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3101. Count Alternating Subarrays 👍

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  long long countAlternatingSubarrays(vector<int>& nums) {
    // dp[i] := the number of alternating subarrays ending in index i
    vector<long> dp(nums.size(), 1);

    for (int i = 1; i < nums.size(); ++i)
      if (nums[i] != nums[i - 1])
        dp[i] += dp[i - 1];

    return accumulate(dp.begin(), dp.end(), 0L);
  }
};

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class Solution {
  public long countAlternatingSubarrays(int[] nums) {
    // dp[i] := the number of alternating subarrays ending in index i
    long[] dp = new long[nums.length];
    Arrays.fill(dp, 1);

    for (int i = 1; i < nums.length; ++i)
      if (nums[i] != nums[i - 1])
        dp[i] += dp[i - 1];

    return Arrays.stream(dp).sum();
  }
}

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class Solution:
  def countAlternatingSubarrays(self, nums: list[int]) -> int:
    # dp[i] := the number of alternating subarrays ending in index i
    dp = [1] * len(nums)

    for i in range(1, len(nums)):
      if nums[i] != nums[i - 1]:
        dp[i] += dp[i - 1]

    return sum(dp)