Test results - Codility

Tasks Details

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1. MinAvgTwoSlice

Find the minimal average of any slice containing at least two elements.

100%

A non-empty array A consisting of N integers is given. A pair of integers (P, Q), such that 0 ≤ P < Q < N, is called a slice of array A (notice that the slice contains at least two elements). The average of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

A[0] = 4 A[1] = 2 A[2] = 2 A[3] = 5 A[4] = 1 A[5] = 5 A[6] = 8

contains the following example slices:

slice (1, 2), whose average is (2 + 2) / 2 = 2;

slice (3, 4), whose average is (5 + 1) / 2 = 3;

slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

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

that, given a non-empty array A consisting of N integers, returns the starting position of the slice with the minimal average. If there is more than one slice with a minimal average, you should return the smallest starting position of such a slice.

For example, given array A such that:

A[0] = 4 A[1] = 2 A[2] = 2 A[3] = 5 A[4] = 1 A[5] = 5 A[6] = 8

the function should return 1, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [2..100,000];

each element of array A is an integer within the range [−10,000..10,000].

Solution

Programming language used Java 8

Time spent on task 5 minutes

Notes

not defined yet

Task timeline

Code: 04:02:58 UTC, java, verify, result: Failed

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 34 35

// 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 int solution(int[] A) {
        // write your code in Java SE 8
        int minAvgIdx=0;
        double minAvg = (A[0]+A[1])/2;
        double currAvg;
        for(int i=0; i< A.length-2; i++) {
            //two number 
            currAvg = (A[i] +A[i+1])/2;
            if(minAvg > currAvg) {
                minAvg = currAvg;
                minAvgIdx = i;
            }
            
            //three number 
            currAvg = (A[i] +A[i+1] +A[i+1])/3;
            if(minAvg > currAvg) {
                minAvg = currAvg;
                minAvgIdx = i;
            }
        }
        currAvg = (A[A.length-2] +A[A.length-1])/2;
        if(minAvg > currAvg) {
            minAvg = currAvg;
            minAvgIdx = A.length-2;
        }
        return  minAvgIdx;
    }
}

Analysis

▶

example
example test

✘

WRONG ANSWER
got 0 expected 1

Code: 04:03:28 UTC, java, verify, result: Failed

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 34 35

// 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 int solution(int[] A) {
        // write your code in Java SE 8
        int minAvgIdx=0;
        double minAvg = (A[0]+A[1])/2;
        double currAvg;
        for(int i=0; i< A.length-2; i++) {
            //two number 
            currAvg = (A[i] +A[i+1])/2;
            if(minAvg > currAvg) {
                minAvg = currAvg;
                minAvgIdx = i;
            }
            
            //three number 
            currAvg = (A[i] +A[i+1] +A[i+2])/3;
            if(minAvg > currAvg) {
                minAvg = currAvg;
                minAvgIdx = i;
            }
        }
        currAvg = (A[A.length-2] +A[A.length-1])/2;
        if(minAvg > currAvg) {
            minAvg = currAvg;
            minAvgIdx = A.length-2;
        }
        return  minAvgIdx;
    }
}

Analysis

▶

example
example test

✘

WRONG ANSWER
got 0 expected 1

Code: 04:06:28 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 34 35

// 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 int solution(int[] A) {
        // write your code in Java SE 8
        int minAvgIdx=0;
        double minAvg = (A[0]+A[1])/2;
        double currAvg;
        for(int i=0; i< A.length-2; i++) {
            //two number 
            currAvg = (double)(A[i] +A[i+1])/2;
            if(minAvg > currAvg) {
                minAvg = currAvg;
                minAvgIdx = i;
            }
            
            //three number 
            currAvg =(double) (A[i] +A[i+1] +A[i+2])/3;
            if(minAvg > currAvg) {
                minAvg = currAvg;
                minAvgIdx = i;
            }
        }
        currAvg =(double) (A[A.length-2] +A[A.length-1])/2;
        if(minAvg > currAvg) {
            minAvg = currAvg;
            minAvgIdx = A.length-2;
        }
        return  minAvgIdx;
    }
}

Analysis

▶

example
example test

✔

Code: 04:06:42 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 34 35

// 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 int solution(int[] A) {
        // write your code in Java SE 8
        int minAvgIdx=0;
        double minAvg = (A[0]+A[1])/2;
        double currAvg;
        for(int i=0; i< A.length-2; i++) {
            //two number 
            currAvg = (double)(A[i] +A[i+1])/2;
            if(minAvg > currAvg) {
                minAvg = currAvg;
                minAvgIdx = i;
            }
            
            //three number 
            currAvg =(double) (A[i] +A[i+1] +A[i+2])/3;
            if(minAvg > currAvg) {
                minAvg = currAvg;
                minAvgIdx = i;
            }
        }
        currAvg =(double) (A[A.length-2] +A[A.length-1])/2;
        if(minAvg > currAvg) {
            minAvg = currAvg;
            minAvgIdx = A.length-2;
        }
        return  minAvgIdx;
    }
}

Analysis

▶

example
example test

✔

Code: 04:06:50 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 34 35

// 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 int solution(int[] A) {
        // write your code in Java SE 8
        int minAvgIdx=0;
        double minAvg = (A[0]+A[1])/2;
        double currAvg;
        for(int i=0; i< A.length-2; i++) {
            //two number 
            currAvg = (double)(A[i] +A[i+1])/2;
            if(minAvg > currAvg) {
                minAvg = currAvg;
                minAvgIdx = i;
            }
            
            //three number 
            currAvg =(double) (A[i] +A[i+1] +A[i+2])/3;
            if(minAvg > currAvg) {
                minAvg = currAvg;
                minAvgIdx = i;
            }
        }
        currAvg =(double) (A[A.length-2] +A[A.length-1])/2;
        if(minAvg > currAvg) {
            minAvg = currAvg;
            minAvgIdx = A.length-2;
        }
        return  minAvgIdx;
    }
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N)

▶

example
example test

✔

▶

double_quadruple
two or four elements

✔

▶

simple1
simple test, the best slice has length 3

✔

▶

simple2
simple test, the best slice has length 3

✔

▶

small_random
random, length = 100

✔

▶

medium_range
increasing, decreasing (legth = ~100) and small functional

✔

▶

medium_random
random, N = ~700

✔

▶

large_ones
numbers from -1 to 1, N = ~100,000

✔

▶

large_random
random, N = ~100,000

✔

▶

extreme_values
all maximal values, N = ~100,000

✔

▶

large_sequence
many seqeneces, N = ~100,000

✔