Given a string s and a character c, find if all occurrences of c appear together in s or not. If the character c does not appear in the string at all, the answer is true.
Examples
Input: s = "1110000323", c = '1' Output: Yes All occurrences of '1' appear together in "1110000323" Input: s = "3231131", c = '1' Output: No All occurrences of 1 are not together Input: s = "abcabc", c = 'c' Output: No All occurrences of 'c' are not together Input: s = "ababcc", c = 'c' Output: Yes All occurrences of 'c' are together
The idea is to traverse given string, as soon as we find an occurrence of c, we keep traversing until we find a character which is not c. We also set a flag to indicate that one more occurrences of c are seen. If we see c again and flag is set, then we return false.
Implementation:
C++
// C++ program to find if all occurrences// of a character appear together in a string.#include <iostream>#include <string>using namespace std;bool checkIfAllTogether(string s, char c){ // To indicate if one or more occurrences // of 'c' are seen or not. bool oneSeen = false; // Traverse given string int i = 0, n = s.length(); while (i < n) { // If current character is same as c, // we first check if c is already seen. if (s[i] == c) { if (oneSeen == true) return false; // If this is very first appearance of c, // we traverse all consecutive occurrences. while (i < n && s[i] == c) i++; // To indicate that character is seen once. oneSeen = true; } else i++; } return true;}// Driver programint main(){ string s = "110029"; if (checkIfAllTogether(s, '1')) cout << "Yes" << endl; else cout << "No" << endl; return 0;} |
Java
// Java program to find if all // occurrences of a character // appear together in a string.import java.io.*;class GFG {static boolean checkIfAllTogether(String s, char c) { // To indicate if one or more // occurrences of 'c' are seen // or not. boolean oneSeen = false; // Traverse given string int i = 0, n = s.length(); while (i < n) { // If current character is // same as c, we first check // if c is already seen. if (s.charAt(i) == c) { if (oneSeen == true) return false; // If this is very first // appearance of c, we // traverse all consecutive // occurrences. while (i < n && s.charAt(i) == c) i++; // To indicate that character // is seen once. oneSeen = true; } else i++; } return true; } // Driver Code public static void main(String[] args) { String s = "110029"; if (checkIfAllTogether(s, '1')) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by Sam007. |
Python3
# Python program to find # if all occurrences# of a character appear# together in a string.# function to find # if all occurrences# of a character appear# together in a string.def checkIfAllTogether(s, c) : # To indicate if one or # more occurrences of # 'c' are seen or not. oneSeen = False # Traverse given string i = 0 n = len(s) while (i < n) : # If current character # is same as c, # we first check # if c is already seen. if (s[i] == c) : if (oneSeen == True) : return False # If this is very first # appearance of c, # we traverse all # consecutive occurrences. while (i < n and s[i] == c) : i = i + 1 # To indicate that character # is seen once. oneSeen = True else : i = i + 1 return True# Driver Codes = "110029";if (checkIfAllTogether(s, '1')) : print ("Yes\n")else : print ("No\n")# This code is contributed by # Manish Shaw (manishshaw1) |
C#
// C# program to find if all occurrences// of a character appear together in a// string.using System;public class GFG { static bool checkIfAllTogether(string s, char c) { // To indicate if one or more // occurrences of 'c' are seen // or not. bool oneSeen = false; // Traverse given string int i = 0, n = s.Length; while (i < n) { // If current character is // same as c, we first check // if c is already seen. if (s[i] == c) { if (oneSeen == true) return false; // If this is very first // appearance of c, we // traverse all consecutive // occurrences. while (i < n && s[i] == c) i++; // To indicate that character // is seen once. oneSeen = true; } else i++; } return true; } // Driver code public static void Main() { string s = "110029"; if (checkIfAllTogether(s, '1')) Console.Write( "Yes" ); else Console.Write( "No" ); }}// This code is contributed by Sam007. |
PHP
<?php// PHP program to find // if all occurrences// of a character appear// together in a string.// function to find // if all occurrences// of a character appear// together in a string.function checkIfAllTogether($s, $c){ // To indicate if one or // more occurrences of // 'c' are seen or not. $oneSeen = false; // Traverse given string $i = 0; $n = strlen($s); while ($i < $n) { // If current character // is same as c, // we first check // if c is already seen. if ($s[$i] == $c) { if ($oneSeen == true) return false; // If this is very first // appearance of c, // we traverse all // consecutive occurrences. while ($i < $n && $s[$i] == $c) $i++; // To indicate that character // is seen once. $oneSeen = true; } else $i++; } return true;}// Driver Code$s = "110029";if (checkIfAllTogether($s, '1')) echo("Yes\n");else echo("No\n");// This code is contributed by Ajit.?> |
Javascript
<script>// Javascript program to find if all // occurrences of a character appear// together in a string.function checkIfAllTogether(s, c){ // To indicate if one or more // occurrences of 'c' are seen // or not. let oneSeen = false; // Traverse given string let i = 0, n = s.length; while (i < n) { // If current character is // same as c, we first check // if c is already seen. if (s[i] == c) { if (oneSeen == true) return false; // If this is very first // appearance of c, we // traverse all consecutive // occurrences. while (i < n && s[i] == c) i++; // To indicate that character // is seen once. oneSeen = true; } else i++; } return true;}// Driver codelet s = "110029"; if (checkIfAllTogether(s, '1')) document.write("Yes");else document.write("No"); // This code is contributed by mukesh07</script> |
Output:
Yes
Complexity Analysis:
- Time Complexity: O(n), where n is the number of characters in the string.
- Auxiliary Space: O(1),
Please suggest if someone has a better solution which is more efficient in terms of space and time.
This article is contributed by Aarti_Rathi.
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