Problem Description

You have an integer array nums, and you need to perform k operations on it using a given multiplier.

For each operation:

1. Find the minimum value in nums
2. If there are multiple occurrences of this minimum value, select the one that appears first (has the smallest index)
3. Multiply this selected value by multiplier and replace it in the array

After performing all k operations, return the modified array.

Example walkthrough:

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

• Operation 1: Minimum is 1 at index 1. After multiplication: [2, 2, 3, 5, 6]
• Operation 2: Minimum is 2 at index 0 (first occurrence). After multiplication: [4, 2, 3, 5, 6]
• Operation 3: Minimum is 2 at index 1. After multiplication: [4, 4, 3, 5, 6]
• Operation 4: Minimum is 3 at index 2. After multiplication: [4, 4, 6, 5, 6]
• Operation 5: Minimum is 4 at index 0 (first occurrence). After multiplication: [8, 4, 6, 5, 6]

The final array would be [8, 4, 6, 5, 6].

The solution uses a min-heap (priority queue) to efficiently track the minimum element and its index. The heap stores tuples of (value, index), which automatically maintains both the minimum value and handles ties by choosing the smallest index first. This approach avoids repeatedly scanning the entire array to find the minimum, making the solution more efficient.

Intuition

The naive approach would be to scan through the entire array for each of the k operations to find the minimum value and its first occurrence. This would take O(n) time for each operation, resulting in O(k * n) total time complexity.

We can optimize this by recognizing that we're repeatedly performing two tasks:

1. Finding the minimum element (with its index)
2. Updating that element and maintaining the sorted order

This pattern suggests using a data structure that efficiently maintains elements in sorted order while supporting quick extraction of the minimum and insertion of updated values. A min-heap is perfect for this scenario because:

• Extracting the minimum takes O(log n) time
• Inserting a new element takes O(log n) time
• The heap naturally maintains the minimum element at the root

The key insight is that we need to track not just the values but also their original indices to handle the "first occurrence" requirement. By storing tuples of (value, index) in the heap, Python's heapq will automatically sort first by value, then by index when values are equal. This elegantly solves the tie-breaking requirement without additional logic.

Since we perform k operations, each involving one heap pop and one heap push (both O(log n)), the total time complexity improves to O(k * log n), which is significantly better than the naive approach when k is large.

The solution modifies the array in-place while using the heap to track which element to modify next, combining the efficiency of heap operations with the simplicity of direct array updates.

Learn more about Math and Heap (Priority Queue) patterns.

Solution Approach

The solution uses a min-heap (priority queue) combined with simulation to efficiently perform the k operations.

Step 1: Initialize the Min-Heap

pq = [(x, i) for i, x in enumerate(nums)]
heapify(pq)

• Create a list of tuples where each tuple contains (value, index) from the original array
• Convert this list into a min-heap using heapify() which takes O(n) time
• The heap will automatically order elements first by value, then by index when values are equal

Step 2: Perform k Operations

for _ in range(k):
    _, i = heappop(pq)
    nums[i] *= multiplier
    heappush(pq, (nums[i], i))

For each operation:

1. Extract the minimum element from the heap using heappop() - this gives us the index i of the minimum value
2. Multiply the value at nums[i] by the multiplier directly in the original array
3. Push the updated value back into the heap as a new tuple (nums[i], i)

Step 3: Return the Result

return nums

After all k operations, the array nums contains the final state and is returned.

Why This Works:

• The heap maintains the invariant that the root always contains the minimum value
• When multiple elements have the same value, Python's tuple comparison ensures the one with the smaller index is selected first
• By modifying nums directly and maintaining the heap separately, we avoid creating unnecessary copies of the array
• Each heap operation (heappop and heappush) takes O(log n) time

Time Complexity: O(n + k * log n) where n is the length of the array

• Building the initial heap: O(n)
• Performing k operations with heap pop and push: O(k * log n)

Space Complexity: O(n) for storing the heap

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Given: nums = [3, 2, 5, 2], k = 3, multiplier = 3

Step 1: Initialize the Min-Heap

We create tuples of (value, index) and build a heap:

• Initial array: [3, 2, 5, 2]
• Tuples: [(3, 0), (2, 1), (5, 2), (2, 3)]
• After heapify: The heap internally arranges these with (2, 1) at the root
  • Note: (2, 1) comes before (2, 3) because when values are equal, the smaller index has priority

Step 2: Perform Operations

Operation 1:

• heappop() returns (2, 1) - the minimum value 2 at index 1
• Multiply: nums[1] = 2 * 3 = 6
• Array becomes: [3, 6, 5, 2]
• heappush((6, 1)) back into the heap
• Heap now has: [(2, 3), (3, 0), (5, 2), (6, 1)] with (2, 3) at root

Operation 2:

• heappop() returns (2, 3) - the minimum value 2 at index 3
• Multiply: nums[3] = 2 * 3 = 6
• Array becomes: [3, 6, 5, 6]
• heappush((6, 3)) back into the heap
• Heap now has: [(3, 0), (5, 2), (6, 1), (6, 3)] with (3, 0) at root

Operation 3:

• heappop() returns (3, 0) - the minimum value 3 at index 0
• Multiply: nums[0] = 3 * 3 = 9
• Array becomes: [9, 6, 5, 6]
• heappush((9, 0)) back into the heap

Step 3: Return Result

Final array: [9, 6, 5, 6]

The key insight is that the heap always gives us the correct minimum element to modify next, handling both the value comparison and index tie-breaking automatically through tuple comparison.

Solution Implementation

Time and Space Complexity

Time Complexity: O((n + k) × log n)

The time complexity breaks down as follows:

• Creating the initial priority queue from the list comprehension takes O(n) time
• The heapify operation transforms the list into a heap in O(n) time
• The main loop runs k iterations, and in each iteration:
  • heappop extracts the minimum element in O(log n) time
  • Multiplying nums[i] is O(1)
  • heappush inserts the updated element back into the heap in O(log n) time
  • Total per iteration: O(log n)
• Overall: O(n) for initialization + O(k × log n) for the loop = O(n + k × log n) = O((n + k) × log n)

Space Complexity: O(n)

The space complexity is determined by:

• The priority queue pq stores n tuples (one for each element in nums), requiring O(n) space
• No other significant auxiliary space is used
• The input array nums is modified in-place and doesn't count toward extra space
• Total auxiliary space: O(n)

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

Common Pitfalls

Pitfall 1: Modifying the Heap Values Instead of the Original Array

The Problem: A common mistake is trying to modify values directly in the heap structure or creating a new array from heap elements at the end, rather than modifying the original nums array during each operation.

Incorrect Approach:

def getFinalState(self, nums: List[int], k: int, multiplier: int) -> List[int]:
    pq = [(x, i) for i, x in enumerate(nums)]
    heapify(pq)
  
    for _ in range(k):
        val, i = heappop(pq)
        val *= multiplier  # This doesn't update anything!
        heappush(pq, (val, i))
  
    # Trying to reconstruct the array from heap - WRONG!
    result = [0] * len(nums)
    while pq:
        val, i = heappop(pq)
        result[i] = val
    return result

Why It Fails:

• The heap contains ALL elements throughout the process, not just the modified ones
• After k operations, the heap still has n elements, but some values are outdated
• Reconstructing from the heap gives incorrect results because old values remain

The Solution: Always modify the original nums array directly and use the heap only for tracking minimum values:

for _ in range(k):
    _, i = heappop(pq)
    nums[i] *= multiplier  # Modify the original array
    heappush(pq, (nums[i], i))  # Push updated value to heap

Pitfall 2: Not Handling the Index Properly When Values Are Equal

The Problem: When implementing without a heap, developers might find all minimum values but fail to select the one with the smallest index.

Incorrect Approach:

def getFinalState(self, nums: List[int], k: int, multiplier: int) -> List[int]:
    for _ in range(k):
        min_val = min(nums)
        # This finds the LAST occurrence, not the first!
        for i in range(len(nums)-1, -1, -1):
            if nums[i] == min_val:
                nums[i] *= multiplier
                break
    return nums

Why It Fails: The backward iteration finds the last occurrence of the minimum value, violating the requirement to select the first occurrence.

The Solution: Either iterate forward to find the first occurrence, or use the heap approach where Python's tuple comparison automatically handles this:

# Correct manual approach
for i in range(len(nums)):
    if nums[i] == min_val:
        nums[i] *= multiplier
        break

# Or use heap with (value, index) tuples
pq = [(x, i) for i, x in enumerate(nums)]

Pitfall 3: Inefficient Repeated Array Scanning

The Problem: Using a naive O(n) search for each operation without utilizing a heap.

Inefficient Approach:

def getFinalState(self, nums: List[int], k: int, multiplier: int) -> List[int]:
    for _ in range(k):
        min_idx = 0
        for i in range(1, len(nums)):
            if nums[i] < nums[min_idx]:
                min_idx = i
        nums[min_idx] *= multiplier
    return nums

Why It's Problematic:

• Time complexity becomes O(k * n) which is inefficient for large k or n
• For k close to n or larger, this becomes significantly slower than the heap approach

The Solution: Use the heap-based approach for O(n + k * log n) complexity, which is much better when k is large or when n is large.