Test results - Codility

Tasks Details

easy

1. StoneWall

Cover "Manhattan skyline" using the minimum number of rectangles.

100%

You are going to build a stone wall. The wall should be straight and N meters long, and its thickness should be constant; however, it should have different heights in different places. The height of the wall is specified by an array H of N positive integers. H[I] is the height of the wall from I to I+1 meters to the right of its left end. In particular, H[0] is the height of the wall's left end and H[N−1] is the height of the wall's right end.

The wall should be built of cuboid stone blocks (that is, all sides of such blocks are rectangular). Your task is to compute the minimum number of blocks needed to build the wall.

Write a function:

class Solution { public int solution(int[] H); }

that, given an array H of N positive integers specifying the height of the wall, returns the minimum number of blocks needed to build it.

For example, given array H containing N = 9 integers:

H[0] = 8 H[1] = 8 H[2] = 5 H[3] = 7 H[4] = 9 H[5] = 8 H[6] = 7 H[7] = 4 H[8] = 8

the function should return 7. The figure shows one possible arrangement of seven blocks.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..100,000];

each element of array H is an integer within the range [1..1,000,000,000].

Solution

Programming language used Java 8

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

Code: 12:21:33 UTC, java, verify, result: Passed

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// you can also use imports, for example:
import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public static int solution(int[] H) {
        // write your code in Java SE 8
        int result = 0;
        Stack<Integer> stack = new Stack<>();
        stack.push(H[0]);
        result ++;
        for (int i = 0; i < H.length; i++) {
            if (H[i] > stack.peek()) {
                stack.push(H[i]);
                result++;
            } else if (H[i] < stack.peek()) {
                while (!stack.empty() && H[i] < stack.peek()) {
                    stack.pop();
                }
                if(!stack.empty() && stack.peek() == H[i]){

                }else{
                    result++;
                }
                stack.push(H[i]);
            }
        }

        return result;
    }
}

Analysis

▶

example

✔

Code: 12:21:38 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

// you can also use imports, for example:
import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public static int solution(int[] H) {
        // write your code in Java SE 8
        int result = 0;
        Stack<Integer> stack = new Stack<>();
        stack.push(H[0]);
        result ++;
        for (int i = 0; i < H.length; i++) {
            if (H[i] > stack.peek()) {
                stack.push(H[i]);
                result++;
            } else if (H[i] < stack.peek()) {
                while (!stack.empty() && H[i] < stack.peek()) {
                    stack.pop();
                }
                if(!stack.empty() && stack.peek() == H[i]){

                }else{
                    result++;
                }
                stack.push(H[i]);
            }
        }

        return result;
    }
}

Analysis

▶

example

✔

Code: 12:21:42 UTC, java, final, score: 100

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

// you can also use imports, for example:
import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public static int solution(int[] H) {
        // write your code in Java SE 8
        int result = 0;
        Stack<Integer> stack = new Stack<>();
        stack.push(H[0]);
        result ++;
        for (int i = 0; i < H.length; i++) {
            if (H[i] > stack.peek()) {
                stack.push(H[i]);
                result++;
            } else if (H[i] < stack.peek()) {
                while (!stack.empty() && H[i] < stack.peek()) {
                    stack.pop();
                }
                if(!stack.empty() && stack.peek() == H[i]){

                }else{
                    result++;
                }
                stack.push(H[i]);
            }
        }

        return result;
    }
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N)

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example

✔

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simple1

✔

▶

simple2

✔

▶

simple3

✔

▶

simple4

✔

▶

boundary_cases

✔

▶

medium1

✔

▶

medium2

✔

▶

medium3

✔

▶

medium4

✔

▶

large_piramid

✔

▶

large_increasing_decreasing

✔

▶

large_up_to_20

✔

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large_up_to_100

✔

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large_max

✔