LeetCode #3301 — MEDIUM

Maximize the Total Height of Unique Towers

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

The Problem

Problem Statement

You are given an array maximumHeight, where maximumHeight[i] denotes the maximum height the i^th tower can be assigned.

Your task is to assign a height to each tower so that:

The height of the i^th tower is a positive integer and does not exceed maximumHeight[i].
No two towers have the same height.

Return the maximum possible total sum of the tower heights. If it's not possible to assign heights, return -1.

Example 1:

Input: maximumHeight = [2,3,4,3]

Output: 10

Explanation:

We can assign heights in the following way: [1, 2, 4, 3].

Example 2:

Input: maximumHeight = [15,10]

Output: 25

Explanation:

We can assign heights in the following way: [15, 10].

Example 3:

Input: maximumHeight = [2,2,1]

Output: -1

Explanation:

It's impossible to assign positive heights to each index so that no two towers have the same height.

Constraints:

1 <= maximumHeight.length <= 10⁵
1 <= maximumHeight[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an array maximumHeight, where maximumHeight[i] denotes the maximum height the ith tower can be assigned. Your task is to assign a height to each tower so that: The height of the ith tower is a positive integer and does not exceed maximumHeight[i]. No two towers have the same height. Return the maximum possible total sum of the tower heights. If it's not possible to assign heights, return -1.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[2,3,4,3]

Example 2

[15,10]

Example 3

[2,2,1]

Step 02

Core Insight

What unlocks the optimal approach

Sort the array <code>maximumHeight</code> in descending order.
After sorting, it can be seen that the maximum height that we can assign to the <code>i<sup>th</sup></code> element is <code>min(maximumHeight[i], maximumHeight[i - 1] - 1)</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 #3301: Maximize the Total Height of Unique Towers
class Solution {
    public long maximumTotalSum(int[] maximumHeight) {
        long ans = 0;
        int mx = 1 << 30;
        Arrays.sort(maximumHeight);
        for (int i = maximumHeight.length - 1; i >= 0; --i) {
            int x = Math.min(maximumHeight[i], mx - 1);
            if (x <= 0) {
                return -1;
            }
            ans += x;
            mx = x;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3301: Maximize the Total Height of Unique Towers
func maximumTotalSum(maximumHeight []int) int64 {
	slices.SortFunc(maximumHeight, func(a, b int) int { return b - a })
	ans := int64(0)
	mx := 1 << 30
	for _, x := range maximumHeight {
		x = min(x, mx-1)
		if x <= 0 {
			return -1
		}
		ans += int64(x)
		mx = x
	}
	return ans
}

# Accepted solution for LeetCode #3301: Maximize the Total Height of Unique Towers
class Solution:
    def maximumTotalSum(self, maximumHeight: List[int]) -> int:
        maximumHeight.sort()
        ans, mx = 0, inf
        for x in maximumHeight[::-1]:
            x = min(x, mx - 1)
            if x <= 0:
                return -1
            ans += x
            mx = x
        return ans

// Accepted solution for LeetCode #3301: Maximize the Total Height of Unique Towers
// 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 #3301: Maximize the Total Height of Unique Towers
// class Solution {
//     public long maximumTotalSum(int[] maximumHeight) {
//         long ans = 0;
//         int mx = 1 << 30;
//         Arrays.sort(maximumHeight);
//         for (int i = maximumHeight.length - 1; i >= 0; --i) {
//             int x = Math.min(maximumHeight[i], mx - 1);
//             if (x <= 0) {
//                 return -1;
//             }
//             ans += x;
//             mx = x;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3301: Maximize the Total Height of Unique Towers
function maximumTotalSum(maximumHeight: number[]): number {
    maximumHeight.sort((a, b) => a - b).reverse();
    let ans: number = 0;
    let mx: number = Infinity;
    for (let x of maximumHeight) {
        x = Math.min(x, mx - 1);
        if (x <= 0) {
            return -1;
        }
        ans += x;
        mx = x;
    }
    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 × log n)

Space

O(log n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.