# 2866. Beautiful Towers II

## Description

You are given a 0-indexed array maxHeights of n integers.

You are tasked with building n towers in the coordinate line. The ith tower is built at coordinate i and has a height of heights[i].

A configuration of towers is beautiful if the following conditions hold:

1. 1 <= heights[i] <= maxHeights[i]
2. heights is a mountain array.

Array heights is a mountain if there exists an index i such that:

• For all 0 < j <= i, heights[j - 1] <= heights[j]
• For all i <= k < n - 1, heights[k + 1] <= heights[k]

Return the maximum possible sum of heights of a beautiful configuration of towers.

Example 1:

Input: maxHeights = [5,3,4,1,1]
Output: 13
Explanation: One beautiful configuration with a maximum sum is heights = [5,3,3,1,1]. This configuration is beautiful since:
- 1 <= heights[i] <= maxHeights[i]
- heights is a mountain of peak i = 0.
It can be shown that there exists no other beautiful configuration with a sum of heights greater than 13.

Example 2:

Input: maxHeights = [6,5,3,9,2,7]
Output: 22
Explanation: One beautiful configuration with a maximum sum is heights = [3,3,3,9,2,2]. This configuration is beautiful since:
- 1 <= heights[i] <= maxHeights[i]
- heights is a mountain of peak i = 3.
It can be shown that there exists no other beautiful configuration with a sum of heights greater than 22.

Example 3:

Input: maxHeights = [3,2,5,5,2,3]
Output: 18
Explanation: One beautiful configuration with a maximum sum is heights = [2,2,5,5,2,2]. This configuration is beautiful since:
- 1 <= heights[i] <= maxHeights[i]
- heights is a mountain of peak i = 2.
Note that, for this configuration, i = 3 can also be considered a peak.
It can be shown that there exists no other beautiful configuration with a sum of heights greater than 18.


Constraints:

• 1 <= n == maxHeights <= 105
• 1 <= maxHeights[i] <= 109

## Solutions

Solution 1: Dynamic Programming + Monotonic Stack

We define $f[i]$ to represent the height sum of the beautiful tower scheme with the last tower as the tallest tower among the first $i+1$ towers. We can get the following state transition equation:

$f[i]= \begin{cases} f[i-1]+heights[i],&\text{if } heights[i]\geq heights[i-1]\\ heights[i]\times(i-j)+f[j],&\text{if } heights[i]<heights[i-1] \end{cases}$

Where $j$ is the index of the first tower to the left of the last tower with a height less than or equal to $heights[i]$. We can use a monotonic stack to maintain this index.

We can use a similar method to find $g[i]$, which represents the height sum of the beautiful tower scheme from right to left with the $i$th tower as the tallest tower. The final answer is the maximum value of $f[i]+g[i]-heights[i]$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $maxHeights$.

• class Solution {
public long maximumSumOfHeights(List<Integer> maxHeights) {
int n = maxHeights.size();
Deque<Integer> stk = new ArrayDeque<>();
int[] left = new int[n];
int[] right = new int[n];
Arrays.fill(left, -1);
Arrays.fill(right, n);
for (int i = 0; i < n; ++i) {
int x = maxHeights.get(i);
while (!stk.isEmpty() && maxHeights.get(stk.peek()) > x) {
stk.pop();
}
if (!stk.isEmpty()) {
left[i] = stk.peek();
}
stk.push(i);
}
stk.clear();
for (int i = n - 1; i >= 0; --i) {
int x = maxHeights.get(i);
while (!stk.isEmpty() && maxHeights.get(stk.peek()) >= x) {
stk.pop();
}
if (!stk.isEmpty()) {
right[i] = stk.peek();
}
stk.push(i);
}
long[] f = new long[n];
long[] g = new long[n];
for (int i = 0; i < n; ++i) {
int x = maxHeights.get(i);
if (i > 0 && x >= maxHeights.get(i - 1)) {
f[i] = f[i - 1] + x;
} else {
int j = left[i];
f[i] = 1L * x * (i - j) + (j >= 0 ? f[j] : 0);
}
}
for (int i = n - 1; i >= 0; --i) {
int x = maxHeights.get(i);
if (i < n - 1 && x >= maxHeights.get(i + 1)) {
g[i] = g[i + 1] + x;
} else {
int j = right[i];
g[i] = 1L * x * (j - i) + (j < n ? g[j] : 0);
}
}
long ans = 0;
for (int i = 0; i < n; ++i) {
ans = Math.max(ans, f[i] + g[i] - maxHeights.get(i));
}
return ans;
}
}

• class Solution {
public:
long long maximumSumOfHeights(vector<int>& maxHeights) {
int n = maxHeights.size();
stack<int> stk;
vector<int> left(n, -1);
vector<int> right(n, n);
for (int i = 0; i < n; ++i) {
int x = maxHeights[i];
while (!stk.empty() && maxHeights[stk.top()] > x) {
stk.pop();
}
if (!stk.empty()) {
left[i] = stk.top();
}
stk.push(i);
}
stk = stack<int>();
for (int i = n - 1; ~i; --i) {
int x = maxHeights[i];
while (!stk.empty() && maxHeights[stk.top()] >= x) {
stk.pop();
}
if (!stk.empty()) {
right[i] = stk.top();
}
stk.push(i);
}
long long f[n], g[n];
for (int i = 0; i < n; ++i) {
int x = maxHeights[i];
if (i && x >= maxHeights[i - 1]) {
f[i] = f[i - 1] + x;
} else {
int j = left[i];
f[i] = 1LL * x * (i - j) + (j != -1 ? f[j] : 0);
}
}
for (int i = n - 1; ~i; --i) {
int x = maxHeights[i];
if (i < n - 1 && x >= maxHeights[i + 1]) {
g[i] = g[i + 1] + x;
} else {
int j = right[i];
g[i] = 1LL * x * (j - i) + (j != n ? g[j] : 0);
}
}
long long ans = 0;
for (int i = 0; i < n; ++i) {
ans = max(ans, f[i] + g[i] - maxHeights[i]);
}
return ans;
}
};

• class Solution:
def maximumSumOfHeights(self, maxHeights: List[int]) -> int:
n = len(maxHeights)
stk = []
left = [-1] * n
for i, x in enumerate(maxHeights):
while stk and maxHeights[stk[-1]] > x:
stk.pop()
if stk:
left[i] = stk[-1]
stk.append(i)
stk = []
right = [n] * n
for i in range(n - 1, -1, -1):
x = maxHeights[i]
while stk and maxHeights[stk[-1]] >= x:
stk.pop()
if stk:
right[i] = stk[-1]
stk.append(i)
f = [0] * n
for i, x in enumerate(maxHeights):
if i and x >= maxHeights[i - 1]:
f[i] = f[i - 1] + x
else:
j = left[i]
f[i] = x * (i - j) + (f[j] if j != -1 else 0)
g = [0] * n
for i in range(n - 1, -1, -1):
if i < n - 1 and maxHeights[i] >= maxHeights[i + 1]:
g[i] = g[i + 1] + maxHeights[i]
else:
j = right[i]
g[i] = maxHeights[i] * (j - i) + (g[j] if j != n else 0)
return max(a + b - c for a, b, c in zip(f, g, maxHeights))


• func maximumSumOfHeights(maxHeights []int) (ans int64) {
n := len(maxHeights)
stk := []int{}
left := make([]int, n)
right := make([]int, n)
for i := range left {
left[i] = -1
right[i] = n
}
for i, x := range maxHeights {
for len(stk) > 0 && maxHeights[stk[len(stk)-1]] > x {
stk = stk[:len(stk)-1]
}
if len(stk) > 0 {
left[i] = stk[len(stk)-1]
}
stk = append(stk, i)
}
stk = []int{}
for i := n - 1; i >= 0; i-- {
x := maxHeights[i]
for len(stk) > 0 && maxHeights[stk[len(stk)-1]] >= x {
stk = stk[:len(stk)-1]
}
if len(stk) > 0 {
right[i] = stk[len(stk)-1]
}
stk = append(stk, i)
}
f := make([]int64, n)
g := make([]int64, n)
for i, x := range maxHeights {
if i > 0 && x >= maxHeights[i-1] {
f[i] = f[i-1] + int64(x)
} else {
j := left[i]
f[i] = int64(x) * int64(i-j)
if j != -1 {
f[i] += f[j]
}
}
}
for i := n - 1; i >= 0; i-- {
x := maxHeights[i]
if i < n-1 && x >= maxHeights[i+1] {
g[i] = g[i+1] + int64(x)
} else {
j := right[i]
g[i] = int64(x) * int64(j-i)
if j != n {
g[i] += g[j]
}
}
}
for i, x := range maxHeights {
ans = max(ans, f[i]+g[i]-int64(x))
}
return
}

func max(a, b int64) int64 {
if a > b {
return a
}
return b
}

• function maximumSumOfHeights(maxHeights: number[]): number {
const n = maxHeights.length;
const stk: number[] = [];
const left: number[] = Array(n).fill(-1);
const right: number[] = Array(n).fill(n);
for (let i = 0; i < n; ++i) {
const x = maxHeights[i];
while (stk.length && maxHeights[stk.at(-1)] > x) {
stk.pop();
}
if (stk.length) {
left[i] = stk.at(-1);
}
stk.push(i);
}
stk.length = 0;
for (let i = n - 1; ~i; --i) {
const x = maxHeights[i];
while (stk.length && maxHeights[stk.at(-1)] >= x) {
stk.pop();
}
if (stk.length) {
right[i] = stk.at(-1);
}
stk.push(i);
}
const f: number[] = Array(n).fill(0);
const g: number[] = Array(n).fill(0);
for (let i = 0; i < n; ++i) {
const x = maxHeights[i];
if (i && x >= maxHeights[i - 1]) {
f[i] = f[i - 1] + x;
} else {
const j = left[i];
f[i] = x * (i - j) + (j >= 0 ? f[j] : 0);
}
}
for (let i = n - 1; ~i; --i) {
const x = maxHeights[i];
if (i + 1 < n && x >= maxHeights[i + 1]) {
g[i] = g[i + 1] + x;
} else {
const j = right[i];
g[i] = x * (j - i) + (j < n ? g[j] : 0);
}
}
let ans = 0;
for (let i = 0; i < n; ++i) {
ans = Math.max(ans, f[i] + g[i] - maxHeights[i]);
}
return ans;
}