For each group of consecutive equal maximums, count the number of subarrays where the group forms the boundary and all elements in between are less.
For each occurrence of the maximum, calculate the number of valid subarrays using the distances to the next greater elements.
Sum up the counts for all such groups.
Code

[{"language":"C++","code":"class Solution {\npublic:\n    long long numberOfSubarrays(vector<int>& nums) {\n        int n = nums.size();\n        vector<int> l(n, -1), r(n, n);\n        stack<int> st;\n        for (int i = 0; i < n; ++i) {\n            while (!st.empty() && nums[st.top()] < nums[i]) {\n                r[st.top()] = i;\n                st.pop();\n            }\n            st.push(i);\n        }\n        while (!st.empty()) st.pop();\n        for (int i = n - 1; i >= 0; --i) {\n            while (!st.empty() && nums[st.top()] <= nums[i]) {\n                l[st.top()] = i;\n                st.pop();\n            }\n            st.push(i);\n        }\n        long long ans = 0;\n        for (int i = 0; i < n;) {\n            int j = i;\n            while (j < n && nums[j] == nums[i]) ++j;\n            for (int k = i; k < j; ++k) {\n                long long left = k - l[k];\n                long long right = r[k] - k;\n                ans += left * right;\n            }\n            i = j;\n        }\n        return ans;\n    }\n};","lang":"cpp"},{"language":"Go","code":"func numberOfSubarrays(nums []int) int64 {\n    n := len(nums)\n    l := make([]int, n)\n    r := make([]int, n)\n    for i := range l {\n        l[i] = -1\n        r[i] = n\n    }\n    st := []int{}\n    for i := 0; i < n; i++ {\n        for len(st) > 0 && nums[st[len(st)-1]] < nums[i] {\n            r[st[len(st)-1]] = i\n            st = st[:len(st)-1]\n        }\n        st = append(st, i)\n    }\n    st = []int{}\n    for i := n - 1; i >= 0; i-- {\n        for len(st) > 0 && nums[st[len(st)-1]] <= nums[i] {\n            l[st[len(st)-1]] = i\n            st = st[:len(st)-1]\n        }\n        st = append(st, i)\n    }\n    var ans int64\n    for i := 0; i < n; {\n        j := i\n        for j < n && nums[j] == nums[i] {\n            j++\n        }\n        for k := i; k < j; k++ {\n            left := int64(k - l[k])\n            right := int64(r[k] - k)\n            ans += left * right\n        }\n        i = j\n    }\n    return ans\n}","lang":"go"},{"language":"Java","code":"class Solution {\n    public long numberOfSubarrays(int[] nums) {\n        int n = nums.length;\n        int[] l = new int[n], r = new int[n];\n        Arrays.fill(l, -1);\n        Arrays.fill(r, n);\n        Deque<Integer> st = new ArrayDeque<>();\n        for (int i = 0; i < n; ++i) {\n            while (!st.isEmpty() && nums[st.peek()] < nums[i]) {\n                r[st.pop()] = i;\n            }\n            st.push(i);\n        }\n        st.clear();\n        for (int i = n - 1; i >= 0; --i) {\n            while (!st.isEmpty() && nums[st.peek()] <= nums[i]) {\n                l[st.pop()] = i;\n            }\n            st.push(i);\n        }\n        long ans = 0;\n        for (int i = 0; i < n;) {\n            int j = i;\n            while (j < n && nums[j] == nums[i]) ++j;\n            for (int k = i; k < j; ++k) {\n                long left = k - l[k];\n                long right = r[k] - k;\n                ans += left * right;\n            }\n            i = j;\n        }\n        return ans;\n    }\n}","lang":"java"},{"language":"Kotlin","code":"class Solution {\n    fun numberOfSubarrays(nums: IntArray): Long {\n        val n = nums.size\n        val l = IntArray(n) { -1 }\n        val r = IntArray(n) { n }\n        val st = ArrayDeque<Int>()\n        for (i in 0 until n) {\n            while (st.isNotEmpty() && nums[st.last()] &#x
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