How do you solve Apply Operations to an Array on LeetCode?

Apply Operations to an Array (LeetCode #2460) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Operations are applied sequentially, not all at once.

What is the brute force solution for Apply Operations to an Array?

The brute force approach for Apply Operations to an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves iterating through the array and applying the operations directly as described. This method is straightforward but inefficient, as it may require multiple passes through the array. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Apply Operations to an Array?

Apply Operations to an Array uses the following concepts: Array, Two Pointers, Simulation. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Apply Operations to an Array?

Explain your thought process clearly while coding. Test your code with edge cases, like arrays with all zeros or all equal elements. Make sure to clarify the problem requirements before jumping into coding.

What are common mistakes when solving Apply Operations to an Array?

Forgetting to handle the last element in the array. Not properly shifting zeros after applying operations.

#2460

Apply Operations to an Array

Easy

ArrayTwo PointersSimulationArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a single pass to apply operations and another pass to shift zeros. This reduces the number of iterations and avoids unnecessary array creation.

⚙️

Algorithm

4 steps

1Step 1: Initialize a pointer j to 0 to track the position of the next non-zero element.
2Step 2: Iterate through the array with index i. If nums[i] equals nums[i + 1], multiply nums[i] by 2 and set nums[i + 1] to 0. Increment j for the next non-zero element.
3Step 3: If nums[i] is not zero, place it at index j and increment j.
4Step 4: After processing, fill the rest of the array from index j to n-1 with zeros.

solution.py14 lines

1def apply_operations(nums):
2    n = len(nums)
3    j = 0
4    for i in range(n - 1):
5        if nums[i] == nums[i + 1]:
6            nums[i] *= 2
7            nums[i + 1] = 0
8    for i in range(n):
9        if nums[i] != 0:
10            nums[j] = nums[i]
11            j += 1
12    for i in range(j, n):
13        nums[i] = 0
14    return nums

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array to apply operations and another pass to shift zeros. The space complexity is O(1) since we modify the array in place without using extra space.

1Operations are applied sequentially, not all at once.
2Shifting zeros to the end requires careful management of indices.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.