Medium
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.
Example 2:
Input: n = 1, queries = [[0,100000]]
Output: [0]
Explanation: Initially array nums = [0], where 0 denotes uncolored elements of the array.
Constraints:
1 <= n <= 105
1 <= queries.length <= 105
queries[i].length == 2
0 <= indexi <= n - 1
1 <= colori <= 105
public class Solution {
public int[] colorTheArray(int n, int[][] q) {
if (n == 1) {
return new int[q.length];
}
int[] ans = new int[q.length];
int[] color = new int[n];
int cnt = 0;
for (int i = 0; i < q.length; i++) {
int ind = q[i][0];
int assColor = q[i][1];
int leftColor = 0;
int rytColor = 0;
if (ind - 1 >= 0) {
leftColor = color[ind - 1];
}
if (ind + 1 < n) {
rytColor = color[ind + 1];
}
if (color[ind] != 0 && leftColor == color[ind]) {
cnt--;
}
if (color[ind] != 0 && rytColor == color[ind]) {
cnt--;
}
color[ind] = assColor;
if (color[ind] != 0 && leftColor == color[ind]) {
cnt++;
}
if (color[ind] != 0 && rytColor == color[ind]) {
cnt++;
}
ans[i] = cnt;
}
return ans;
}
}