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 : 3Input : s = “oooxoooo”
Output : 4
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 <bits/stdc++.h>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 codeint 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 stringpublic 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 stringimport sysimport 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 codeif __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 stringusing 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
<?php// PHP program to Find time taken for signal // to reach all positions in a string// Returns time needed for signal // to traverse through complete string.function maxLength($s, $n){ $right = 0; $left = 0; $coun = 0; $max_length = PHP_INT_MIN; $s = $s . '1'; // for the calculation of last index // for strings like oxoooo, xoxxoooo .. for ($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 $coun = (int)ceil((double)$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
<script>// Javascript program to Find time taken for signal// to reach all positions in a string// Returns time needed for signal to traverse// through complete string.function maxLength( s, n){ var right = 0, left = 0; var coun = 0, max_length = Number.MIN_VALUE; s = s + '1'; // for the calculation of last index // for strings like oxoooo, xoxxoooo .. for (var 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 = Math.ceil(coun / (right + left)); max_length = Math.max(max_length, coun); } coun = 0; } } return max_length;}var s = "oooxoooooooooxooo";var n = s.length;document.write( maxLength(s, n)); // This code is contributed by SoumikMondal</script> |
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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