There is a solution with linear complexity using Queues.

public int maxSubarrayLength(int[] nums) {
        int n = nums.length;
        Deque<Integer> stack = new ArrayDeque<>();

        for (int i = n - 1; i >= 0; i--) {
            if (stack.isEmpty() || nums[i] < nums[stack.peek()]) {
                stack.push(i);
            }
        }

        int ans = 0;
        int m = Integer.MIN_VALUE;

        for (int i = 0; i < n; i++) {
            while (!stack.isEmpty() && stack.peek() <= i) {
                stack.pop();
            }

            if (nums[i] > m) {
                m = nums[i];

                while (!stack.isEmpty() && m > nums[stack.peek()]) {
                    ans = Math.max(ans, stack.pop() - i + 1);
                }
            }
        }

        return ans;
    }