##### Welcome to Subscribe On Youtube

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;
}
};

• 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
}