Find the maximum width of a ramp in A.  If one doesn’t exist, return 0.

Example 1:

Input: [6,0,8,2,1,5] 
Output: 4 
Explanation:  The maximum width ramp is achieved at (i, j) = (1, 5): A[1] = 0 and A[5] = 5. 

Example 2:

Input: [9,8,1,0,1,9,4,0,4,1] 
Output: 7 
Explanation:  The maximum width ramp is achieved at (i, j) = (2, 9): A[2] = 1 and A[9] = 1. 

Note:

1. 2 <= A.length <= 50000
2. 0 <= A[i] <= 50000

Solution: Stack

1. Using a stack to store start candidates’ (decreasing order) index
2. Scan from right to left, compare the current number with the one on the top of the stack, pop if greater.

e.g.
A = [6,0,8,2,1,5]
stack = [0, 1] => [6, 0]
cur: A[5] = 5, stack.top = A[1] = 0, ramp = 5, stack.pop()
cur: A[4] = 1, stack.top = A[0] = 6
cur: A[3] = 2, stack.top = A[0] = 6
cur: A[2] = 8, stack.top = A[0] = 6, ramp = 2, stack.pop()
stack.isEmpty() => END

// Author: Huahua, running time: 44 ms
class Solution {
public:
  int maxWidthRamp(vector<int>& A) {
    stack<int> s;
    for (int i = 0; i < A.size(); ++i)
      if (s.empty() || A[i] < A[s.top()])
        s.push(i);
    int ans = 0;
    for (int i = A.size() - 1; i >= 0; --i)
      while (!s.empty() && A[i] >= A[s.top()]) {
        ans = max(ans, i - s.top());
        s.pop();
      }
    return ans;
  }
};

# Author: Huahua, running time: 92 ms
class Solution:
  def maxWidthRamp(self, A):
    s = []
    for i in range(len(A)):
      if not s or A[i] < A[s[-1]]: s.append(i)
    ans = 0
    for i in range(len(A) - 1, -1, -1):
      while s and A[i] >= A[s[-1]]:
        ans = max(ans, i - s.pop())        
    return ans

The post 花花酱 LeetCode 962. Maximum Width Ramp appeared first on Huahua's Tech Road.