# 2672. Number of Adjacent Elements With the Same Color

## Description

There is a 0-indexed array nums of length n. Initially, all elements are uncolored (has a value of 0).

You are given a 2D integer array queries where queries[i] = [indexi, colori].

For each query, you color the index indexi with the color colori in the array nums.

Return an array answer of the same length as queries where answer[i] is the number of adjacent elements with the same color after the ith query.

More formally, answer[i] is the number of indices j, such that 0 <= j < n - 1 and nums[j] == nums[j + 1] and nums[j] != 0 after 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 nums = [0,0,0,0], where 0 denotes uncolored elements of the array.
- After the 1st query nums = [2,0,0,0]. The count of adjacent elements with the same color is 0.
- After the 2nd query nums = [2,2,0,0]. The count of adjacent elements with the same color is 1.
- After the 3rd query nums = [2,2,0,1]. The count of adjacent elements with the same color is 1.
- After the 4th query nums = [2,1,0,1]. The count of adjacent elements with the same color is 0.
- After the 5th query nums = [2,1,1,1]. The count of adjacent elements with the same color is 2.


Example 2:

Input: n = 1, queries = [[0,100000]]
Output: [0]
Explanation: Initially array nums = [0], where 0 denotes uncolored elements of the array.
- After the 1st query nums = [100000]. The count of adjacent elements 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

## Solutions

• class Solution {
public int[] colorTheArray(int n, int[][] queries) {
int m = queries.length;
int[] nums = new int[n];
int[] ans = new int[m];
for (int k = 0, x = 0; k < m; ++k) {
int i = queries[k][0], c = queries[k][1];
if (i > 0 && nums[i] > 0 && nums[i - 1] == nums[i]) {
--x;
}
if (i < n - 1 && nums[i] > 0 && nums[i + 1] == nums[i]) {
--x;
}
if (i > 0 && nums[i - 1] == c) {
++x;
}
if (i < n - 1 && nums[i + 1] == c) {
++x;
}
ans[k] = x;
nums[i] = c;
}
return ans;
}
}

• class Solution {
public:
vector<int> colorTheArray(int n, vector<vector<int>>& queries) {
vector<int> nums(n);
vector<int> ans;
int x = 0;
for (auto& q : queries) {
int i = q[0], c = q[1];
if (i > 0 && nums[i] > 0 && nums[i - 1] == nums[i]) {
--x;
}
if (i < n - 1 && nums[i] > 0 && nums[i + 1] == nums[i]) {
--x;
}
if (i > 0 && nums[i - 1] == c) {
++x;
}
if (i < n - 1 && nums[i + 1] == c) {
++x;
}
ans.push_back(x);
nums[i] = c;
}
return ans;
}
};

• class Solution:
def colorTheArray(self, n: int, queries: List[List[int]]) -> List[int]:
nums = [0] * n
ans = [0] * len(queries)
x = 0
for k, (i, c) in enumerate(queries):
if i > 0 and nums[i] and nums[i - 1] == nums[i]:
x -= 1
if i < n - 1 and nums[i] and nums[i + 1] == nums[i]:
x -= 1
if i > 0 and nums[i - 1] == c:
x += 1
if i < n - 1 and nums[i + 1] == c:
x += 1
ans[k] = x
nums[i] = c
return ans


• func colorTheArray(n int, queries [][]int) (ans []int) {
nums := make([]int, n)
x := 0
for _, q := range queries {
i, c := q[0], q[1]
if i > 0 && nums[i] > 0 && nums[i-1] == nums[i] {
x--
}
if i < n-1 && nums[i] > 0 && nums[i+1] == nums[i] {
x--
}
if i > 0 && nums[i-1] == c {
x++
}
if i < n-1 && nums[i+1] == c {
x++
}
ans = append(ans, x)
nums[i] = c
}
return
}

• function colorTheArray(n: number, queries: number[][]): number[] {
const nums: number[] = new Array(n).fill(0);
const ans: number[] = [];
let x = 0;
for (const [i, c] of queries) {
if (i > 0 && nums[i] > 0 && nums[i - 1] == nums[i]) {
--x;
}
if (i < n - 1 && nums[i] > 0 && nums[i + 1] == nums[i]) {
--x;
}
if (i > 0 && nums[i - 1] == c) {
++x;
}
if (i < n - 1 && nums[i + 1] == c) {
++x;
}
ans.push(x);
nums[i] = c;
}
return ans;
}