Test results - Codility

Tasks Details

easy

1. TapeEquilibrium

Minimize the value |(A[0] + ... + A[P-1]) - (A[P] + ... + A[N-1])|.

100%

Not assessed

A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.

Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].

The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|

In other words, it is the absolute difference between the sum of the first part and the sum of the second part.

For example, consider array A such that:

A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3

We can split this tape in four places:

P = 1, difference = |3 − 10| = 7

P = 2, difference = |4 − 9| = 5

P = 3, difference = |6 − 7| = 1

P = 4, difference = |10 − 3| = 7

Write a function:

function solution(A);

that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.

For example, given:

A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3

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 [−1,000..1,000].

Solution

Programming language used JavaScript

Time spent on task 23 minutes

Notes

not defined yet

Task timeline

Code: 19:16:01 UTC, js, verify, result: Passed

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function solution(A) {
    // write your code in JavaScript 1.6
    
    var p = 1;
    var sumPartOne = A[p - 1];
    var sumPartTwo = sumUpArray(A.slice(p, A.length));
    var diff = Math.abs(sumPartOne - sumPartTwo);
    
    for(p; p < A.length - 1; p++) {
        sumPartOne += A[p];
        sumPartTwo -= A[p];
        var tempDiff = Math.abs(sumPartOne - sumPartTwo);
        if(tempDiff < diff) {
            diff = tempDiff;
        }
    }
    
    return diff;
}

function sumUpArray(A) {
    var sum = 0;
    
    for(var i = 0; i < A.length; i++) {
        sum += A[i];
    }
    
    return sum;
}

Analysis

▶

example
example test

✔

Code: 19:16:15 UTC, js, verify, result: Passed

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function solution(A) {
    // write your code in JavaScript 1.6
    
    var p = 1;
    var sumPartOne = A[p - 1];
    var sumPartTwo = sumUpArray(A.slice(p, A.length));
    var diff = Math.abs(sumPartOne - sumPartTwo);
    
    for(p; p < A.length - 1; p++) {
        sumPartOne += A[p];
        sumPartTwo -= A[p];
        var tempDiff = Math.abs(sumPartOne - sumPartTwo);
        if(tempDiff < diff) {
            diff = tempDiff;
        }
    }
    
    return diff;
}

function sumUpArray(A) {
    var sum = 0;
    
    for(var i = 0; i < A.length; i++) {
        sum += A[i];
    }
    
    return sum;
}

Analysis

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example
example test

✔

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test_data1

✔

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test_data0

✔

Code: 19:16:35 UTC, js, 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

function solution(A) {
    // write your code in JavaScript 1.6
    
    var p = 1;
    var sumPartOne = A[p - 1];
    var sumPartTwo = sumUpArray(A.slice(p, A.length));
    var diff = Math.abs(sumPartOne - sumPartTwo);
    
    for(p; p < A.length - 1; p++) {
        sumPartOne += A[p];
        sumPartTwo -= A[p];
        var tempDiff = Math.abs(sumPartOne - sumPartTwo);
        if(tempDiff < diff) {
            diff = tempDiff;
        }
    }
    
    return diff;
}

function sumUpArray(A) {
    var sum = 0;
    
    for(var i = 0; i < A.length; i++) {
        sum += A[i];
    }
    
    return sum;
}

Analysis

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example
example test

✔

▶

test_data1

✔

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test_data0

✔

Code: 19:17:03 UTC, js, 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

function solution(A) {
    // write your code in JavaScript 1.6
    
    var p = 1;
    var sumPartOne = A[p - 1];
    var sumPartTwo = sumUpArray(A.slice(p, A.length));
    var diff = Math.abs(sumPartOne - sumPartTwo);
    
    for(p; p < A.length - 1; p++) {
        sumPartOne += A[p];
        sumPartTwo -= A[p];
        var tempDiff = Math.abs(sumPartOne - sumPartTwo);
        if(tempDiff < diff) {
            diff = tempDiff;
        }
    }
    
    return diff;
}

function sumUpArray(A) {
    var sum = 0;
    
    for(var i = 0; i < A.length; i++) {
        sum += A[i];
    }
    
    return sum;
}

Analysis

▶

example
example test

✔

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test_data1

✔

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test_data0

✔

Code: 19:17:07 UTC, js, final, score: 100

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function solution(A) {
    // write your code in JavaScript 1.6
    
    var p = 1;
    var sumPartOne = A[p - 1];
    var sumPartTwo = sumUpArray(A.slice(p, A.length));
    var diff = Math.abs(sumPartOne - sumPartTwo);
    
    for(p; p < A.length - 1; p++) {
        sumPartOne += A[p];
        sumPartTwo -= A[p];
        var tempDiff = Math.abs(sumPartOne - sumPartTwo);
        if(tempDiff < diff) {
            diff = tempDiff;
        }
    }
    
    return diff;
}

function sumUpArray(A) {
    var sum = 0;
    
    for(var i = 0; i < A.length; i++) {
        sum += A[i];
    }
    
    return sum;
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N)

▶

example
example test

✔

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double
two elements

✔

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simple_positive
simple test with positive numbers, length = 5

✔

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simple_negative
simple test with negative numbers, length = 5

✔

▶

small_random
random small, length = 100

✔

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small_range
range sequence, length = ~1,000

✔

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small
small elements

✔

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medium_random1
random medium, numbers from 0 to 100, length = ~10,000

✔

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medium_random2
random medium, numbers from -1,000 to 50, length = ~10,000

✔

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large_ones
large sequence, numbers from -1 to 1, length = ~100,000

✔

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large_random
random large, length = ~100,000

✔

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large_sequence
large sequence, length = ~100,000

✔

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large_extreme
large test with maximal and minimal values, length = ~100,000

✔