962 - Maximum Width Ramp

A ramp in an integer array nums is a pair (i, j) for which i < j and nums[i] <= nums[j]. The width of such a ramp is j - i.

Given an integer array nums, return the maximum width of a ramp in nums. If there is no ramp in nums, return 0.

 Example 1:

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

Example 2:

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

 Constraints:

• 2 <= nums.length <= 5 * 104
• 0 <= nums[i] <= 5 * 104

C++

• class Solution {
  public:
      int maxWidthRamp(vector<int>& nums) {
          int n = nums.size();
          stack<int> stk;
          for (int i = 0; i < n; ++i) {
              if (stk.empty() || nums[stk.top()] > nums[i]) stk.push(i);
          }
          int ans = 0;
          for (int i = n - 1; i; --i) {
              while (!stk.empty() && nums[stk.top()] <= nums[i]) {
                  ans = max(ans, i - stk.top());
                  stk.pop();
              }
              if (stk.empty()) break;
          }
          return ans;
      }
  };

PYTHON

• class Solution:
      def maxWidthRamp(self, nums):
          n = len(nums)
          stk = []
          for i in range(n):
              if not stk or nums[stk[-1]] > nums[i]:
                  stk.append(i)
          
          ans = 0
          for i in range(n - 1, 0, -1):
              while stk and nums[stk[-1]] <= nums[i]:
                  ans = max(ans, i - stk[-1])
                  stk.pop()
              if not stk:
                  break
                  
          return ans

JAVA

• class Solution {
      public int maxWidthRamp(int[] nums) {
          int n = nums.length;
          Deque<Integer> stk = new ArrayDeque<>();
          for (int i = 0; i < n; ++i) {
              if (stk.isEmpty() || nums[stk.peek()] > nums[i]) {
                  stk.push(i);
              }
          }
          int ans = 0;
          for (int i = n - 1; i >= 0; --i) {
              while (!stk.isEmpty() && nums[stk.peek()] <= nums[i]) {
                  ans = Math.max(ans, i - stk.pop());
              }
              if (stk.isEmpty()) {
                  break;
              }
          }
          return ans;
      }
  }