LeetCode #2866 — MEDIUM

Beautiful Towers II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

You are tasked with building n towers in the coordinate line. The i^th 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 <= heights[i] <= maxHeights[i]
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.length <= 10⁵
1 <= maxHeights[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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 <= heights[i] <= maxHeights[i] 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Stack

Example 1

[5,3,4,1,1]

Example 2

[6,5,3,9,2,7]

Example 3

[3,2,5,5,2,3]

Core Insight

What unlocks the optimal approach

Try all the possible indices <code>i</code> as the peak.
Let <code>left[i]</code> be the maximum sum of heights for the prefix <code>0, …, i</code> when index <code>i</code> is the peak.
Let <code>right[i]</code> be the maximum sum of heights for suffix <code>i, …, (n - 1)</code> when <code>i</code> is the peak
Compute values of <code>left[i]</code> from left to right using DP. For each <code>i</code> from <code>0</code> to <code>n - 1</code>, <code>left[i] = maxHeights * (i - j) + answer[j]</code>, where <code>j</code> is the rightmost index to the left of <code>i</code> such that <code>maxHeights[j] < maxHeights[i] </code>.
For each <code>i</code> from <code>n - 1</code> to <code>0</code>, <code>right[i] = maxHeights * (j - i) + answer[j]</code>, where <code>j</code> is the leftmost index to the right of <code>i</code> such that <code>maxHeights[j] < maxHeights[i] </code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2866: Beautiful Towers II
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;
    }
}

// Accepted solution for LeetCode #2866: Beautiful Towers II
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
}

# Accepted solution for LeetCode #2866: Beautiful Towers II
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))

// Accepted solution for LeetCode #2866: Beautiful Towers II
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2866: Beautiful Towers II
// 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;
//     }
// }

// Accepted solution for LeetCode #2866: Beautiful Towers II
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.