Check if the string contains consecutive letters and each letter occurs exactly once
Given string str. The task is to check if the string contains consecutive letters and each letter occurs exactly once.
Examples:
Input: str = “fced”
Output: Yes
The string contains ‘c’, ‘d’, ‘e’ and ‘f’ which are consecutive letters.Input: str = “xyz”
Output: YesInput: str = “abd”
Output: No
Approach:
The following steps can be followed to solve the problem:
- Sort the given string in ascending order.
- Check if s[i]-s[i-1]==1, for every index i from 1 to n-1.
- If the condition holds for every index, print “Yes”, else print “No”.
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to check if // the condition holds bool check(string s) { // Get the length of the string int l = s.length(); // sort the given string sort(s.begin(), s.end()); // Iterate for every index and // check for the condition for ( int i = 1; i < l; i++) { // If are not consecutive if (s[i] - s[i - 1] != 1) return false ; } return true ; } // Driver code int main() { // 1st example string str = "dcef" ; if (check(str)) cout << "Yes\n" ; else cout << "No\n" ; // 2nd example str = "xyza" ; if (check(str)) cout << "Yes\n" ; else cout << "No\n" ; return 0; } |
Java
// Java program to implement // the above approach import java.util.*; class GfG { // Function to check if // the condition holds static boolean check( char s[]) { // Get the length of the string int l = s.length; // sort the given string Arrays.sort(s); // Iterate for every index and // check for the condition for ( int i = 1 ; i < l; i++) { // If are not consecutive if (s[i] - s[i - 1 ] != 1 ) return false ; } return true ; } // Driver code public static void main(String[] args) { // 1st example String str = "dcef" ; if (check(str.toCharArray()) == true ) System.out.println( "Yes" ); else System.out.println( "No" ); // 2nd example String str1 = "xyza" ; if (check(str1.toCharArray()) == true ) System.out.println( "Yes" ); else System.out.println( "No" ); } } |
Python3
# Python3 program to implement # the above approach # Function to check if # the condition holds def check(s): # Get the length of the string l = len (s) # sort the given string s = ''.join( sorted (s)) # Iterate for every index and # check for the condition for i in range ( 1 , l): # If are not consecutive if ord (s[i]) - ord (s[i - 1 ]) ! = 1 : return False return True # Driver code if __name__ = = "__main__" : # 1st example string = "dcef" if check(string): print ( "Yes" ) else : print ( "No" ) # 2nd example string = "xyza" if check(string): print ( "Yes" ) else : print ( "No" ) # This code is contributed by Rituraj Jain |
C#
// C# program to implement // the above approach using System; using System.Collections; class GfG { // Function to check if // the condition holds static bool check( char [] s) { // Get the length of the string int l = s.Length; // sort the given string Array.Sort(s); // Iterate for every index and // check for the condition for ( int i = 1; i < l; i++) { // If are not consecutive if (s[i] - s[i - 1] != 1) return false ; } return true ; } // Driver code public static void Main() { // 1st example string str = "dcef" ; if (check(str.ToCharArray()) == true ) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); // 2nd example String str1 = "xyza" ; if (check(str1.ToCharArray()) == true ) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed by Ryuga |
Javascript
<script> // Javascript program to implement // the above approach // Function to check if // the condition holds function check(s) { // Get the length of the string let l = s.length; // sort the given string s.sort(); // Iterate for every index and // check for the condition for (let i = 1; i < l; i++) { // If are not consecutive if ((s[i].charCodeAt() - s[i - 1].charCodeAt()) != 1) return false ; } return true ; } // 1st example let str = "dcef" ; if (check(str.split( '' )) == true ) document.write( "Yes" + "</br>" ); else document.write( "No" + "</br>" ); // 2nd example let str1 = "xyza" ; if (check(str1.split( '' )) == true ) document.write( "Yes" ); else document.write( "No" ); // This code is contributed by mukesh07. </script> |
Yes No
Time complexity: O(N logN)
Auxiliary Space: O(1)
Efficient approach:
- Find max and min ASCII values of the string’s characters
- Find the sum of the ASCII values of all the characters from the string
- So if a sequence of characters are a(ASCII = 96) to d(ASCII = 99) then, the result expected sum should be (sum from 0 to 99) minus (sum of 0 to 95)
- Mathematical Equation:
MAX_VALUE*(MAX_VALUE+1)/2 - (MIN_VALUE-1)*((MIN_VALUE-1)+1)/2
- Check whether the calculated sum and the expected sum is equal or not
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include<bits/stdc++.h> using namespace std; bool check(string str) { int min = INT_MAX; int max = -INT_MAX; int sum = 0; // For all the characters of the string for ( int i = 0; i < str.size(); i++) { // Find the ascii value of the character int ascii = str[i]; // Check if its a valid character, // if not then return false if (ascii < 96 || ascii > 122) return false ; // Calculate sum of all the // characters ascii values sum += ascii; // Find minimum ascii value // from the string if (min > ascii) min = ascii; // Find maximum ascii value // from the string if (max < ascii) max = ascii; } // To get the previous element // of the minimum ASCII value min -= 1; // Take the expected sum // from the above equation int eSum = ((max * (max + 1)) / 2) - ((min * (min + 1)) / 2); // Check if the expected sum is // equals to the calculated sum or not return sum == eSum; } // Driver code int main() { // 1st example string str = "dcef" ; if (check(str)) cout << ( "Yes" ); else cout << ( "No" ); // 2nd example string str1 = "xyza" ; if (check(str1)) cout << ( "\nYes" ); else cout << ( "\nNo" ); } // This code is contributed by amreshkumar3 |
Java
// Java program to implement // the above approach public class GFG { public static boolean check(String str) { int min = Integer.MAX_VALUE; int max = Integer.MIN_VALUE; int sum = 0 ; // for all the characters of the string for ( int i = 0 ; i < str.length(); i++) { // find the ascii value of the character int ascii = ( int )str.charAt(i); // check if its a valid character, // if not then return false if (ascii < 96 || ascii > 122 ) return false ; // calculate sum of all the // characters ascii values sum += ascii; // find minimum ascii value // from the string if (min > ascii) min = ascii; // find maximum ascii value // from the string if (max < ascii) max = ascii; } // To get the previous element // of the minimum ASCII value min -= 1 ; // take the expected sum // from the above equation int eSum = ((max * (max + 1 )) / 2 ) - ((min * (min + 1 )) / 2 ); // check if the expected sum is // equals to the calculated sum or not return sum == eSum; } // Driver code public static void main(String[] args) { // 1st example String str = "dcef" ; if (check(str)) System.out.println( "Yes" ); else System.out.println( "No" ); // 2nd example String str1 = "xyza" ; if (check(str1)) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by Arijit Basu(ArijitXfx) |
Python3
# Python3 program to implement # the above approach import sys def check( str ): min = sys.maxsize max = - sys.maxsize - 1 sum = 0 # For all the characters of the string for i in range ( len ( str )): # Find the ascii value of the character ascii = str [i] # Check if its a valid character, # if not then return false if ( ord (ascii) < 96 or ord (ascii) > 122 ): return False # Calculate sum of all the # characters ascii values sum + = ord (ascii) # Find minimum ascii value # from the string if ( min > ord (ascii)): min = ord (ascii) # Find maximum ascii value # from the string if ( max < ord (ascii)): max = ord (ascii) # To get the previous element # of the minimum ASCII value min - = 1 # Take the expected sum # from the above equation eSum = ((( max * ( max + 1 )) / / 2 ) - (( min * ( min + 1 )) / / 2 )) # Check if the expected sum is # equals to the calculated sum or not return sum = = eSum # Driver code if __name__ = = '__main__' : # 1st example str = "dcef" if (check( str )): print ( "Yes" ) else : print ( "No" ) # 2nd example str1 = "xyza" if (check(str1)): print ( "Yes" ) else : print ( "No" ) # This code is contributed by mohit kumar 29 |
C#
// C# program to implement // the above approach using System; class GFG { static bool check( string str) { int min = Int32.MaxValue; int max = Int32.MinValue; int sum = 0; // for all the characters of the string for ( int i = 0; i < str.Length; i++) { // find the ascii value of the character int ascii = ( int )str[i]; // check if its a valid character, // if not then return false if (ascii < 96 || ascii > 122) return false ; // calculate sum of all the // characters ascii values sum += ascii; // find minimum ascii value // from the string if (min > ascii) min = ascii; // find maximum ascii value // from the string if (max < ascii) max = ascii; } // To get the previous element // of the minimum ASCII value min -= 1; // take the expected sum // from the above equation int eSum = ((max * (max + 1)) / 2) - ((min * (min + 1)) / 2); // check if the expected sum is // equals to the calculated sum or not return sum == eSum; } // Driver code static void Main() { // 1st example string str = "dcef" ; if (check(str)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); // 2nd example string str1 = "xyza" ; if (check(str1)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed by divyesh072019. |
Javascript
<script> // javascript program to implement // the above approach function check( str) { var min = Number.MAX_VALUE; var max = Number.MIN_VALUE; var sum = 0; // for all the characters of the string for (i = 0; i < str.length; i++) { // find the ascii value of the character var ascii = parseInt( str.charCodeAt(i)); // check if its a valid character, // if not then return false if (ascii < 96 || ascii > 122) return false ; // calculate sum of all the // characters ascii values sum += ascii; // find minimum ascii value // from the string if (min > ascii) min = ascii; // find maximum ascii value // from the string if (max < ascii) max = ascii; } // To get the previous element // of the minimum ASCII value min -= 1; // take the expected sum // from the above equation var eSum = parseInt((max * (max + 1)) / 2) - ((min * (min + 1)) / 2); // check if the expected sum is // equals to the calculated sum or not return sum == eSum; } // Driver code // 1st example var str = "dcef" ; if (check(str)) document.write( "Yes<br/>" ); else document.write( "No<br/>" ); // 2nd example var str1 = "xyza" ; if (check(str1)) document.write( "Yes" ); else document.write( "No" ); // This code contributed by Rajput-Ji </script> |
Yes No
Time Complexity: O(N)
Auxiliary Space: O(1)
Approach (Using set):
In this approach, we use an unordered set to keep track of the letters we’ve seen so far. We iterate through the string s, and for each character c in the string, we check if it has already been seen in the set. If it has, this means that the string does not contain each letter exactly once, so we return false. If it has not been seen, we insert the letter into the set.
After we’ve finished iterating through the string and checking for duplicate letters, we use the min_element and max_element functions from the algorithm library to get the minimum and maximum characters in the string, respectively. We then check if the difference between the maximum and minimum characters plus one is equal to the length of the string. If it is, this means that the string contains consecutive letters, so we return true. Otherwise, we return false.
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to check if // the condition holds bool check(string s) { // use unordered set to keep track of seen letters unordered_set< char > seen; for ( char c : s) { // check if letter has already been seen if (seen.count(c)) { return false ; } seen.insert(c); // insert letter into set } // get minimum character in string char min_char = *min_element(s.begin(), s.end()); // get maximum character in string char max_char = *max_element(s.begin(), s.end()); // check if consecutive letters return (max_char - min_char + 1 == s.length()); } // Driver code int main() { // 1st example string str = "dcef" ; if (check(str)) cout << "Yes\n" ; else cout << "No\n" ; // 2nd example str = "xyza" ; if (check(str)) cout << "Yes\n" ; else cout << "No\n" ; return 0; } |
Java
import java.util.HashSet; public class Main { // Function to check if the condition holds static boolean check(String s) { // Use HashSet to keep track of seen letters HashSet<Character> seen = new HashSet<>(); for ( char c : s.toCharArray()) { // Check if letter has already been seen if (seen.contains(c)) { return false ; } seen.add(c); // Insert letter into set } // Get minimum character in string char minChar = s.charAt( 0 ); for ( char c : s.toCharArray()) { minChar = ( char ) Math.min(minChar, c); } // Get maximum character in string char maxChar = s.charAt( 0 ); for ( char c : s.toCharArray()) { maxChar = ( char ) Math.max(maxChar, c); } // Check if consecutive letters return (maxChar - minChar + 1 == s.length()); } // Driver code public static void main(String[] args) { // 1st example String str = "dcef" ; if (check(str)) System.out.println( "Yes" ); else System.out.println( "No" ); // 2nd example str = "xyza" ; if (check(str)) System.out.println( "Yes" ); else System.out.println( "No" ); // This code is contributed by Shivam Tiwari } } |
Python3
# Function to check if # the condition holds def check(s): # use set to keep track of seen letters seen = set () for c in s: # check if letter has already been seen if c in seen: return False seen.add(c) # insert letter into set # get minimum character in string min_char = min (s) # get maximum character in string max_char = max (s) # check if consecutive letters return ( ord (max_char) - ord (min_char) + 1 = = len (s)) # Driver code if __name__ = = '__main__' : # 1st example str = "dcef" if (check( str )): print ( "Yes" ) else : print ( "No" ) # 2nd example str = "xyza" if (check( str )): print ( "Yes" ) else : print ( "No" ) |
C#
using System; using System.Collections.Generic; using System.Linq; public class GFG { // Function to check if the condition holds public static bool Check( string s) { // Use HashSet<char> to keep track of seen letters HashSet< char > seen = new HashSet< char >(); foreach ( char c in s) { // Check if letter has already been seen if (seen.Contains(c)) { return false ; } seen.Add(c); // Insert letter into set } // Get minimum character in string char min_char = s.Min(); // Get maximum character in string char max_char = s.Max(); // Check if consecutive letters return (max_char - min_char + 1 == s.Length); } // Driver code public static void Main() { // 1st example string str = "dcef" ; if (Check(str)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); // 2nd example str = "xyza" ; if (Check(str)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); // This code is contributed by Shivam Tiwari } } |
Javascript
// Function to check if the condition holds function check(s) { // Use Set to keep track of seen letters let seen = new Set(); for (let c of s) { // Check if letter has already been seen if (seen.has(c)) { return false ; } seen.add(c); // Insert letter into set } // Get minimum character in string let minChar = s.charAt(0); for (let c of s) { minChar = Math.min(minChar, c); } // Get maximum character in string let maxChar = s.charAt(0); for (let c of s) { maxChar = Math.max(maxChar, c); } // Check if consecutive letters return (maxChar.charCodeAt(0) - minChar.charCodeAt(0) + 1 === s.length); } // Driver code // 1st example let str = "dcef" ; if (check(str)) console.log( "Yes" ); else console.log( "No" ); // 2nd example str = "xyza" ; if (check(str)) console.log( "Yes" ); else console.log( "No" ); // This code is contributed by Shivam Tiwari |
Yes No
Time Complexity: O(n), where n is the length of the input string.
Space Complexity: O(k), where k is the number of distinct characters in the string.
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