Leetcode 962. Maximum Width Ramp

Intuition:
Keep a decresing stack.
For each number,
binary search the first smaller number in the stack.

When the number is smaller the the last,
push it into the stack.

Python

    def maxWidthRamp(self, A):
        stack = []
        res = 0
        for i in range(len(A))[::-1]:
            if not stack or A[i] > stack[-1][0]:
                stack.append([A[i], i])
            else:
                j = stack[bisect.bisect(stack, [A[i], i])][1]
                res = max(res, j - i)
        return res

Improve the above idea.
Will add some explanation after dinner. :)

Java

    public int maxWidthRamp(int[] A) {
        Stack s = new Stack<>();
        int res = 0, n = A.length;
        for (int i = 0; i < n; ++i)
            if (s.empty() || A[s.peek()] > A[i])
                s.add(i);
        for (int i = n - 1; i >= 0; --i)
            while (!s.empty() && A[s.peek()] <= A[i])
                res = Math.max(res, i - s.pop());
        return res;
    }

C++

    int maxWidthRamp(vector& A) {
        stack s;
        int res = 0, n = A.size();
        for (int i = 0; i < n; ++i)
            if (s.empty() || A[s.top()] > A[i])
                s.push(i);
        for (int i = n - 1; i >= 0; --i)
            while (!s.empty() && A[s.top()] <= A[i])
                res = max(res, i - s.top()), s.pop();
        return res;
    }

Python:

    def maxWidthRamp(self, A):
        s = []
        res = 0
        for i, a in enumerate(A):
            if not s or A[s[-1]] > a:
                s.append(i)
        for j in range(len(A))[::-1]:
            while s and A[s[-1]] <= A[j]:
                res = max(res, j - s.pop())
        return res