How do you solve Merge Adjacent Equal Elements on LeetCode?

Merge Adjacent Equal Elements (LeetCode #3834) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a stack simplifies the merging process.

What is the time complexity of Merge Adjacent Equal Elements?

The optimal solution for Merge Adjacent Equal Elements runs in O(n) time and uses O(n) space. Each element is processed once, leading to linear time complexity. The stack may store all elements in the worst case.

What is the brute force solution for Merge Adjacent Equal Elements?

The brute force approach for Merge Adjacent Equal Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. Repeatedly scan the array for adjacent equal elements and merge them. This process continues until no more merges are possible. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Merge Adjacent Equal Elements?

Merge Adjacent Equal Elements uses the following concepts: Array, Stack, Simulation. The recommended patterns to study are: Stack, Array.

What are the interview tips for Merge Adjacent Equal Elements?

Explain your thought process clearly. Discuss edge cases like empty arrays. Practice coding on a whiteboard or online editor.

What are common mistakes when solving Merge Adjacent Equal Elements?

Forgetting to check the stack before merging. Not handling the case where the stack is empty.

#3834

Merge Adjacent Equal Elements

Medium

ArrayStackSimulationStackArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Use a stack to efficiently manage merges. Push elements onto the stack, and if the top of the stack equals the current element, pop and merge them.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty stack.
2Step 2: Iterate through each element in the array.
3Step 3: If the stack is not empty and the top equals the current element, pop the stack and push their sum; otherwise, push the current element.

solution.py8 lines

1def merge(nums):
2    stack = []
3    for num in nums:
4        if stack and stack[-1] == num:
5            stack[-1] += num
6        else:
7            stack.append(num)
8    return stack

ℹ

Complexity note: Each element is processed once, leading to linear time complexity. The stack may store all elements in the worst case.

1Using a stack simplifies the merging process.
2Merging from left to right ensures we handle the earliest pairs first.

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