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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 <= 1051 <= queries.length <= 105queries[i].length == 20 <= indexi <= n - 11 <= colori <= 105
Solutions
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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; }