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Programming language:
Spoken language:

Two non-empty zero-indexed 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), beyond input storage (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 zero-indexed 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), beyond input storage (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 zero-indexed 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), beyond input storage (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));

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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