recognizing that any number ending in k can be written as 10x + k for some non-negative integer x. If we have n such numbers summing to num, then num - n*k must be divisible by 10 (since it represents the sum of all the 10x parts). The code finds the smallest n where this condition holds.

Intuition

Let's think about what numbers we can use. Any number ending with digit k can be written as 10x + k where x is some non-negative integer. For example, if k = 3, we have numbers like 3 (when x = 0), 13 (when x = 1), 23 (when x = 2), and so on.

Now, if we use n such numbers to sum up to num, we can write:

• (10x₁ + k) + (10x₂ + k) + ... + (10xₙ + k) = num
• Rearranging: 10(x₁ + x₂ + ... + xₙ) + n*k = num
• Therefore: 10(x₁ + x₂ + ... + xₙ) = num - n*k

For this equation to have a valid solution, num - n*k must be divisible by 10 (since the left side is always a multiple of 10). Also, num - n*k must be non-negative because the sum x₁ + x₂ + ... + xₙ cannot be negative.

The key insight is that we want the minimum n. So we start checking from n = 1 and incrementally increase it. The first n that makes num - n*k both non-negative and divisible by 10 is our answer.

Why does this give us the minimum? Because once we find a valid n, we know we can distribute the value (num - n*k)/10 among our n numbers. We could use larger values of n, but that would mean using more numbers, which contradicts our goal of finding the minimum set size.

Special case: If num = 0, we need zero numbers (empty set), so we return 0 immediately.

Learn more about Greedy, Math and Dynamic Programming patterns.

Solution Approach

The implementation follows directly from our mathematical insight that num - n*k must be divisible by 10 for a valid solution.

First, we handle the special case where num = 0. An empty set has a sum of 0, so we return 0 immediately.

For non-zero values of num, we iterate through possible set sizes from 1 to num:

for i in range(1, num + 1):
    if (t := num - k * i) >= 0 and t % 10 == 0:
        return i

In each iteration with set size i:

1. Calculate t = num - k * i, which represents the remaining value after subtracting i instances of the units digit k
2. Check two conditions:
   • t >= 0: Ensures we haven't subtracted too much (the remaining value must be non-negative)
   • t % 10 == 0: Ensures the remaining value is divisible by 10

When both conditions are met, we've found our answer. The value i represents the minimum number of integers needed where each ends with digit k and their sum equals num.

Why does this work? Because when t is divisible by 10, we can distribute the value t/10 among our i numbers as the tens and higher digits. For example, if we need i = 3 numbers ending in k, and t = 30, we could use numbers like (10 + k), (10 + k), (10 + k), distributing the 30 as three 10s.

If no valid i is found after checking all possibilities from 1 to num, we return -1, indicating it's impossible to form such a set.

The algorithm has a time complexity of O(num) in the worst case, as we might need to check up to num iterations. The space complexity is O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's walk through an example with num = 58 and k = 9.

We need to find the minimum number of positive integers, each ending with digit 9, that sum to 58.

Starting with our approach, we check different set sizes i:

When i = 1:

• Calculate t = num - k * i = 58 - 9 * 1 = 49
• Check: Is t >= 0? Yes (49 ≥ 0)
• Check: Is t % 10 == 0? No (49 % 10 = 9)
• Not valid, continue.

When i = 2:

• Calculate t = num - k * i = 58 - 9 * 2 = 40
• Check: Is t >= 0? Yes (40 ≥ 0)
• Check: Is t % 10 == 0? Yes (40 % 10 = 0)
• Valid! Return i = 2

So the minimum size is 2. How do we construct such a set?

Since t = 40 and t/10 = 4, we need to distribute 4 among our 2 numbers as the tens digit. We could use:

• 9 (which is 010 + 9) and 49 (which is 410 + 9)
• Or 19 (which is 110 + 9) and 39 (which is 310 + 9)
• Or 29 (which is 210 + 9) and 29 (which is 210 + 9)

All these sets have exactly 2 numbers ending in 9 that sum to 58. The algorithm correctly identifies that 2 is the minimum possible size.

Solution Implementation

Time and Space Complexity

The time complexity is O(num), where num is the input value. The loop iterates from 1 to num in the worst case, performing constant-time operations in each iteration (arithmetic operations and modulo checks). Since the loop can run up to num times, the time complexity is linear with respect to the value of num.

The space complexity is O(1). The algorithm only uses a constant amount of extra space for the loop variable i and the temporary variable t, regardless of the input size. No additional data structures are created that scale with the input.

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

Common Pitfalls

1. Forgetting the Zero Edge Case

A common mistake is not handling num = 0 correctly. When the target sum is 0, the answer should be 0 (empty set), not -1 or 1.

Incorrect approach:

def minimumNumbers(self, num: int, k: int) -> int:
    for i in range(1, num + 1):  # Starting from 1, missing the 0 case
        if (num - k * i) >= 0 and (num - k * i) % 10 == 0:
            return i
    return -1

Solution: Always check for num = 0 first and return 0 immediately.

2. Incorrect Loop Boundary

Another pitfall is using an incorrect upper bound for the loop. Some might think to optimize by using a smaller boundary like num // k, but this can miss valid solutions.

Incorrect approach:

def minimumNumbers(self, num: int, k: int) -> int:
    if num == 0:
        return 0
    for i in range(1, num // k + 1):  # Wrong boundary - too restrictive
        if (num - k * i) >= 0 and (num - k * i) % 10 == 0:
            return i
    return -1

Why it's wrong: When k = 1 and num = 10, we need 10 elements (each being 1), but num // k + 1 = 11 would work. However, if k = 3 and num = 10, we might need up to 10 elements theoretically, but num // k + 1 = 4 would cut off potential solutions.

Solution: Use range(1, num + 1) to ensure all possibilities are checked. The early return ensures efficiency.

3. Misunderstanding the Divisibility Check

Some might incorrectly check if num % 10 == k instead of checking if (num - k * i) % 10 == 0.

Incorrect approach:

def minimumNumbers(self, num: int, k: int) -> int:
    if num == 0:
        return 0
    if num % 10 != k:  # Wrong logic - too restrictive
        return -1
    # ... rest of the code

Why it's wrong: This assumes we can only solve the problem if num itself ends with k, which is incorrect. For example, num = 58 and k = 9 has a valid solution {9, 49}, even though 58 doesn't end with 9.

Solution: The correct check is (num - k * i) % 10 == 0 for each potential set size i.

4. Integer Overflow Concerns (Language-Specific)

In languages with fixed integer sizes, calculating k * i might cause overflow for large values.

Potential issue in other languages:

# In languages like C++ or Java with int32
remaining = num - k * i  # Could overflow if k * i is very large

Solution: In Python, this isn't an issue due to arbitrary precision integers. In other languages, consider using long/int64 types or add additional boundary checks to prevent overflow.