Given a sequence of characters, print consecutive sequence of characters in a line, otherwise
print it in a new line.
Examples:
Input : ABCXYZACCD
Output : ABC
XYZ
A
C
CD
Input : ABCZYXACCD
Output: ABC
ZYX
A
C
CD
The idea is to traverse string from left to right. For every traversed character, print it in a line if it is consecutive to previous one, else print a new line character.
C++
// C++ program to print consecutive characters // together in a line. #include <iostream> using namespace std; void print(string str) { cout << str[0]; for (int i=1; str[i]!='\0'; i++) { if ((str[i] == str[i-1]+1) || (str[i] == str[i-1]-1)) cout << str[i]; else cout << "\n" << str[i];; } } // Driver code int main() { string str = "ABCXYZACCD"; print(str); return 0; } |
Java
// Java program to print consecutive characters // together in a line. class GFG { static void print(String str1) { char str[] = str1.toCharArray(); System.out.print(str[0]); for (int i = 1; i < str.length; i++) { if ((str[i] == str[i - 1] + 1) || (str[i] == str[i - 1] - 1)) { System.out.print(str[i]); } else { System.out.print("\n" + str[i]); } } } // Driver code public static void main(String[] args) { String str = "ABCXYZACCD"; print(str); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program to print consecutive characters # together in a line. def _print(string): print(string[0], end = "") for i in range(1, len(string)): if (ord(string[i]) == ord(string[i - 1]) + 1 or ord(string[i]) == ord(string[i - 1]) - 1): print(string[i], end = "") else: print() print(string[i], end = "") # Driver Code if __name__ == "__main__": string = "ABCXYZACCD" _print(string) # This code is contributed by # sanjeev2552 |
C#
// C# program to print consecutive characters // together in a line. using System; class GFG { static void print(String str1) { char []str = str1.ToCharArray(); Console.Write(str[0]); for (int i = 1; i < str.Length; i++) { if ((str[i] == str[i - 1] + 1) || (str[i] == str[i - 1] - 1)) { Console.Write(str[i]); } else { Console.Write("\n" + str[i]); } } } // Driver code public static void Main() { String str = "ABCXYZACCD"; print(str); } } // This code is contributed by PrinciRaj1992 |
Output:
ABC XYZ A C CD
Time Complexity : O(n)
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