3264. Final Array State After K Multiplication Operations I - Detailed Explanation

In the Java solution, we use a PriorityQueue<int[]> with a custom comparator to sort by the first element (the value) and then by the second element (the index) to break ties. We always update the result array so that it reflects the latest values (though we don't need to use it for finding the min, it’s useful for the final answer).

Complexity Analysis: Building the heap takes O(n) (or O(n log n) if you push elements one by one). Each of the k operations involves two heap operations (pop and push) that each take O(log n) time. Overall time complexity is O(n + k log n). For small n and k, this is similar to the other approaches, but for larger values it is much more scalable. For example, if n = 10^5 and k = 10^5, this approach would be on the order of 10^5 + 10^5 * log(10^5) ≈ 10^5 + 10^5 * ~17 ≈ 1.8×10^6 operations, which is quite feasible. In contrast, brute force would be 10^10 operations, and sorting each time would be 10^5 * (10^5 log 10^5) which is even larger. Space complexity is O(n) to store the heap of n elements (plus a copy of the array for result).

This heap-based approach is optimal for this problem’s typical constraints, and it cleanly handles the tie-breaking by index.

Common Mistakes

• Ignoring the “first occurrence” rule: A frequent mistake is to pick any minimum or the last occurrence instead of the first. For instance, if nums = [2, 2, 9] and you have one operation, you must pick the element at index 0 (the first 2), not index 1. Failing to enforce this can lead to the wrong final state. Always ensure that if there’s a tie for minimum value, the lowest index is chosen.

• Not updating the array or data structure correctly: After multiplying the smallest element, some forget to put the new value back in the structure that tracks minima. If you use a heap or sorted list, you must remove the old value and insert the updated value. If you use an array in place, make sure you update the array element after each operation. Forgetting to do so will make subsequent operations work with outdated values.

• Using the wrong data structure: Some might try using a max-heap by accident or use a data structure that doesn’t easily give the min. Ensure to use a min-heap or sort ascending, not descending. Also, if implementing the heap approach, not including the index in the heap can be problematic. For example, if you only store values in the heap, you lose track of which specific element was multiplied – this matters if there are duplicate values.

• Off-by-one errors in loops: Make sure to run exactly k operations. For instance, a for loop from 0 to k-1 or a while loop that decrements k each time are common patterns. If you accidentally run one extra iteration or one fewer, the result will be incorrect.

• Assuming multiplier changes order monotonically: It’s intuitive that multiplying the smallest number by a value ≥ 1 will increase it (or leave it the same if multiplier = 1). However, don’t assume the smallest element stays the same element throughout all operations. The smallest element’s position can change as you update values. Some might incorrectly always target the same index or think the array becomes sorted after operations – you must actually simulate each operation (or use a mathematical insight for special cases).

Edge Cases to Watch Out For

• All elements equal: If every element in nums is the same, then the first element will be selected repeatedly until it possibly becomes larger than the others. For example, nums=[5,5,5], k=2, multiplier=2 → after 1st op: [10,5,5] (first element became 10, others 5), after 2nd op: [10,10,5] (now second element was the smallest 5). The algorithm should correctly move on to the next index once the first is no longer the smallest. If multiplier=1 in such a case, the first element will be chosen every time (but its value stays the same), effectively wasting all operations on the first element – the final array remains unchanged.

• multiplier = 1: This is a special scenario – multiplying by 1 means values never change. No matter how many operations you perform, the array will remain the same. The code should handle this gracefully (likely by still “performing” k operations, but each operation doesn’t change anything). Make sure this doesn’t cause an infinite loop or other logic issues. For instance, in the heap approach, you’ll keep removing and adding the same smallest element; it should not get stuck incorrectly (our heap method handles it because it will keep removing and adding the same value, which is fine).

• Array of size 1: If nums has just one element, that element will be selected every time. After k operations, it will be nums[0] * (multiplier^k) (multiplied k times by the multiplier). Ensure your solution handles a single-element array (all approaches above do handle this).

• Minimum value at multiple positions: We discussed this, but to reiterate, when the smallest value appears in multiple places, only the leftmost one should be affected until its value possibly changes. The others remain untouched until they themselves become the first in line (if ever). Test cases where the array has repeats of the same number (especially the smallest number) are important.

• No operations (k = 0): Although the constraints say k >= 1, if you ever encounter a scenario with zero operations, the output should just be the original array unchanged. This is trivially handled by our loops (they won’t execute if k is 0).

Alternative Variations

• Final Array State After K Multiplication Operations II: LeetCode actually has a follow-up (Problem 3266) which introduces a twist – after performing the operations, you have to take each number modulo some value (likely to keep numbers within limits). The core logic of selecting the minimum k times remains similar, but when applying a modulus you must be careful. One common pitfall is applying the modulus during the operations in the heap, which can mess up the ordering. The correct approach for the follow-up is usually to perform the multiplications conceptually and only apply the modulo at the end (or to adjust comparisons if applying modulo in between). The constraints in the follow-up are larger (k can be very large), so it also requires a more mathematical approach to avoid iterating k times fully. For instance, if k is huge, each element will be multiplied a certain number of times (e.g., ⌊k/n⌋ times each, plus some remainder operations), which can be calculated without simulating every single operation.

• Multiplier is negative or zero: The current problem restricts multiplier >= 1. If we allowed other values:

  • If multiplier = 0, then after the first operation, the smallest element becomes 0, which would then likely remain the smallest for all subsequent operations (since all values are non-negative here). The array would end up with that position being 0 and no other changes after the first op (because 0 * 0 stays 0). If multiple operations, it just keeps that element 0.
  • If multiplier could be negative, the behavior changes drastically: multiplying by -1, for example, would make the smallest element become a (potentially) large negative number, which would then certainly be the new smallest (if other numbers were positive or smaller in magnitude). It might then keep picking the same element repeatedly if -1, causing it to flip sign each time (for an odd number of operations it ends negative, even number it ends back positive). Problems like “Maximize Sum after K Negations” (see related problems below) explore a scenario similar to multiplier = -1 but with an objective to maximize sum.
  • If multiplier > 1 (as in our problem) the smallest tends to increase or stay equal, so typically the role of "smallest" rotates among the elements.
  • If multiplier = 1, as noted, nothing changes.

• Different selection criteria: A variation could be to pick the maximum element each time instead of the minimum (and maybe multiply or divide it). The approach would then use a max-heap instead of a min-heap, but the idea is analogous. In fact, some problems ask for repeatedly taking the largest element and doing something (e.g., divide it or subtract something from it).

• Different operation on the element: Instead of multiplication, a variation might involve adding a number to the smallest element, or subtracting, etc. The greedy approach with a heap would still apply: always target the element that meets the criteria (min or max) and update it.

• K distributed among elements: Another interpretation is if we had to choose k elements (not necessarily distinct indices) to multiply by a factor (instead of sequential operations). That would be a different problem altogether, more like choosing k elements to maximize/minimize something.

In summary, the pattern of repeatedly selecting and updating the extreme (min or max) element shows up in many variations – the key is often using a heap or another efficient structure to handle it.

Related LeetCode Problems

• 2530. Maximal Score After Applying K Operations

• 238. Product of Array Except Self

• 1868. Product of Two Run-Length Encoded Arrays