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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;
}

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