LeetCode Problem 3091

Difficulty: 🟡 Medium
Topics: Math, Greedy, Enumeration

Solution

Problem Understanding

The problem gives us a single positive integer k. We start with an array containing exactly one element:

nums = [1]

We are allowed to perform two kinds of operations any number of times:

1. Increase the value of any existing element by 1
2. Duplicate any existing element and append the duplicate to the array

Our goal is to make the total sum of the array greater than or equal to k, while using the minimum possible number of operations.

The key detail is that duplication copies the current value of an element. This means increasing a number before duplicating it can be much more efficient than repeatedly increasing many small elements later.

For example, suppose we transform [1] into [4] using three increment operations. After that, every duplication adds another 4 to the sum in a single operation. This is much more efficient than duplicating smaller numbers.

The input consists of a single integer:

1 <= k <= 10^5

The constraint is small enough that we can try multiple candidate strategies through enumeration, but large enough that brute-force state simulation would become impractical.

There are several important edge cases:

• If k = 1, the initial array already satisfies the condition, so the answer is 0.
• Very small values such as k = 2 or k = 3 can expose off-by-one mistakes in formulas.
• Cases where k is not divisible evenly by the chosen element value require ceiling division.
• A naive greedy strategy that always duplicates or always increments does not always produce the optimal answer.

Approaches

Brute Force Approach

A brute-force strategy would attempt to simulate every possible sequence of operations. At each step, we could either increment an existing element or duplicate one of the elements.

This produces an enormous branching factor because:

• Every element can potentially be incremented
• Every element can potentially be duplicated
• The number of elements grows over time

The number of possible states grows exponentially, making exhaustive search infeasible even for moderate values of k.

Even if we compress states by storing only sums and maximum values, the search space remains far too large.

The brute-force method is correct because it explores every possible operation sequence and eventually finds the minimum one, but it is computationally impractical.

Optimal Observation

The key insight is that the exact distribution of numbers in the array does not matter as much as:

• The value we build up to
• How many copies we create of that value

Suppose we spend some operations increasing the original 1 into a value x.

Starting from 1, reaching x requires:

x - 1

increment operations.

Now we have:

[x]

If we duplicate this value d times, we end up with:

d + 1

copies of x.

The final sum becomes:

x * (d + 1)

We need:

x * (d + 1) >= k

For a fixed x, the minimum number of duplications needed is:

ceil(k / x) - 1

So the total operations become:

(x - 1) + (ceil(k / x) - 1)

We can simply enumerate all possible values of x and take the minimum.

Since k <= 10^5, checking all values from 1 to k is completely efficient.

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	Exponential	Exponential	Explores all operation sequences
Optimal	O(k)	O(1)	Enumerates possible target values

Algorithm Walkthrough

1. Handle the trivial case where k == 1.

The initial array already has sum 1, so no operations are needed. 2. Enumerate every possible target value x.

Here, x represents the value we first build using increment operations before performing duplications. 3. Compute the increment cost.

Since we start from 1, converting it into x requires:

x - 1

operations. 4. Compute how many copies of x are needed.

If every element has value x, then we need at least:

ceil(k / x)

copies. 5. Compute the duplication cost.

We already have one copy after the increments, so we need:

ceil(k / x) - 1

duplications. 6. Compute the total operations.

total = (x - 1) + (ceil(k / x) - 1)

7. Track the minimum answer across all candidate values of x.
8. Return the minimum total operations found.

Why it works

The strategy works because duplication preserves value. Once we decide the value x we want to build, the best possible configuration is to duplicate that value repeatedly, since every duplication adds the maximum possible sum increase for one operation.

Any optimal solution can therefore be represented as:

• First increasing a value to some target x
• Then duplicating it enough times

By enumerating all possible x, we guarantee that we examine the globally optimal balance between increment operations and duplication operations.

Python Solution

class Solution:
    def minOperations(self, k: int) -> int:
        if k == 1:
            return 0

        answer = float("inf")

        for value in range(1, k + 1):
            increments = value - 1

            copies_needed = (k + value - 1) // value
            duplications = copies_needed - 1

            total_operations = increments + duplications

            answer = min(answer, total_operations)

        return answer

The implementation directly follows the mathematical observation developed earlier.

The variable value represents the target number we build before duplicating. Since we start from 1, the number of increment operations needed is value - 1.

To determine how many copies are necessary, we compute ceiling division using:

(k + value - 1) // value

This gives the minimum number of copies whose total sum reaches at least k.

Because we already possess one copy after the increment phase, the number of duplication operations is one less than the total copies required.

For every candidate value, we compute the total operations and keep the minimum answer.

Go Solution

func minOperations(k int) int {
    if k == 1 {
        return 0
    }

    answer := int(1e9)

    for value := 1; value <= k; value++ {
        increments := value - 1

        copiesNeeded := (k + value - 1) / value
        duplications := copiesNeeded - 1

        totalOperations := increments + duplications

        if totalOperations < answer {
            answer = totalOperations
        }
    }

    return answer
}

The Go implementation mirrors the Python logic very closely.

Go does not provide built-in infinity values for integers, so we initialize answer with a sufficiently large number such as 1e9.

Integer division in Go automatically truncates toward zero, which allows the same ceiling-division formula:

(k + value - 1) / value

Since the constraints are only up to 10^5, integer overflow is not a concern.

Worked Examples

Example 1

Input:

k = 11

We enumerate possible target values.

Target Value x	Increment Ops	Copies Needed	Duplication Ops	Total Ops
1	0	11	10	10
2	1	6	5	6
3	2	4	3	5
4	3	3	2	5
5	4	3	2	6
6	5	2	1	6

The minimum value is 5.

One optimal construction is:

[1]

Increment three times:

[4]

Duplicate twice:

[4, 4, 4]

Final sum:

12

Total operations:

3 + 2 = 5

Example 2

Input:

k = 1

The initial array already has sum 1.

No operations are required.

Output:

0

Complexity Analysis

Measure	Complexity	Explanation
Time	O(k)	We enumerate all candidate target values from 1 to k
Space	O(1)	Only a few variables are used

The algorithm performs a constant amount of work for each possible target value. Since there are exactly k candidates, the total runtime is linear in k.

The memory usage remains constant because we do not allocate any additional data structures proportional to the input size.

Test Cases

solution = Solution()

assert solution.minOperations(1) == 0   # already satisfies condition
assert solution.minOperations(2) == 1   # one duplication is enough
assert solution.minOperations(3) == 2   # increment once, duplicate once
assert solution.minOperations(4) == 2   # duplicate twice
assert solution.minOperations(5) == 3   # optimal balance
assert solution.minOperations(11) == 5  # provided example
assert solution.minOperations(12) == 5  # exact multiple after duplication
assert solution.minOperations(15) == 6  # larger mixed strategy
assert solution.minOperations(100) == 18  # stress moderate value
assert solution.minOperations(100000) > 0  # upper constraint boundary

Test	Why
k = 1	Validates zero-operation edge case
k = 2	Smallest non-trivial input
k = 3	Checks mixed increment and duplication behavior
k = 4	Ensures repeated duplication works
k = 5	Validates balancing operations
k = 11	Confirms provided example
k = 12	Tests exact divisibility
k = 15	Tests larger mixed strategy
k = 100	Moderate stress test
k = 100000	Upper constraint boundary

Edge Cases

Case 1: k = 1

This is the most important boundary condition. The initial array already contains [1], whose sum is exactly 1. A buggy implementation might still perform operations unnecessarily or incorrectly enter the enumeration loop.

The implementation handles this immediately with:

if k == 1:
    return 0

Case 2: Ceiling Division Requirements

Suppose k = 11 and we choose x = 4.

We need:

ceil(11 / 4) = 3

copies.

Using normal floor division would incorrectly compute only 2 copies, producing a sum of 8, which is insufficient.

The implementation correctly uses ceiling division:

(k + value - 1) // value

Case 3: Large Values Near the Constraint Limit

For values close to 10^5, brute-force state simulation would become infeasible due to exponential growth in possibilities.

The optimized solution avoids this completely by reducing the problem to mathematical enumeration. Even at the maximum constraint, the algorithm performs only 100000 iterations, which is easily fast enough.