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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

Find a symmetry point of a string, if any.

Programming language:
Spoken language:

编写一个函数

int solution(char *S);

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

Copyright 2009–2020 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

编写一个函数

int solution(string &S);

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

Copyright 2009–2020 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

编写一个函数

class Solution { public int solution(string S); }

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

Copyright 2009–2020 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

编写一个函数

func Solution(S string) int

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

class Solution { public int solution(String S); }

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

class Solution { public int solution(String S); }

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

function solution(S);

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

fun solution(S: String): Int

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

function solution(S)

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

int solution(NSString *S);

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

function solution(S: PChar): longint;

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

function solution($S);

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

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

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

def solution(S)

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

def solution(s)

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

object Solution { def solution(s: String): Int }

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

public func solution(_ S : inout String) -> Int

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

编写一个函数

Private Function solution(S As String) As Integer

从给出的字符串 S 中，找到并返回这样一个字符的下标（下标从 0 开始算）， 使得这个字符左边的子字符串，刚好与右边的子字符串相反 （但如果这样的字符不存在的话，返回 −1）。

例如，给出这样一个字符串

"`racecar`"

你的函数应该返回 3，因为对于下标为 3 的字符 e， 其左边相邻的子字符串是 "`rac`"， 而右边相邻的子字符串是 "`car`"。

注： 与空字符串（长度为 0 的字符串）相反的还是一个空字符串。

假定:

- S 长度范围 [0..2,000,000].

Write a function:

int solution(char *S);

that, given a string S, returns the index (counting from 0) of a character such that the part of the string to the left of that character is a reversal of the part of the string to its right. The function should return −1 if no such index exists.

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

the function should return 3, because the substring to the left of the character "`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

int solution(string &S);

that, given a string S, returns the index (counting from 0) of a character such that the part of the string to the left of that character is a reversal of the part of the string to its right. The function should return −1 if no such index exists.

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

the function should return 3, because the substring to the left of the character "`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

class Solution { public int solution(string S); }

that, given a string S, returns the index (counting from 0) of a character such that the part of the string to the left of that character is a reversal of the part of the string to its right. The function should return −1 if no such index exists.

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

the function should return 3, because the substring to the left of the character "`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

func Solution(S string) int

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

class Solution { public int solution(String S); }

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

class Solution { public int solution(String S); }

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

function solution(S);

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

fun solution(S: String): Int

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

function solution(S)

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

int solution(NSString *S);

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

function solution(S: PChar): longint;

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

function solution($S);

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

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

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

def solution(S)

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

def solution(s)

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

object Solution { def solution(s: String): Int }

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

public func solution(_ S : inout String) -> Int

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

Write a function:

Private Function solution(S As String) As Integer

*Note:* reversing an empty string (i.e. a string whose length is zero) gives an empty string.

For example, given a string:

"`racecar`"

`e`" at index 3 is "`rac`", and the one to the right is "`car`".

Given a string:

"`x`"

the function should return 0, because both substrings are empty.

Write an ** efficient** algorithm for the following assumptions:

- the length of S is within the range [0..2,000,000].

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