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Programming language:
Spoken language:

Two non-empty arrays A and B, each consisting of N integers, are given. Four functions are defined based on these arrays:

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

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

that, given two arrays A and B consisting of N integers each, returns the minimum value of S(X) where X can be any real number.

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

- expected worst-case time complexity is O(N*log(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.

Two non-empty arrays A and B, each consisting of N integers, are given. Four functions are defined based on these arrays:

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

double solution(vector<int> &A, vector<int> &B);

that, given two arrays A and B consisting of N integers each, returns the minimum value of S(X) where X can be any real number.

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

- expected worst-case time complexity is O(N*log(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.

Two non-empty arrays A and B, each consisting of N integers, are given. Four functions are defined based on these arrays:

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

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

that, given two arrays A and B consisting of N integers each, returns the minimum value of S(X) where X can be any real number.

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

- expected worst-case time complexity is O(N*log(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.

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

func Solution(A []int, B []int) float64

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

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

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

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

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

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

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

function solution(A, B);

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

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

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

function solution(A, B)

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

- expected worst-case time complexity is O(N*log(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.

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

double solution(NSMutableArray *A, NSMutableArray *B);

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

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

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

function solution(A: array of longint; B: array of longint; N: longint): double;

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

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

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

function solution($A, $B);

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

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

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

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

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

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

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

def solution(A, B)

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

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

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

def solution(a, b)

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

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

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

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

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

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

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

public func solution(inout A : [Int], inout _ B : [Int]) -> Float64

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

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

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

public func solution(_ A : inout [Int], _ B : inout [Int]) -> Float64

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

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

F(X,K) = A[K]*X + B[K] U(X) = max{ F(X,K) : 0 ≤ K < N }D(X) = min{ F(X,K) : 0 ≤ K < N }S(X) = U(X) − D(X)

Write a function:

Private Function solution(A As Integer(), B As Integer()) As Double

For example, given the following arrays A and B consisting of three elements each:

the function should return 0.5 because:

U(X) = −1*X + 3 if X ≤ 1 U(X) = 0*X + 2 if 1 ≤ X ≤ 2 U(X) = 1*X + 0 if 2 ≤ X

and:

D(X) = 1*X + 0 if X ≤ 1.5 D(X) = −1*X + 3 if 1.5 ≤ X

so for X = 1.5, function S(X) is equal to 0.5 and this is the minimum value of this function.

Assume that:

- N is an integer within the range [1..100,000];
- each element of arrays A, B is an integer within the range [−1,000..1,000].

Complexity:

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

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