Description
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 * 1040 <= nums[i] <= 5 * 104
Solutions
Solution 1: Monotonic Stack
According to the problem, we can find that the subsequence formed by all possible nums[i] must be monotonically decreasing. Why is that? Let's prove it by contradiction.
Suppose there exist i1<i2 and nums[i1]≤nums[i2], then actually nums[i2] cannot be a candidate value, because nums[i1] is more to the left and would be a better value. Therefore, the subsequence formed by nums[i] must be monotonically decreasing, and i must start from 0.
We use a monotonically decreasing stack stk (from bottom to top) to store all possible nums[i]. Then we traverse j starting from the right boundary. If we encounter nums[stk.top()]≤nums[j], it means a ramp is formed at this point. We cyclically pop the top elements from the stack and update ans.
The time complexity is O(n) and the space complexity is O(n), where n represents the length of nums.
PythonJavaC++GoTypeScriptJavaScript
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
function maxWidthRamp(nums: number[]): number {
let [ans, n] = [0, nums.length];
const stk: number[] = [];
for (let i = 0; i < n - 1; i++) {
if (stk.length === 0 || nums[stk.at(-1)!] > nums[i]) {
stk.push(i);
}
}
for (let i = n - 1; i >= 0; i--) {
while (stk.length && nums[stk.at(-1)!] <= nums[i]) {
ans = Math.max(ans, i - stk.pop()!);
}
if (stk.length === 0) break;
}
return ans;
}(code-box)
function maxWidthRamp(nums) {
let [ans, n] = [0, nums.length];
const stk = [];
for (let i = 0; i < n - 1; i++) {
if (stk.length === 0 || nums[stk.at(-1)] > nums[i]) {
stk.push(i);
}
}
for (let i = n - 1; i >= 0; i--) {
while (stk.length && nums[stk.at(-1)] <= nums[i]) {
ans = Math.max(ans, i - stk.pop());
}
if (stk.length === 0) break;
}
return ans;
}(code-box)
