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Approach\nThe approach used in the code involves a depth-first search (DFS) algorithm. The dfs function takes three parameters: i represents the current index, 
# Approach\nThe approach used in the code involves a depth-first search (DFS) algorithm. The dfs function takes three parameters: i represents the current index, l represents the lower bound, and h represents the upper bound.\n\nThe function checks if the upper bound h is one greater than the lower bound l. If so, it means there are no more elements to consider, and the function returns 1.\n\nOtherwise, the function checks if the value at index i is between l and h. If it is, it recursively calls dfs twice: once with the left subarray (from l to A[i]) and once with the right subarray (from A[i] to h). It also calculates the number of combinations between the left and right subarrays using math.comb and multiplies it with the recursive results.\n\nIf the value at index i is not between l and h, the function makes a recursive call with the same index but updates only the upper bound.\n\nFinally, the function returns the result of the recursive calls subtracted by 1.\n\nThe numOfWays function initializes some variables and then calls dfs with the initial parameters.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: The time complexity depends on the number of recursive calls made. Each recursive call reduces the problem size, so the time complexity can be expressed as O(2^n), where n is the length of the input list A.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: The space complexity is determined by the depth of the recursive calls, which can go up to n. Therefore, the space complexity is O(n).\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def numOfWays(self, A):\n        n = len(A)\n        mod = 10 ** 9 + 7\n\n        def dfs(i, l, h):\n            if h == l + 1:\n                return 1\n            if l < A[i] < h:\n                return (dfs(i + 1, l, A[i]) * dfs(i + 1, A[i], h) * math.comb(h - l - 2, A[i] - l - 1)) % mod\n            return dfs(i + 1, l, h)\n            \n        return dfs(0, 0, n + 1) - 1\n\n```\n\n### *Kindly Upvote\u270C\uD83C\uDFFC*
['Python3']