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# Find time taken for signal to reach all positions in a string

• Difficulty Level : Medium
• Last Updated : 06 Dec, 2022

Given a string of ‘x’ and ‘o’. From each ‘x’ a signal originates which travels in both direction. It takes one unit of time for the signal to travel to the next cell. If a cell contains ‘o’, the signal changes it to ‘x’ . Task is to compute the time taken to convert the array into a string of signals.

A string of signals contains only ‘x’ in it.

Examples:

Input : s = “oooxooooxooo”
Output : 3

Input : s = “oooxoooo”
Output :

Source: OYO Rooms Interview Set 2

The idea is to find the length of the longest contiguous ‘o’. Also check, if ‘x’ is present on the right side and left side of contiguous ‘o’ then the time period will be reduced to half the maximum length else the time period will be equal to the maximum length.

Below is the implementation of the above approach.

## C++

 `// C++ program to Find time taken for signal``// to reach all positions in a string``#include ``using` `namespace` `std;` `// Returns time needed for signal to traverse``// through complete string.``int` `maxLength(string s, ``int` `n)``{``    ``int` `right = 0, left = 0;``    ``int` `coun = 0, max_length = INT_MIN;` `    ``s = s + ``'1'``; ``// for the calculation of last index``    ``// for strings like oxoooo, xoxxoooo ..` `    ``for` `(``int` `i = 0; i <= n; i++) {``        ``if` `(s[i] == ``'o'``)``            ``coun++;` `        ``else` `{` `            ``// if coun is greater than max_length``            ``if` `(coun > max_length) {` `                ``right = 0;``                ``left = 0;` `                ``// if 'x' is present at the right side``                ``// of max_length``                ``if` `(s[i] == ``'x'``)``                    ``right = 1;` `                ``// if 'x' is present at left side of``                ``// max_length``                ``if` `(((i - coun) > 0) && (s[i - coun - 1] == ``'x'``))``                    ``left = 1;` `                ``// We use ceiling function to handle odd``                ``// number 'o's``                ``coun = ``ceil``((``double``)coun / (right + left));` `                ``max_length = max(max_length, coun);``            ``}``            ``coun = 0;``        ``}``    ``}` `    ``return` `max_length;``}` `// Driver code``int` `main()``{``    ``string s = ``"oooxoooooooooxooo"``;``    ``int` `n = s.size();``    ``cout << maxLength(s, n);``    ``return` `0;``}`

## Java

 `// Java program to Find time taken for signal``// to reach all positions in a string``public` `class` `GFG {` `    ``// Returns time needed for signal to traverse``    ``// through complete string.``    ``static` `int` `maxLength(String s, ``int` `n)``    ``{``        ``int` `right = ``0``, left = ``0``;``        ``int` `coun = ``0``, max_length = Integer.MIN_VALUE;` `        ``s = s + ``'1'``; ``// for the calculation of last index``        ``// for strings like oxoooo, xoxxoooo ..` `        ``for` `(``int` `i = ``0``; i <= n; i++) {``            ``if` `(s.charAt(i) == ``'o'``)``                ``coun++;` `            ``else` `{` `                ``// if coun is greater than max_length``                ``if` `(coun > max_length) {` `                    ``right = ``0``;``                    ``left = ``0``;` `                    ``// if 'x' is present at the right side``                    ``// of max_length``                    ``if` `(s.charAt(i) == ``'x'``)``                        ``right = ``1``;` `                    ``// if 'x' is present at left side of``                    ``// max_length``                    ``if` `(((i - coun) > ``0``) && (s.charAt(i - coun - ``1``) == ``'x'``))``                        ``left = ``1``;` `                    ``// We use ceiling function to handle odd``                    ``// number 'o's``                    ``coun = (``int``)Math.ceil((``double``)coun / (right + left));` `                    ``max_length = Math.max(max_length, coun);``                ``}``                ``coun = ``0``;``            ``}``        ``}` `        ``return` `max_length;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``String s = ``"oooxoooooooooxooo"``;``        ``int` `n = s.length();``        ``System.out.println(maxLength(s, n));``    ``}``    ``// This code is contributed by ANKITRAI1``}`

## Python3

 `# Python3 program to Find time taken for signal``# to reach all positions in a string``import` `sys``import` `math` `# Returns time needed for signal``# to traverse through complete string.``def` `maxLength(s, n):``    ``right ``=` `0``    ``left ``=` `0``    ``coun ``=` `0``    ``max_length ``=` `-``(sys.maxsize``-``1``)``    ` `    ``# for the calculation of last index``    ``s ``=` `s``+``'1'``    ` `    ``# for strings like oxoooo, xoxxoooo ``    ``for` `i ``in` `range``(``0``, n ``+` `1``):``        ``if` `s[i]``=``=``'o'``:``            ``coun``+``=` `1``        ``else``:``            ` `            ``# if coun is greater than``            ``# max_length``            ``if` `coun>max_length:``                ``right ``=` `0``                ``left ``=` `0``                ` `                ``# if 'x' is present at the right side``                ``# of max_length``                ``if` `s[i]``=``=``'x'``:``                    ``right ``=` `1``                    ` `                ``# if 'x' is present at left side of``                ``# max_length``                ``if` `i``-``coun>``0` `and` `s[i``-``coun``-``1``] ``=``=` `'x'``:``                    ``left ``=` `1``                    ` `                ``# We use ceiling function to handle odd``                ``# number 'o's``                ``coun ``=` `math.ceil(``float``(coun``/``(right ``+` `left)))``                ``max_length ``=` `max``(max_length, coun)``            ``coun ``=` `0` `    ``return` `max_length` `# Driver code``if` `__name__``=``=``'__main__'``:``    ``s ``=` `"oooxoooooooooxooo"``    ``n ``=` `len``(s)``    ``print``(maxLength(s, n))` `# This code is contributed by``# Shrikant13`

## C#

 `// C# program to Find time taken``// for signal to reach all``// positions in a string``using` `System;` `class` `GFG {` `    ``// Returns time needed for signal``    ``// to traverse through complete string.``    ``static` `int` `maxLength(``string` `s, ``int` `n)``    ``{``        ``int` `right = 0, left = 0;``        ``int` `coun = 0, max_length = ``int``.MinValue;` `        ``s = s + ``'1'``; ``// for the calculation of last index` `        ``// for strings like oxoooo, xoxxoooo ..``        ``for` `(``int` `i = 0; i <= n; i++) {``            ``if` `(s[i] == ``'o'``)``                ``coun++;` `            ``else` `{` `                ``// if coun is greater than max_length``                ``if` `(coun > max_length) {``                    ``right = 0;``                    ``left = 0;` `                    ``// if 'x' is present at the right``                    ``// side of max_length``                    ``if` `(s[i] == ``'x'``)``                        ``right = 1;` `                    ``// if 'x' is present at left side``                    ``// of max_length``                    ``if` `(((i - coun) > 0) && (s[i - coun - 1] == ``'x'``))``                        ``left = 1;` `                    ``// We use ceiling function to``                    ``// handle odd number 'o's``                    ``coun = (``int``)Math.Ceiling((``double``)coun / (right + left));` `                    ``max_length = Math.Max(max_length, coun);``                ``}``                ``coun = 0;``            ``}``        ``}` `        ``return` `max_length;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``string` `s = ``"oooxoooooooooxooo"``;``        ``int` `n = s.Length;``        ``Console.Write(maxLength(s, n));``    ``}``}` `// This code is contributed``// by ChitraNayal`

## PHP

 ` ``\$max_length``)``            ``{` `                ``\$right` `= 0;``                ``\$left` `= 0;` `                ``// if 'x' is present at the right side``                ``// of max_length``                ``if` `(``\$s``[``\$i``] == ``'x'``)``                    ``\$right` `= 1;` `                ``// if 'x' is present at left side of``                ``// max_length``                ``if` `(((``\$i` `- ``\$coun``) > 0) &&``                    ``(``\$s``[``\$i` `- ``\$coun` `- 1] == ``'x'``))``                    ``\$left` `= 1;` `                ``// We use ceiling function to handle odd``                ``// number 'o's``                ``\$coun` `= (int)``ceil``((double)``\$coun` `/ (``\$right` `+``                                                   ``\$left``));` `                ``\$max_length` `= max(``\$max_length``, ``\$coun``);``            ``}``            ``\$coun` `= 0;``        ``}``    ``}` `    ``return` `\$max_length``;``}` `// Driver code``\$s` `= ``"oooxoooooooooxooo"``;``\$n` `= ``strlen``(``\$s``);``echo``(maxLength(``\$s``, ``\$n``));` `// This code is contributed by Code_Mech`

## Javascript

 ``

Output

`5`

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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