Description
You are given an array heights of n integers representing the number of bricks in n consecutive towers. Your task is to remove some bricks to form a mountain-shaped tower arrangement. In this arrangement, the tower heights are non-decreasing, reaching a maximum peak value with one or multiple consecutive towers and then non-increasing.
Return the maximum possible sum of heights of a mountain-shaped tower arrangement.
Example 1:
Input: heights = [5,3,4,1,1]
Output: 13
Explanation:
We remove some bricks to make heights = [5,3,3,1,1], the peak is at index 0.
Example 2:
Input: heights = [6,5,3,9,2,7]
Output: 22
Explanation:
We remove some bricks to make heights = [3,3,3,9,2,2], the peak is at index 3.
Example 3:
Input: heights = [3,2,5,5,2,3]
Output: 18
Explanation:
We remove some bricks to make heights = [2,2,5,5,2,2], the peak is at index 2 or 3.
Constraints:
1 <= n == heights.length <= 1031 <= heights[i] <= 109
Solutions
Solution 1: Enumeration
We can enumerate each tower as the tallest tower, each time expanding to the left and right, calculating the height of each other position, and then accumulating to get the height sum t. The maximum of all height sums is the answer.
The time complexity is O(n2), and the space complexity is O(1). Here, n is the length of the array maxHeights.
PythonJavaC++GoTypeScript
class Solution:
def maximumSumOfHeights(self, maxHeights: List[int]) -> int:
ans, n = 0, len(maxHeights)
for i, x in enumerate(maxHeights):
y = t = x
for j in range(i - 1, -1, -1):
y = min(y, maxHeights[j])
t += y
y = x
for j in range(i + 1, n):
y = min(y, maxHeights[j])
t += y
ans = max(ans, t)
return ans(code-box)
class Solution {
public long maximumSumOfHeights(List<Integer> maxHeights) {
long ans = 0;
int n = maxHeights.size();
for (int i = 0; i < n; ++i) {
int y = maxHeights.get(i);
long t = y;
for (int j = i - 1; j >= 0; --j) {
y = Math.min(y, maxHeights.get(j));
t += y;
}
y = maxHeights.get(i);
for (int j = i + 1; j < n; ++j) {
y = Math.min(y, maxHeights.get(j));
t += y;
}
ans = Math.max(ans, t);
}
return ans;
}
}(code-box)
class Solution {
public:
long long maximumSumOfHeights(vector<int>& maxHeights) {
long long ans = 0;
int n = maxHeights.size();
for (int i = 0; i < n; ++i) {
long long t = maxHeights[i];
int y = t;
for (int j = i - 1; ~j; --j) {
y = min(y, maxHeights[j]);
t += y;
}
y = maxHeights[i];
for (int j = i + 1; j < n; ++j) {
y = min(y, maxHeights[j]);
t += y;
}
ans = max(ans, t);
}
return ans;
}
};(code-box)
func maximumSumOfHeights(maxHeights []int) (ans int64) {
n := len(maxHeights)
for i, x := range maxHeights {
y, t := x, x
for j := i - 1; j >= 0; j-- {
y = min(y, maxHeights[j])
t += y
}
y = x
for j := i + 1; j < n; j++ {
y = min(y, maxHeights[j])
t += y
}
ans = max(ans, int64(t))
}
return
}(code-box)
function maximumSumOfHeights(maxHeights: number[]): number {
let ans = 0;
const n = maxHeights.length;
for (let i = 0; i < n; ++i) {
const x = maxHeights[i];
let [y, t] = [x, x];
for (let j = i - 1; ~j; --j) {
y = Math.min(y, maxHeights[j]);
t += y;
}
y = x;
for (let j = i + 1; j < n; ++j) {
y = Math.min(y, maxHeights[j]);
t += y;
}
ans = Math.max(ans, t);
}
return ans;
}(code-box)
Solution 2: Dynamic Programming + Monotonic Stack
Solution 1 is sufficient to pass this problem, but the time complexity is relatively high. We can use "Dynamic Programming + Monotonic Stack" to optimize the enumeration process.
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],&if heights[i]≥ heights[i-1]\
heights[i]×(i-j)+f[j],&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 ith 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.
PythonJavaC++GoTypeScript
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))(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
