Beautiful Towers I

Beautiful Towers I - Problem

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.

Input & Output

Example 1 — Basic Mountain

$ Input: heights = [5,3,4,1,1]

› Output: 13

💡 Note: Best mountain with peak at index 0: [5,3,3,1,1] with sum 13

Example 2 — Single Tower

$ Input: heights = [6]

› Output: 6

💡 Note: Only one tower, so the mountain is just [6] with sum 6

Example 3 — Decreasing Array

$ Input: heights = [5,4,3,2,1]

› Output: 15

💡 Note: Already mountain-shaped with peak at start: [5,4,3,2,1] sum = 15

Constraints

1 ≤ heights.length ≤ 10³
1 ≤ heights[i] ≤ 10⁶

Visualization

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Asked in

G Google 25 a Amazon 20 M Microsoft 15

The key insight is to try each position as a potential mountain peak and calculate the maximum sum possible. For each peak, we can only remove bricks (not add), so heights form non-decreasing sequence to the peak and non-increasing from the peak. Best approach is Two-Pass Optimization with Time: O(n²), Space: O(1).

Common Approaches

✓ Brute Force - Try Every Peak

⏱️ Time: O(n³) Space: O(1)

For each possible peak position, simulate building the mountain by determining the maximum height at each position while maintaining the mountain property.

Two-Pass Optimization

⏱️ Time: O(n²) Space: O(n)

Use two passes to pre-compute the maximum height each tower can have when approached from the left and right, then combine them for each potential peak position.

Brute Force - Try Every Peak — Algorithm Steps

Try each index as potential peak
For left side: ensure non-decreasing up to peak
For right side: ensure non-increasing from peak
Calculate total sum for this configuration

Visualization

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Step-by-Step Walkthrough

Choose Peak

Select each index as potential peak

Shape Left

Make left side non-decreasing to peak

Shape Right

Make right side non-increasing from peak

Calculate Sum

Sum all heights and track maximum

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

long long solution(int* heights, int n) {
    long long maxSum = 0;
    
    for (int peak = 0; peak < n; peak++) {
        int* mountain = (int*)calloc(n, sizeof(int));
        mountain[peak] = heights[peak];
        
        // Fill left side (non-decreasing to peak)
        for (int i = peak - 1; i >= 0; i--) {
            mountain[i] = (heights[i] < mountain[i + 1]) ? heights[i] : mountain[i + 1];
        }
        
        // Fill right side (non-increasing from peak)
        for (int i = peak + 1; i < n; i++) {
            mountain[i] = (heights[i] < mountain[i - 1]) ? heights[i] : mountain[i - 1];
        }
        
        long long sum = 0;
        for (int i = 0; i < n; i++) {
            sum += mountain[i];
        }
        if (sum > maxSum) {
            maxSum = sum;
        }
        
        free(mountain);
    }
    
    return maxSum;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int heights[1000];
    int n = 0;
    
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '[' && token[0] != ']') {
            heights[n++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    printf("%lld\n", solution(heights, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

For each of n peaks, we scan left and right sides, each taking O(n) time

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables for calculations

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try Every Peak — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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