How do you solve Number of Adjacent Elements With the Same Color on LeetCode?

Number of Adjacent Elements With the Same Color (LeetCode #2672) 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: Only the neighbors of the modified index need to be checked for adjacent pairs.

What is the time complexity of Number of Adjacent Elements With the Same Color?

The optimal solution for Number of Adjacent Elements With the Same Color runs in O(n) time and uses O(n) space. This complexity is efficient because we only check the affected neighbors (up to 2) for each query, leading to a linear time complexity overall.

What is the brute force solution for Number of Adjacent Elements With the Same Color?

The brute force approach for Number of Adjacent Elements With the Same Color is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will iterate through the entire array after each query to count the number of adjacent pairs with the same color. This is straightforward but inefficient as it checks all pairs every time. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Adjacent Elements With the Same Color?

Number of Adjacent Elements With the Same Color uses the following concepts: Array. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Adjacent Elements With the Same Color?

Always think about how to minimize the amount of work done after each query. Explain your thought process clearly; it shows your understanding of the problem. Practice similar problems to recognize patterns in data structures and algorithms.

What are common mistakes when solving Number of Adjacent Elements With the Same Color?

Not considering the boundaries of the array when checking neighbors. Forgetting to update the count correctly after changing colors.

#2672

Number of Adjacent Elements With the Same Color

Medium

ArrayHash MapArray

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)

Instead of counting adjacent pairs from scratch after each query, we can keep track of only the affected pairs (the pairs adjacent to the changed index). This reduces unnecessary computations.

⚙️

Algorithm

3 steps

1Step 1: Initialize an array colors of size n and a variable to keep track of the number of adjacent pairs.
2Step 2: For each query, check the colors of the neighboring elements before and after the update to adjust the count of adjacent pairs.
3Step 3: Update the color and append the current count to the result.

solution.py16 lines

1def countAdjacentPairs(n, queries):
2    colors = [0] * n
3    count = 0
4    result = []
5    for index, color in queries:
6        if index > 0 and colors[index] == colors[index - 1]:
7            count -= 1
8        if index < n - 1 and colors[index] == colors[index + 1]:
9            count -= 1
10        colors[index] = color
11        if index > 0 and colors[index] == colors[index - 1]:
12            count += 1
13        if index < n - 1 and colors[index] == colors[index + 1]:
14            count += 1
15        result.append(count)
16    return result

ℹ

Complexity note: This complexity is efficient because we only check the affected neighbors (up to 2) for each query, leading to a linear time complexity overall.

1Only the neighbors of the modified index need to be checked for adjacent pairs.
2Keeping track of the count allows for efficient updates rather than recalculating from scratch.

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