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Lesson 90

Tasks from Indeed Prime 2015 challenge

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Tasks from Indeed Prime 2016 College Coders challenge

Lesson 99

Future training

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Find the minimal average of any slice containing at least two elements.

Programming language:
Spoken language:

A non-empty array A consisting of N integers is given. A pair of integers (P, Q), such that 0 ≤ P < Q < N, is called a *slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

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

that, given a non-empty array A consisting of N integers, returns the starting position of the slice with the minimal average. If there is more than one slice with a minimal average, you should return the smallest starting position of such a slice.

For example, given array A such that:

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 [−10,000..10,000].

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. A pair of integers (P, Q), such that 0 ≤ P < Q < N, is called a *slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

int solution(vector<int> &A);

that, given a non-empty array A consisting of N integers, returns the starting position of the slice with the minimal average. If there is more than one slice with a minimal average, you should return the smallest starting position of such a slice.

For example, given array A such that:

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 [−10,000..10,000].

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. A pair of integers (P, Q), such that 0 ≤ P < Q < N, is called a *slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

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

that, given a non-empty array A consisting of N integers, returns the starting position of the slice with the minimal average. If there is more than one slice with a minimal average, you should return the smallest starting position of such a slice.

For example, given array A such that:

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 [−10,000..10,000].

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.

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

func Solution(A []int) int

For example, given array A such that:

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 [−10,000..10,000].

Complexity:

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

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

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

For example, given array A such that:

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 [−10,000..10,000].

Complexity:

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

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

function solution(A);

For example, given array A such that:

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 [−10,000..10,000].

Complexity:

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

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

function solution(A)

For example, given array A such that:

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 [−10,000..10,000].

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.

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

int solution(NSMutableArray *A);

For example, given array A such that:

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 [−10,000..10,000].

Complexity:

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

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

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

For example, given array A such that:

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 [−10,000..10,000].

Complexity:

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

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

function solution($A);

For example, given array A such that:

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 [−10,000..10,000].

Complexity:

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

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

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

For example, given array A such that:

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 [−10,000..10,000].

Complexity:

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

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

def solution(A)

For example, given array A such that:

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 [−10,000..10,000].

Complexity:

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

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

def solution(a)

For example, given array A such that:

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 [−10,000..10,000].

Complexity:

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

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

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

For example, given array A such that:

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 [−10,000..10,000].

Complexity:

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

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

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

For example, given array A such that:

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 [−10,000..10,000].

Complexity:

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

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

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

For example, given array A such that:

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 [−10,000..10,000].

Complexity:

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

*slice* of array A (notice that the slice contains at least two elements). The *average* of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

contains the following example slices:

- slice (1, 2), whose average is (2 + 2) / 2 = 2;
- slice (3, 4), whose average is (5 + 1) / 2 = 3;
- slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

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

For example, given array A such that:

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 [−10,000..10,000].

Complexity:

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

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