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1063. Number of Valid Subarrays 👍

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  int validSubarrays(vector<int>& nums) {
    // For each `num` in `nums`, each element x in the stack can be the leftmost
    // element s.t. [x, num] forms a valid subarray, so the size of the stack is
    // the number of valid subarrays ending in the current number.
    //
    // e.g. nums = [1, 3, 2]
    // num = 1, stack = [1] -> valid subarray is [1]
    // num = 3, stack = [1, 3] -> valid subarrays are [1, 3], [3]
    // num = 2, stack = [1, 2] -> valid subarrays are [1, 3, 2], [2]
    int ans = 0;
    stack<int> stack;

    for (const int num : nums) {
      while (!stack.empty() && stack.top() > num)
        stack.pop();
      stack.push(num);
      ans += stack.size();
    }

    return ans;
  }
};

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class Solution {
  public int validSubarrays(int[] nums) {
    // For each `num` in `nums`, each element x in the stack can be the leftmost
    // element s.t. [x, num] forms a valid subarray, so the size of the stack is
    // the number of valid subarrays ending in the current number.
    //
    // e.g. nums = [1, 3, 2]
    // num = 1, stack = [1] -> valid subarray is [1]
    // num = 3, stack = [1, 3] -> valid subarrays are [1, 3], [3]
    // num = 2, stack = [1, 2] -> valid subarrays are [1, 3, 2], [2]
    int ans = 0;
    Deque<Integer> stack = new ArrayDeque<>();

    for (final int num : nums) {
      while (!stack.isEmpty() && stack.peek() > num)
        stack.pop();
      stack.push(num);
      ans += stack.size();
    }

    return ans;
  }
}

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class Solution:
  def validSubarrays(self, nums: list[int]) -> int:
    # For each `num` in `nums`, each element x in the stack can be the leftmost
    # element s.t. [x, num] forms a valid subarray, so the size of the stack is
    # the number of valid subarrays ending in the current number.
    #
    # e.g. nums = [1, 3, 2]
    # num = 1, stack = [1] -> valid subarray is [1]
    # num = 3, stack = [1, 3] -> valid subarrays are [1, 3], [3]
    # num = 2, stack = [1, 2] -> valid subarrays are [1, 3, 2], [2]
    ans = 0
    stack = []

    for num in nums:
      while stack and stack[-1] > num:
        stack.pop()
      stack.append(num)
      ans += len(stack)

    return ans