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

In this guide, you will get 2672. Number of Adjacent Elements With the Same Color LeetCode Solution with the best time and space complexity. The solution to Number of Adjacent Elements With the Same Color problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Number of Adjacent Elements With the Same Color solution in C++
4. Number of Adjacent Elements With the Same Color solution in Java
5. Number of Adjacent Elements With the Same Color solution in Python
6. Additional Resources

Problem Statement of Number of Adjacent Elements With the Same Color

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] = [indexi, colori]. For the ith query:

Set colors[indexi] to colori.
Count adjacent pairs in colors set to the same color (regardless of colori).

Return an array answer of the same length as queries where answer[i] is the answer to the ith query.

Example 1:

Input: n = 4, queries = [[0,2],[1,2],[3,1],[1,1],[2,1]]
Output: [0,1,1,0,2]
Explanation:

Initially array colors = [0,0,0,0], where 0 denotes uncolored elements of the array.
After the 1st query colors = [2,0,0,0]. The count of adjacent pairs with the same color is 0.
After the 2nd query colors = [2,2,0,0]. The count of adjacent pairs with the same color is 1.
After the 3rd query colors = [2,2,0,1]. The count of adjacent pairs with the same color is 1.
After the 4th query colors = [2,1,0,1]. The count of adjacent pairs with the same color is 0.
After the 5th query colors = [2,1,1,1]. The count of adjacent pairs with the same color is 2.

Example 2:

Input: n = 1, queries = [[0,100000]]
Output: [0]
Explanation:
After the 1st query colors = [100000]. The count of adjacent pairs with the same color is 0.

Constraints:

1 <= n <= 105
1 <= queries.length <= 105
queries[i].length == 2
0 <= indexi <= n – 1
1 <= colori <= 105

Complexity Analysis

• Time Complexity: O(n + q)
• Space Complexity: O(n + q)

2672. Number of Adjacent Elements With the Same Color LeetCode Solution in C++

class Solution {
 public:
  vector<int> colorTheArray(int n, vector<vector<int>>& queries) {
    vector<int> ans;
    vector<int> arr(n);
    int sameColors = 0;

    for (const vector<int>& query : queries) {
      const int i = query[0];
      const int color = query[1];
      if (i + 1 < n) {
        if (arr[i + 1] > 0 && arr[i + 1] == arr[i])
          --sameColors;
        if (arr[i + 1] == color)
          ++sameColors;
      }
      if (i > 0) {
        if (arr[i - 1] > 0 && arr[i - 1] == arr[i])
          --sameColors;
        if (arr[i - 1] == color)
          ++sameColors;
      }
      arr[i] = color;
      ans.push_back(sameColors);
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

2672. Number of Adjacent Elements With the Same Color LeetCode Solution in Java

class Solution {
  public int[] colorTheArray(int n, int[][] queries) {
    int[] ans = new int[queries.length];
    int[] arr = new int[n];
    int sameColors = 0;

    for (int i = 0; i < queries.length; ++i) {
      final int j = queries[i][0];
      final int color = queries[i][1];
      if (j + 1 < n) {
        if (arr[j + 1] > 0 && arr[j + 1] == arr[j])
          --sameColors;
        if (arr[j + 1] == color)
          ++sameColors;
      }
      if (j > 0) {
        if (arr[j - 1] > 0 && arr[j - 1] == arr[j])
          --sameColors;
        if (arr[j - 1] == color)
          ++sameColors;
      }
      arr[j] = color;
      ans[i] = sameColors;
    }

    return ans;
  }
}
// code provided by PROGIEZ

2672. Number of Adjacent Elements With the Same Color LeetCode Solution in Python

class Solution:
  def colorTheArray(self, n: int, queries: list[list[int]]) -> list[int]:
    ans = []
    arr = [0] * n
    sameColors = 0

    for i, color in queries:
      if i + 1 < n:
        if arr[i + 1] > 0 and arr[i + 1] == arr[i]:
          sameColors -= 1
        if arr[i + 1] == color:
          sameColors += 1
      if i > 0:
        if arr[i - 1] > 0 and arr[i - 1] == arr[i]:
          sameColors -= 1
        if arr[i - 1] == color:
          sameColors += 1
      arr[i] = color
      ans.append(sameColors)

    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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