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Check if the string contains consecutive letters and each letter occurs exactly once
  • Difficulty Level : Medium
  • Last Updated : 10 May, 2021

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: Yes

Input: 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>
Output: 
Yes
No

 

Time complexity: O(N logN)

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 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 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 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 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 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>
Output: 
Yes
No

 

Time Complexity: O(N) 
 

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