Test results - Codility

Tasks Details

medium

1. MinMaxDivision

Divide array A into K blocks and minimize the largest sum of any block.

41%

33%

50%

Task description

You are given integers K, M and a non-empty array A consisting of N integers. Every element of the array is not greater than M.

You should divide this array into K blocks of consecutive elements. The size of the block is any integer between 0 and N. Every element of the array should belong to some block.

The sum of the block from X to Y equals A[X] + A[X + 1] + ... + A[Y]. The sum of empty block equals 0.

The large sum is the maximal sum of any block.

For example, you are given integers K = 3, M = 5 and array A such that:

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

The array can be divided, for example, into the following blocks:

[2, 1, 5, 1, 2, 2, 2], [], [] with a large sum of 15;

[2], [1, 5, 1, 2], [2, 2] with a large sum of 9;

[2, 1, 5], [], [1, 2, 2, 2] with a large sum of 8;

[2, 1], [5, 1], [2, 2, 2] with a large sum of 6.

The goal is to minimize the large sum. In the above example, 6 is the minimal large sum.

Write a function:

def solution(K, M, A)

that, given integers K, M and a non-empty array A consisting of N integers, returns the minimal large sum.

For example, given K = 3, M = 5 and array A such that:

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

the function should return 6, as explained above.

Write an efficient algorithm for the following assumptions:

N and K are integers within the range [1..100,000];

M is an integer within the range [0..10,000];

each element of array A is an integer within the range [0..M].

Solution

Programming language used Python

Time spent on task 2 minutes

Notes

not defined yet

Task timeline

‌

Code: 12:07:03 UTC, py, verify, result: Failed

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def blocksNo(A, maxBlock):
    # Initially set the A[0] being an individual block

    blocksNumber = 1    # The number of blocks, that A could
                        # be divided to with the restriction
                        # that, the sum of each block is less
                        # than or equal to maxBlock
    preBlockSum = A[0]

    for element in A[1:]:
        # Try to extend the previous block
        if preBlockSum + element > maxBlock:
            # Fail to extend the previous block, because
            # of the sum limitation maxBlock
            preBlockSum = element
            blocksNumber += 1
        else:
            preBlockSum += element

    return blocksNumber

def solution(K, A):
    blocksNeeded = 0    # Given the restriction on the sum of
                        # each block, how many blocks could 
                        # the original A be divided to?
    resultLowerBound = max(A)
    resultUpperBound = sum(A)
    result = 0          # Minimal large sum

    # Handle two special cases
    if K == 1:      return resultUpperBound
    if K >= len(A): return resultLowerBound

    # Binary search the result
    while resultLowerBound <= resultUpperBound:
        resultMaxMid = (resultLowerBound + resultUpperBound) / 2
        blocksNeeded = blocksNo(A, resultMaxMid)
        if blocksNeeded <= K:
            # With large sum being resultMaxMid or resultMaxMid-,
            # we need blocksNeeded/blocksNeeded- blocks. While we
            # have some unused blocks (K - blocksNeeded), We could
            # try to use them to decrease the large sum.
            resultUpperBound = resultMaxMid - 1
            result = resultMaxMid
        else:
            # With large sum being resultMaxMid or resultMaxMid-,
            # we need to use more than K blocks. So resultMaxMid
            # is impossible to be our answer.
            resultLowerBound = resultMaxMid + 1

    return result

Code: 12:07:33 UTC, py, verify, result: Passed

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def blocksNo(A, maxBlock):
    # Initially set the A[0] being an individual block

    blocksNumber = 1    # The number of blocks, that A could
                        # be divided to with the restriction
                        # that, the sum of each block is less
                        # than or equal to maxBlock
    preBlockSum = A[0]

    for element in A[1:]:
        # Try to extend the previous block
        if preBlockSum + element > maxBlock:
            # Fail to extend the previous block, because
            # of the sum limitation maxBlock
            preBlockSum = element
            blocksNumber += 1
        else:
            preBlockSum += element

    return blocksNumber

def solution(K, M, A):
    blocksNeeded = 0    # Given the restriction on the sum of
                        # each block, how many blocks could 
                        # the original A be divided to?
    resultLowerBound = M
    resultUpperBound = sum(A)
    result = 0          # Minimal large sum

    # Handle two special cases
    if K == 1:      return resultUpperBound
    if K >= len(A): return resultLowerBound

    # Binary search the result
    while resultLowerBound <= resultUpperBound:
        resultMaxMid = (resultLowerBound + resultUpperBound) / 2
        blocksNeeded = blocksNo(A, resultMaxMid)
        if blocksNeeded <= K:
            # With large sum being resultMaxMid or resultMaxMid-,
            # we need blocksNeeded/blocksNeeded- blocks. While we
            # have some unused blocks (K - blocksNeeded), We could
            # try to use them to decrease the large sum.
            resultUpperBound = resultMaxMid - 1
            result = resultMaxMid
        else:
            # With large sum being resultMaxMid or resultMaxMid-,
            # we need to use more than K blocks. So resultMaxMid
            # is impossible to be our answer.
            resultLowerBound = resultMaxMid + 1

    return result

Code: 12:07:39 UTC, py, verify, result: Passed

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def blocksNo(A, maxBlock):
    # Initially set the A[0] being an individual block

    blocksNumber = 1    # The number of blocks, that A could
                        # be divided to with the restriction
                        # that, the sum of each block is less
                        # than or equal to maxBlock
    preBlockSum = A[0]

    for element in A[1:]:
        # Try to extend the previous block
        if preBlockSum + element > maxBlock:
            # Fail to extend the previous block, because
            # of the sum limitation maxBlock
            preBlockSum = element
            blocksNumber += 1
        else:
            preBlockSum += element

    return blocksNumber

def solution(K, M, A):
    blocksNeeded = 0    # Given the restriction on the sum of
                        # each block, how many blocks could 
                        # the original A be divided to?
    resultLowerBound = M
    resultUpperBound = sum(A)
    result = 0          # Minimal large sum

    # Handle two special cases
    if K == 1:      return resultUpperBound
    if K >= len(A): return resultLowerBound

    # Binary search the result
    while resultLowerBound <= resultUpperBound:
        resultMaxMid = (resultLowerBound + resultUpperBound) / 2
        blocksNeeded = blocksNo(A, resultMaxMid)
        if blocksNeeded <= K:
            # With large sum being resultMaxMid or resultMaxMid-,
            # we need blocksNeeded/blocksNeeded- blocks. While we
            # have some unused blocks (K - blocksNeeded), We could
            # try to use them to decrease the large sum.
            resultUpperBound = resultMaxMid - 1
            result = resultMaxMid
        else:
            # With large sum being resultMaxMid or resultMaxMid-,
            # we need to use more than K blocks. So resultMaxMid
            # is impossible to be our answer.
            resultLowerBound = resultMaxMid + 1

    return result

Code: 12:07:43 UTC, py, final, score: 41

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def blocksNo(A, maxBlock):
    # Initially set the A[0] being an individual block

    blocksNumber = 1    # The number of blocks, that A could
                        # be divided to with the restriction
                        # that, the sum of each block is less
                        # than or equal to maxBlock
    preBlockSum = A[0]

    for element in A[1:]:
        # Try to extend the previous block
        if preBlockSum + element > maxBlock:
            # Fail to extend the previous block, because
            # of the sum limitation maxBlock
            preBlockSum = element
            blocksNumber += 1
        else:
            preBlockSum += element

    return blocksNumber

def solution(K, M, A):
    blocksNeeded = 0    # Given the restriction on the sum of
                        # each block, how many blocks could 
                        # the original A be divided to?
    resultLowerBound = M
    resultUpperBound = sum(A)
    result = 0          # Minimal large sum

    # Handle two special cases
    if K == 1:      return resultUpperBound
    if K >= len(A): return resultLowerBound

    # Binary search the result
    while resultLowerBound <= resultUpperBound:
        resultMaxMid = (resultLowerBound + resultUpperBound) / 2
        blocksNeeded = blocksNo(A, resultMaxMid)
        if blocksNeeded <= K:
            # With large sum being resultMaxMid or resultMaxMid-,
            # we need blocksNeeded/blocksNeeded- blocks. While we
            # have some unused blocks (K - blocksNeeded), We could
            # try to use them to decrease the large sum.
            resultUpperBound = resultMaxMid - 1
            result = resultMaxMid
        else:
            # With large sum being resultMaxMid or resultMaxMid-,
            # we need to use more than K blocks. So resultMaxMid
            # is impossible to be our answer.
            resultLowerBound = resultMaxMid + 1

    return result

Analysis summary

The following issues have been detected: wrong answers.

Analysis

▶

example
example test

✔

▶

extreme_single
single elements

✘

WRONG ANSWER
got 100 expected 42

▶

extreme_double
single and double elements

✘

WRONG ANSWER
got 10 expected 4

▶

extreme_min_max
maximal / minimal values

✔

▶

simple1
simple tests

✔

▶

simple2
simple tests

✘

WRONG ANSWER
got 1000 expected 999

▶

tiny_random_ones
random values {0, 1}, N = 100

✘

WRONG ANSWER
got 0 expected 3

▶

small_random_ones
random values {0, 1}, N = 100

✘

WRONG ANSWER
got 0 expected 27

▶

medium_zeros
many zeros and 99 in the middle, length = 15,000

✘

WRONG ANSWER
got 100 expected 99

▶

medium_random
random values {1, 100}, N = 20,000

✔

▶

large_random
random values {0, ..., MAX_INT}, N = 100,000

✔

▶

large_random_ones
random values {0, 1}, N = 100,000

✘

WRONG ANSWER
got 10000 expected 452

▶

all_the_same
all the same values, N = 100,000

✔