Welcome to Subscribe On Youtube

3208. Alternating Groups II

Description

There is a circle of red and blue tiles. You are given an array of integers colors and an integer k. The color of tile i is represented by colors[i]:

  • colors[i] == 0 means that tile i is red.
  • colors[i] == 1 means that tile i is blue.

An alternating group is every k contiguous tiles in the circle with alternating colors (each tile in the group except the first and last one has a different color from its left and right tiles).

Return the number of alternating groups.

Note that since colors represents a circle, the first and the last tiles are considered to be next to each other.

 

Example 1:

Input: colors = [0,1,0,1,0], k = 3

Output: 3

Explanation:

Alternating groups:

Example 2:

Input: colors = [0,1,0,0,1,0,1], k = 6

Output: 2

Explanation:

Alternating groups:

Example 3:

Input: colors = [1,1,0,1], k = 4

Output: 0

Explanation:

 

Constraints:

  • 3 <= colors.length <= 105
  • 0 <= colors[i] <= 1
  • 3 <= k <= colors.length

Solutions

Solution 1: Single Pass

We can unfold the ring into an array of length $2n$ and then traverse this array from left to right. We use a variable $\textit{cnt}$ to record the current length of the alternating group. If we encounter the same color, we reset $\textit{cnt}$ to $1$; otherwise, we increment $\textit{cnt}$. If $\textit{cnt} \ge k$ and the current position $i$ is greater than or equal to $n$, then we have found an alternating group, and we increment the answer by one.

After the traversal, we return the answer.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{colors}$. The space complexity is $O(1)$.

  • class Solution {
        public int numberOfAlternatingGroups(int[] colors, int k) {
            int n = colors.length;
            int ans = 0, cnt = 0;
            for (int i = 0; i < n << 1; ++i) {
                if (i > 0 && colors[i % n] == colors[(i - 1) % n]) {
                    cnt = 1;
                } else {
                    ++cnt;
                }
                ans += i >= n && cnt >= k ? 1 : 0;
            }
            return ans;
        }
    }
    
  • class Solution {
    public:
        int numberOfAlternatingGroups(vector<int>& colors, int k) {
            int n = colors.size();
            int ans = 0, cnt = 0;
            for (int i = 0; i < n << 1; ++i) {
                if (i && colors[i % n] == colors[(i - 1) % n]) {
                    cnt = 1;
                } else {
                    ++cnt;
                }
                ans += i >= n && cnt >= k ? 1 : 0;
            }
            return ans;
        }
    };
    
  • class Solution:
        def numberOfAlternatingGroups(self, colors: List[int], k: int) -> int:
            n = len(colors)
            ans = cnt = 0
            for i in range(n << 1):
                if i and colors[i % n] == colors[(i - 1) % n]:
                    cnt = 1
                else:
                    cnt += 1
                ans += i >= n and cnt >= k
            return ans
    
    
  • func numberOfAlternatingGroups(colors []int, k int) (ans int) {
    	n := len(colors)
    	cnt := 0
    	for i := 0; i < n<<1; i++ {
    		if i > 0 && colors[i%n] == colors[(i-1)%n] {
    			cnt = 1
    		} else {
    			cnt++
    		}
    		if i >= n && cnt >= k {
    			ans++
    		}
    	}
    	return
    }
    
  • function numberOfAlternatingGroups(colors: number[], k: number): number {
        const n = colors.length;
        let [ans, cnt] = [0, 0];
        for (let i = 0; i < n << 1; ++i) {
            if (i && colors[i % n] === colors[(i - 1) % n]) {
                cnt = 1;
            } else {
                ++cnt;
            }
            ans += i >= n && cnt >= k ? 1 : 0;
        }
        return ans;
    }
    
    

All Problems

All Solutions