LeetCode Problem 3264

Difficulty: 🟢 Easy
Topics: Array, Math, Heap (Priority Queue), Simulation

Solution

LeetCode 3264 - Final Array State After K Multiplication Operations I

Problem Understanding

This problem asks us to repeatedly modify an array according to a very specific rule.

We are given:

• An integer array nums
• An integer k, representing how many operations to perform
• An integer multiplier

For each of the k operations, we must find the smallest value currently present in the array. If the minimum value appears multiple times, we must choose the one with the smallest index, meaning the first occurrence from left to right.

Once that element is selected, we replace it with:

$$x \times multiplier$$

After performing exactly k operations, we return the resulting array.

The important detail is that the array changes after every operation. Therefore, the minimum element for the next operation must be determined using the updated array, not the original one.

The constraints are very small:

• nums.length <= 100
• k <= 10

This tells us that even relatively inefficient solutions are acceptable because the total amount of work is tiny. In fact, a straightforward simulation is more than fast enough.

Several edge cases are worth noting:

• Multiple elements may share the same minimum value. We must always choose the first occurrence.
• multiplier can be 1, meaning values never change.
• The minimum element after multiplication may still remain the minimum element and be selected again.
• Arrays may contain only one element, causing the same element to be updated repeatedly.
• Since all values are positive integers, we never need to worry about negative numbers affecting minimum comparisons.

Approaches

Brute Force

The most direct approach is to simulate exactly what the problem describes.

For each of the k operations:

1. Scan the entire array to find the first occurrence of the minimum value.
2. Multiply that element by multiplier.
3. Continue to the next operation.

This approach is obviously correct because it follows the problem statement literally.

Finding the minimum requires scanning all n elements. Since we perform this operation k times, the total complexity is O(k * n).

Optimal Approach

A more advanced solution uses a min-heap.

The heap stores pairs:

• Current value
• Original index

The heap ordering is:

1. Smaller value first
2. Smaller index first when values are equal

This exactly matches the problem's selection rule.

For each operation:

1. Remove the minimum element from the heap.
2. Multiply its value.
3. Update the array.
4. Insert the new value back into the heap.

This avoids repeatedly scanning the entire array.

Although the constraints are small enough that the brute force solution is perfectly acceptable, the heap solution demonstrates a more scalable approach and is typically considered the optimal implementation.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(k × n)	O(1)	Repeatedly scan array to find minimum
Optimal	O((n + k) log n)	O(n)	Uses a min-heap to retrieve the minimum efficiently

Algorithm Walkthrough

We will use a min-heap that stores (value, index) pairs.

Step 1

Create a heap containing every element of the array together with its index.

The index is stored because ties must be broken by choosing the first occurrence.

Step 2

Convert the collection into a valid min-heap.

The heap will always place the smallest value at the top. If two values are equal, the smaller index comes first.

Step 3

Repeat the following process exactly k times.

Remove the top element from the heap. This element represents the current minimum value according to the problem rules.

Step 4

Multiply the extracted value by multiplier.

This produces the new value that should replace the selected array element.

Step 5

Update the corresponding position in nums.

This keeps the array synchronized with the heap.

Step 6

Insert the updated (newValue, index) pair back into the heap.

This ensures future operations consider the modified value.

Step 7

After all k operations have been completed, return the updated array.

Why it works

The heap always contains exactly one entry for every array element. Because entries are ordered first by value and then by index, the top of the heap always corresponds to the element the problem requires us to choose. After modifying an element, we immediately insert its updated value back into the heap, preserving this property. Therefore every operation matches the problem statement exactly, which guarantees the final array is correct.

Python Solution

from typing import List
import heapq

class Solution:
    def getFinalState(self, nums: List[int], k: int, multiplier: int) -> List[int]:
        heap = [(value, index) for index, value in enumerate(nums)]
        heapq.heapify(heap)

        for _ in range(k):
            value, index = heapq.heappop(heap)

            value *= multiplier
            nums[index] = value

            heapq.heappush(heap, (value, index))

        return nums

The implementation begins by building a heap containing every value together with its index. Python's tuple comparison naturally provides the desired ordering because tuples are compared lexicographically. This means (value, index) automatically sorts by value first and index second.

The loop executes exactly k times. During each iteration, the smallest element is removed from the heap, multiplied by multiplier, and written back into the array.

After updating the array, the modified value is inserted back into the heap so future iterations operate on the current array state.

Once all operations have been performed, the updated array is returned.

Go Solution

package main

import "container/heap"

type Node struct {
	value int
	index int
}

type MinHeap []Node

func (h MinHeap) Len() int {
	return len(h)
}

func (h MinHeap) Less(i, j int) bool {
	if h[i].value != h[j].value {
		return h[i].value < h[j].value
	}
	return h[i].index < h[j].index
}

func (h MinHeap) Swap(i, j int) {
	h[i], h[j] = h[j], h[i]
}

func (h *MinHeap) Push(x interface{}) {
	*h = append(*h, x.(Node))
}

func (h *MinHeap) Pop() interface{} {
	old := *h
	n := len(old)
	item := old[n-1]
	*h = old[:n-1]
	return item
}

func getFinalState(nums []int, k int, multiplier int) []int {
	h := &MinHeap{}

	for i, value := range nums {
		*h = append(*h, Node{
			value: value,
			index: i,
		})
	}

	heap.Init(h)

	for k > 0 {
		node := heap.Pop(h).(Node)

		node.value *= multiplier
		nums[node.index] = node.value

		heap.Push(h, node)
		k--
	}

	return nums
}

The Go solution follows exactly the same logic as the Python version. Since Go does not provide a built in priority queue, we implement the heap.Interface from the container/heap package.

The Less method enforces the required ordering by comparing values first and indices second. This guarantees that ties are resolved according to the problem statement.

Because the constraints are small, integer overflow is not a concern for this problem.

Worked Examples

Example 1

Input

nums = [2,1,3,5,6]
k = 5
multiplier = 2

Initial heap:

(1,1) (2,0) (3,2) (5,3) (6,4)

Operation	Selected	Updated Array
1	1 at index 1	[2,2,3,5,6]
2	2 at index 0	[4,2,3,5,6]
3	2 at index 1	[4,4,3,5,6]
4	3 at index 2	[4,4,6,5,6]
5	4 at index 0	[8,4,6,5,6]

Final answer:

[8,4,6,5,6]

Example 2

Input

nums = [1,2]
k = 3
multiplier = 4

Operation	Selected	Updated Array
1	1 at index 0	[4,8]
2	2 at index 1	[4,8]
3	4 at index 0	[16,8]

A more detailed trace is:

Operation	Heap Minimum	New Value	Array State
Start	(1,0)	-	[1,2]
1	(1,0)	4	[4,2]
2	(2,1)	8	[4,8]
3	(4,0)	16	[16,8]

Final answer:

[16,8]

Complexity Analysis

Measure	Complexity	Explanation
Time	O((n + k) log n)	Heap construction plus k pop/push operations
Space	O(n)	Heap stores one entry per array element

Building the heap requires O(n) time. Each of the k operations performs one extraction and one insertion, both costing O(log n). Therefore the total complexity is O(n + k log n), commonly written as O((n + k) log n).

The heap stores all array elements, requiring O(n) extra space.

Test Cases

sol = Solution()

assert sol.getFinalState([2, 1, 3, 5, 6], 5, 2) == [8, 4, 6, 5, 6]  # Example 1

assert sol.getFinalState([1, 2], 3, 4) == [16, 8]  # Example 2

assert sol.getFinalState([5], 3, 2) == [40]  # Single element updated repeatedly

assert sol.getFinalState([1, 1, 1], 1, 2) == [2, 1, 1]  # First minimum chosen

assert sol.getFinalState([1, 1, 1], 2, 2) == [2, 2, 1]  # Tie breaking continues

assert sol.getFinalState([3, 4, 5], 2, 1) == [3, 4, 5]  # Multiplier equals 1

assert sol.getFinalState([1, 100], 10, 2) == [128, 200]  # Same minimum selected multiple times

assert sol.getFinalState([2, 2, 2, 2], 4, 3) == [6, 6, 6, 6]  # All values equal

assert sol.getFinalState([100], 10, 5) == [976562500]  # Maximum-style repeated growth

assert sol.getFinalState([1, 2, 3], 1, 5) == [5, 2, 3]  # Single operation

Test	Why
[2,1,3,5,6], k=5, m=2	Official example
[1,2], k=3, m=4	Official example
[5], k=3, m=2	Single-element array
[1,1,1], k=1, m=2	Verifies first-occurrence tie breaking
[1,1,1], k=2, m=2	Repeated tie handling
[3,4,5], k=2, m=1	Values never change
[1,100], k=10, m=2	Minimum can be selected many times
[2,2,2,2], k=4, m=3	All elements initially identical
[100], k=10, m=5	Large repeated multiplication
[1,2,3], k=1, m=5	Minimal operation count

Edge Cases

Multiple Minimum Values

When several elements share the same minimum value, selecting the wrong occurrence will produce an incorrect result. For example, in [1,1,1], the first operation must modify index 0, not index 1 or 2. The heap solution stores both value and index, and compares indices when values are equal, guaranteeing correct tie breaking.

Multiplier Equal to One

If multiplier = 1, every multiplication leaves the selected value unchanged. A buggy implementation might assume values always change and accidentally enter incorrect logic. The heap approach works naturally because the updated value is simply reinserted unchanged.

Single Element Array

When the array contains only one element, that same element must be selected in every operation. Some implementations accidentally assume there are multiple candidates. The heap always contains exactly one entry, so repeated updates are handled correctly.

The Same Element Remains the Minimum

After multiplication, an element may still be the smallest value in the array. For example, with nums = [1, 100] and multiplier = 2, the first element remains the minimum for several operations. The algorithm correctly reinserts the updated value into the heap, allowing it to be selected again whenever it is still the smallest element.

All Elements Equal

When every element starts with the same value, the first-occurrence rule becomes critical. For example, with [2,2,2,2], the first operation must update index 0, the second operation index 1, and so on. The (value, index) ordering in the heap ensures this behavior exactly matches the problem specification.