LeetCode 1063: Number of Valid Subarrays Solution

Master LeetCode problem 1063 (Number of Valid Subarrays), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1063. Number of Valid Subarrays

Hard

Array Stack Monotonic Stack

Problem Explanation

Explanation:

Given an array of integers, we need to find the number of valid subarrays. A subarray is considered valid if all the elements in the subarray are greater than or equal to a given target value.

To solve this problem, we can use a monotonic stack approach. We iterate through the array and maintain a stack that stores the indices of elements in increasing order. When we encounter an element that is less than the target value, we pop elements from the stack until we find an element that is greater than or equal to the target value. For each valid subarray ending at the current index, we can calculate the number of valid subarrays by considering the elements between the current index and the top of the stack.

At the end of the iteration, we can calculate the total number of valid subarrays by considering the remaining elements in the stack.

Algorithm:

Initialize a stack to store indices and a variable to store the result.
Iterate through the array:
- If the stack is not empty and the current element is less than the target value, pop elements from the stack and calculate the number of valid subarrays.
- For each valid subarray, update the result.
- Push the current index onto the stack.
After the iteration, calculate the number of valid subarrays for the remaining elements in the stack.
Return the total number of valid subarrays.

Time Complexity: O(n), where n is the number of elements in the array. Space Complexity: O(n), where n is the number of elements in the array.

: :

Solution Code

class Solution {
    public int validSubarrays(int[] nums) {
        Stack<Integer> stack = new Stack<>();
        int result = 0;
        
        for (int i = 0; i < nums.length; i++) {
            while (!stack.isEmpty() && nums[i] < nums[stack.peek()]) {
                result += i - stack.pop();
            }
            stack.push(i);
        }
        
        while (!stack.isEmpty()) {
            result += nums.length - stack.pop();
        }
        
        return result;
    }
}

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