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Formatted question description: https://leetcode.ca/all/962.html

962. Maximum Width Ramp (Medium)

Given an array A of integers, a ramp is a tuple (i, j) for which i < j and A[i] <= A[j].  The width of such a ramp is j - i.

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

 

Related Topics:
Array

Solution 1.

Consider the array ends with 4, 1, 2. For the a number n in front:

• If n <= 2, we use the last 2 to form the ramp. The last 1 won’t be used.
• If 2 < n <= 4, we use the last 4 to form the ramp.

So we can traverse from rear to front, and save the numbers in ascending order:

numbers: [2, 4]
indexes: [N - 1, N - 3]

Then we traverse from front to rear, use lower_bound to find the smallest number greater than the current A[i]. The corresponding index can be used form a ramp.

// OJ: https://leetcode.com/problems/maximum-width-ramp/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int maxWidthRamp(vector<int>& A) {
        int N = A.size(), ans = 0;
        vector<int> v, index;
        for (int i = N - 1; i >= 0; --i) {
            if (v.empty() || v.back() < A[i]) {
                v.push_back(A[i]);
                index.push_back(i);
            }
        }
        for (int i = 0; i < N; ++i) {
            int ind = lower_bound(v.begin(), v.end(), A[i]) - v.begin();
            ans = max(ans, index[ind] - i);
        }
        return ans;
    }
};

• Java
• C++
• Python
• Go

• class Solution {
      public int maxWidthRamp(int[] A) {
          if (A == null || A.length < 2)
              return 0;
          int maxWidth = 0;
          Stack<Integer> numStack = new Stack<Integer>();
          Stack<Integer> indicesStack = new Stack<Integer>();
          numStack.push(A[0]);
          indicesStack.push(0);
          int length = A.length;
          for (int i = 1; i < length; i++) {
              int num = A[i];
              if (numStack.isEmpty() || num < numStack.peek()) {
                  numStack.push(num);
                  indicesStack.push(i);
              } else {
                  Stack<Integer> tempNumStack = new Stack<Integer>();
                  Stack<Integer> tempIndicesStack = new Stack<Integer>();
                  while (!numStack.isEmpty() && num >= numStack.peek()) {
                      tempNumStack.push(numStack.pop());
                      tempIndicesStack.push(indicesStack.pop());
                  }
                  int prevIndex = tempIndicesStack.peek();
                  maxWidth = Math.max(maxWidth, i - prevIndex);
                  while (!tempNumStack.isEmpty()) {
                      numStack.push(tempNumStack.pop());
                      indicesStack.push(tempIndicesStack.pop());
                  }
              }
          }
          return maxWidth;
      }
  }
  
  ############
  
  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;
      }
  }
  

• // OJ: https://leetcode.com/problems/maximum-width-ramp/
  // Time: O(NlogN)
  // Space: O(N)
  class Solution {
  public:
      int maxWidthRamp(vector<int>& A) {
          int N = A.size(), ans = 0;
          vector<int> v, index;
          for (int i = N - 1; i >= 0; --i) {
              if (v.empty() || v.back() < A[i]) {
                  v.push_back(A[i]);
                  index.push_back(i);
              }
          }
          for (int i = 0; i < N; ++i) {
              int ind = lower_bound(v.begin(), v.end(), A[i]) - v.begin();
              ans = max(ans, index[ind] - i);
          }
          return ans;
      }
  };
  

• class Solution:
      def maxWidthRamp(self, nums: List[int]) -> int:
          stk = []
          for i, v in enumerate(nums):
              if not stk or nums[stk[-1]] > v:
                  stk.append(i)
          ans = 0
          for i in range(len(nums) - 1, -1, -1):
              while stk and nums[stk[-1]] <= nums[i]:
                  ans = max(ans, i - stk.pop())
              if not stk:
                  break
          return ans
  
  ############
  
  class Solution(object):
      def maxWidthRamp(self, A):
          """
          :type A: List[int]
          :rtype: int
          """
          N = len(A)
          stack = []
          res = 0
          for i, a in enumerate(A):
              if not stack or stack[-1][1] > a:
                  stack.append((i, a))
              else:
                  x = len(stack) - 1
                  while x >= 0 and stack[x][1] <= a:
                      res = max(res, i - stack[x][0])
                      x -= 1
          return res
  

• func maxWidthRamp(nums []int) int {
  	n := len(nums)
  	stk := []int{}
  	for i, v := range nums {
  		if len(stk) == 0 || nums[stk[len(stk)-1]] > v {
  			stk = append(stk, i)
  		}
  	}
  	ans := 0
  	for i := n - 1; i >= 0; i-- {
  		for len(stk) > 0 && nums[stk[len(stk)-1]] <= nums[i] {
  			ans = max(ans, i-stk[len(stk)-1])
  			stk = stk[:len(stk)-1]
  		}
  		if len(stk) == 0 {
  			break
  		}
  	}
  	return ans
  }
  
  func max(a, b int) int {
  	if a > b {
  		return a
  	}
  	return b
  }
  

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