Building Shadow Calculator - Problem

You are given an array of integers representing the heights of buildings in a row. The buildings are illuminated by a light source positioned to the left of all buildings, casting shadows to the right.

Each building casts a shadow with length equal to its height. However, if a building is blocked by a taller building to its left, it won't cast any shadow (the taller building's shadow covers it completely).

Calculate the total length of all visible shadows cast to the right.

Example: For buildings [3, 1, 4, 1, 5], only buildings with heights 3, 4, and 5 cast shadows (heights 3→4→5 are increasing from left). The two buildings with height 1 are blocked. Total shadow length = 3 + 4 + 5 = 12.

Input & Output

Example 1 — Basic Case

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

› Output: 12

💡 Note: Building 3 casts shadow (no taller building to left). Building 1 is blocked by 3. Building 4 casts shadow (4 > 3). Building 1 is blocked by 4. Building 5 casts shadow (5 > 4). Total: 3 + 4 + 5 = 12

Example 2 — All Increasing

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

› Output: 15

💡 Note: Each building is taller than all previous ones, so all cast shadows: 1 + 2 + 3 + 4 + 5 = 15

Example 3 — All Decreasing

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

› Output: 5

💡 Note: Only the first building (height 5) casts a shadow. All others are blocked by it. Total: 5

Constraints

1 ≤ heights.length ≤ 10⁴
1 ≤ heights[i] ≤ 10⁹

Visualization

Tap to expand

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that only buildings taller than all previous buildings can cast shadows. Track the maximum height seen so far - if current height exceeds it, add to total and update maximum. Best approach uses single pass with maximum tracking. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check Every Building's Blockers

⏱️ Time: O(n²) Space: O(1)

For every building, scan all buildings to its left to find if there's a taller one that would block its shadow. If no taller building exists to the left, add this building's height to the total shadow length.

Single Pass with Maximum Tracking

⏱️ Time: O(n) Space: O(1)

Iterate through buildings once, maintaining the maximum height encountered. A building casts a shadow only if its height is greater than the current maximum. When a building casts a shadow, update the maximum height.

Brute Force - Check Every Building's Blockers — Algorithm Steps

Iterate through each building from left to right
For each building, check all previous buildings
If no previous building is taller or equal, add height to total
Return total shadow length

Visualization

Tap to expand

Step-by-Step Walkthrough

Select Building

Pick current building to check

Scan Left

Check all previous buildings for taller ones

Add or Skip

If no blocker found, add height to total

Code -

solution.c — C

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

int solution(int* heights, int n) {
    int totalShadow = 0;
    
    for (int i = 0; i < n; i++) {
        int isBlocked = 0;
        // Check all buildings to the left
        for (int j = 0; j < i; j++) {
            if (heights[j] >= heights[i]) {
                isBlocked = 1;
                break;
            }
        }
        
        // If not blocked, add to shadow
        if (!isBlocked) {
            totalShadow += heights[i];
        }
    }
    
    return totalShadow;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON array manually
    int heights[100];
    int n = 0;
    char* ptr = strchr(line, '[');
    if (ptr) {
        ptr++;
        char* token = strtok(ptr, ",]");
        while (token != NULL && n < 100) {
            heights[n++] = atoi(token);
            token = strtok(NULL, ",]");
        }
    }
    
    int result = solution(heights, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops: for each of n buildings, check up to n-1 previous buildings

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for loop counters and total sum

✓ Linear Space

23.4K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Building Shadow Calculator - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check Every Building's Blockers — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler