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2444. Count Subarrays With Fixed Bounds 👍

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  long long countSubarrays(vector<int>& nums, int minK, int maxK) {
    long long ans = 0;
    int j = -1;
    int prevMinKIndex = -1;
    int prevMaxKIndex = -1;

    for (int i = 0; i < nums.size(); ++i) {
      if (nums[i] < minK || nums[i] > maxK)
        j = i;
      if (nums[i] == minK)
        prevMinKIndex = i;
      if (nums[i] == maxK)
        prevMaxKIndex = i;
      // Any index k in [j + 1, min(prevMinKIndex, prevMaxKIndex)] can be the
      // start of the subarray s.t. nums[k..i] satisfies the conditions.
      ans += max(0, min(prevMinKIndex, prevMaxKIndex) - j);
    }

    return ans;
  }
};
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class Solution {
  public long countSubarrays(int[] nums, int minK, int maxK) {
    long ans = 0;
    int j = -1;
    int prevMinKIndex = -1;
    int prevMaxKIndex = -1;

    for (int i = 0; i < nums.length; ++i) {
      if (nums[i] < minK || nums[i] > maxK)
        j = i;
      if (nums[i] == minK)
        prevMinKIndex = i;
      if (nums[i] == maxK)
        prevMaxKIndex = i;
      // Any index k in [j + 1, min(prevMinKIndex, prevMaxKIndex)] can be the
      // start of the subarray s.t. nums[k..i] satisfies the conditions.
      ans += Math.max(0, Math.min(prevMinKIndex, prevMaxKIndex) - j);
    }

    return ans;
  }
}
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class Solution:
  def countSubarrays(self, nums: List[int], minK: int, maxK: int) -> int:
    ans = 0
    j = -1
    prevMinKIndex = -1
    prevMaxKIndex = -1

    for i, num in enumerate(nums):
      if num < minK or num > maxK:
        j = i
      if num == minK:
        prevMinKIndex = i
      if num == maxK:
        prevMaxKIndex = i
      # Any index k in [j + 1, min(prevMinKIndex, prevMaxKIndex)] can be the
      # start of the subarray s.t. nums[k..i] satisfies the conditions.
      ans += max(0, min(prevMinKIndex, prevMaxKIndex) - j)

    return ans