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AVAILABLE LESSONS:

Lesson 1

Iterations

Lesson 2

Arrays

Lesson 3

Time Complexity

Lesson 4

Counting Elements

Lesson 5

Prefix Sums

Lesson 6

Sorting

Lesson 7

Stacks and Queues

Lesson 8

Leader

Lesson 9

Maximum slice problem

Lesson 10

Prime and composite numbers

Lesson 11

Sieve of Eratosthenes

Lesson 12

Euclidean algorithm

Lesson 13

Fibonacci numbers

Lesson 14

Binary search algorithm

Lesson 15

Caterpillar method

Lesson 16

Greedy algorithms

Lesson 17

Dynamic programming

Lesson 90

Tasks from Indeed Prime 2015 challenge

Lesson 91

Tasks from Indeed Prime 2016 challenge

Lesson 92

Tasks from Indeed Prime 2016 College Coders challenge

Lesson 99

Future training

painless

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

Programming language:
Spoken language:

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:

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:

int solution(int A[], int N);

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(N) (not counting the storage required for input arguments).

Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

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:

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:

int solution(vector<int> &A);

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(N) (not counting the storage required for input arguments).

Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

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:

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:

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

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(N) (not counting the storage required for input arguments).

Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

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

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

For example, consider array A such that:

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:

func Solution(A []int) int

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

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

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

For example, consider array A such that:

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:

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

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

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

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

For example, consider array A such that:

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:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

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

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

For example, consider array A such that:

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:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

Note: All arrays in this task are zero-indexed, unlike the common Lua convention. You can use `#A` to get the length of the array A.

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

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

For example, consider array A such that:

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:

int solution(NSMutableArray *A);

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

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

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

For example, consider array A such that:

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: array of longint; N: longint): longint;

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

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

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

For example, consider array A such that:

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:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

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

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

For example, consider array A such that:

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:

sub solution { my (@A)=@_; ... }

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

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

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

For example, consider array A such that:

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:

def solution(A)

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

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

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

For example, consider array A such that:

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:

def solution(a)

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

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

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

For example, consider array A such that:

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:

object Solution { def solution(a: Array[Int]): Int }

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

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

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

For example, consider array A such that:

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:

public func solution(inout A : [Int]) -> Int

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

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

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

For example, consider array A such that:

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:

public func solution(_ A : inout [Int]) -> Int

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

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

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

For example, consider array A such that:

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:

Private Function solution(A As Integer()) As Integer

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

For example, given:

the function should return 1, as explained above.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);

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