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

The optimal solution has a time complexity of O(q).

What data structures are used to solve Number of Adjacent Elements With the Same Color?

This problem uses Array. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Number of Adjacent Elements With the Same Color?

Insight 1: The problem requires maintaining a running count of adjacent elements with the same color as the array is updated. Insight 2: The color update at an index can only affect the adjacent pairs involving that index (index-1, index) and (index, index+1).

Which companies ask LeetCode 2672. Number of Adjacent Elements With the Same Color in interviews?

This problem is frequently asked at Capital One, Roblox, Visa.

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Number of Adjacent Elements With the Same Color - LeetCode 2672 Solution

Number of Adjacent Elements With the Same Color - Complete Solution Guide

Number of Adjacent Elements With the Same Color is LeetCode problem 2672, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an integer n representing an array colors of length n where all elements are set to 0's meaning uncolored . You are also given a 2D integer array queries where queries[i] = [index i , color i ] . For the i th query : Set colors[index i ] to color i . Count the number of adjacent pairs in colors which have the same color (regardless of color i ). Return an array answer of the same length as queries where answer[i] is the answer to the i th query. Example 1: Input: n = 4, queries = [

Detailed Explanation

The problem presents an array `colors` of length `n`, initialized with all elements set to 0, representing an uncolored state. A series of `queries` is given, where each query consists of an `index` and a `color`. For each query, the element at the specified `index` in the `colors` array is updated with the given `color`. After each update, the goal is to count the number of adjacent pairs of elements in the `colors` array that have the same color. The final output is an array `answer` where `answer[i]` represents the count of adjacent pairs with the same color after processing the `i`-th query.

Solution Approach

The solution maintains an array `colors` representing the colors at each index and a variable `current_count` to store the number of adjacent pairs with the same color. For each query, before updating the color at the given `index`, we check if the current color at that `index` is non-zero. If it is, we check if it forms an adjacent pair on the left and right and decrement the `current_count` accordingly. Then, the color at the given `index` is updated to the new color. Finally, we check if the new color at the given `index` forms an adjacent pair on the left and right and increment the `current_count` accordingly. The updated `current_count` is added to the answer array.

Step-by-Step Algorithm

Step 1: Initialize an array `colors` of size `n` with all elements set to 0.
Step 2: Initialize `current_count` to 0 to keep track of the number of adjacent pairs with the same color.
Step 3: Iterate through each query in the `queries` array.
Step 4: For each query, extract the `index` and `new_color`.
Step 5: Store the old color at the current `index` to the `old_color` variable.
Step 6: If the `old_color` and `new_color` are the same, append the `current_count` to the `answer` array and continue to the next query.
Step 7: If `old_color` is not 0, check if there is an adjacent element on the left (if index > 0) with the same `old_color`, if so decrement `current_count`.
Step 8: If `old_color` is not 0, check if there is an adjacent element on the right (if index < n-1) with the same `old_color`, if so decrement `current_count`.
Step 9: Update the color at the given `index` in the `colors` array to the `new_color`.
Step 10: Check if there is an adjacent element on the left (if index > 0) with the same `new_color`, if so increment `current_count`.
Step 11: Check if there is an adjacent element on the right (if index < n-1) with the same `new_color`, if so increment `current_count`.
Step 12: Append the `current_count` to the `answer` array.
Step 13: Return the `answer` array.

Key Insights

Insight 1: The problem requires maintaining a running count of adjacent elements with the same color as the array is updated.
Insight 2: The color update at an index can only affect the adjacent pairs involving that index (index-1, index) and (index, index+1).
Insight 3: Recomputing the entire adjacent pair count after each query would be inefficient. The running count must be updated incrementally.

Complexity Analysis

Time Complexity: O(q)

Space Complexity: O(n)

Topics

This problem involves: Array.

Companies

Asked at: Capital One, Roblox, Visa.

Number of Adjacent Elements With the Same Color | LeetCode 2672

Medium

Array

Time: O(q)Space: O(n)

Solution Code

Available in:

class Solution {
    public int[] colorTheArray(int n, int[][] queries) {
        int[] colors = new int[n];
        int current_count = 0;
        int[] answer = new int[queries.length];

        for (int i = 0; i < queries.length; i++) {
            int index = queries[i][0];
            int new_color = queries[i][1];

            int old_color = colors[index];

            if (old_color == new_color) {
                answer[i] = current_count;
                continue;
            }

            if (old_color != 0) {
                if (index > 0 && colors[index - 1] == old_color) {
                    current_count -= 1;
                }
                if (index < n - 1 && colors[index + 1] == old_color) {
                    current_count -= 1;
                }
            }

            colors[index] = new_color;

            if (index > 0 && colors[index - 1] == new_color) {
                current_count += 1;
            }
            if (index < n - 1 && colors[index + 1] == new_color) {
                current_count += 1;
            }

            answer[i] = current_count;
        }

        return answer;
    }
}

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Problem Info

Description

Problem 2672: Number of Adjacent Elements With the Same Color - Detailed explanation and solution approaches.

Topics

Array

Companies

Capital OneRobloxVisa

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:2672

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

13 steps

1
Step 1: Initialize an array `colors` of size `n` with all elements set to 0.
2
Step 2: Initialize `current_count` to 0 to keep track of the number of adjacent pairs with the same color.
3
Step 3: Iterate through each query in the `queries` array.
4
Step 4: For each query, extract the `index` and `new_color`.
5
Step 5: Store the old color at the current `index` to the `old_color` variable.
6
Step 6: If the `old_color` and `new_color` are the same, append the `current_count` to the `answer` array and continue to the next query.
7
Step 7: If `old_color` is not 0, check if there is an adjacent element on the left (if index > 0) with the same `old_color`, if so decrement `current_count`.
8
Step 8: If `old_color` is not 0, check if there is an adjacent element on the right (if index < n-1) with the same `old_color`, if so decrement `current_count`.
9
Step 9: Update the color at the given `index` in the `colors` array to the `new_color`.
10
Step 10: Check if there is an adjacent element on the left (if index > 0) with the same `new_color`, if so increment `current_count`.
11
Step 11: Check if there is an adjacent element on the right (if index < n-1) with the same `new_color`, if so increment `current_count`.
12
Step 12: Append the `current_count` to the `answer` array.
13
Step 13: Return the `answer` array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(q)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by iterating through the queries. The space complexity is determined by the `colors` array and the `answer` array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

After each query, iterate through the `colors` array and count the number of adjacent pairs with the same color.

Time: O(n*q), where n is the size of the colors array, and q is the number of queries. We must iterate n times for each query.Space: O(n), where n is the size of the colors array.

Trade-offs: This approach is much slower than the current solution, but requires less complex code.

✨

Key Insights

1Insight 1: The problem requires maintaining a running count of adjacent elements with the same color as the array is updated.
2Insight 2: The color update at an index can only affect the adjacent pairs involving that index (index-1, index) and (index, index+1).
3Insight 3: Recomputing the entire adjacent pair count after each query would be inefficient. The running count must be updated incrementally.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: `n = 1`. In this case, there are no adjacent pairs, so the answer will always be 0, regardless of the queries. The code handles this correctly.
Edge case 2: Consecutive identical queries at the same index. The code handles this correctly by only updating the `current_count` when `old_color` != `new_color`.
Edge case 3: Queries with the same index are performed consecutively. The `old_color` is correctly handled in this case because it stores the updated `colors` array and the algorithm will treat the new `color` with the old color at that index.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Recomputing the adjacent pairs count after each query by iterating over the entire array. This leads to a time complexity of O(n*q).

✗Common mistake 2: Not handling the case where the old color and the new color are the same. This can lead to incorrect counts.

✗Common mistake 3: Forgetting to check the boundaries (index > 0 and index < n-1) when accessing adjacent elements which causes index out of bounds errors.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Imagine a sensor network where nodes have associated color. These colors are changing dynamically according to some events and we need to track adjacent sensor nodes with the same color for certain operations or maintenance.

🏦Real-world scenario 2: Consider a system monitoring network segments and assigning colors to segments based on their current status (e.g., green for healthy, red for critical). This algorithm can be used to quickly determine if there are clusters of segments with the same status based on color adjacency.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 31, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Number of Adjacent Elements With the Same Color"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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