Answer:
- letter_counts = {}
- string1 = "I have a dream"
-
- string1 = string1.lower()
-
- for x in string1:
- if(x == " "):
- continue
- if x not in letter_counts:
- letter_counts[x] = 1
- else:
- letter_counts[x] += 1
-
- print(letter_counts)
Explanation:
The solution is written in Python 3.
Firstly create a letter_count dictionary (Line 1)
Next, create a sample string and assign it to string1 variable and convert all the characters to lowercase using lower method (Line 4).
Create a for loop to traverse through each character in string1 and check if the current character is a single space, just skip to the next iteration (Line 7 -8). If the current character is not found in letter_counts dictionary, set the initial count value 1 to x property of letter_counts. Otherwise increment the x property value by one (Line 9 -12).
After completion of loop, print the letter_count dictionary. We shall get the sample output {'i': 1, 'h': 1, 'a': 3, 'v': 1, 'e': 2, 'd': 1, 'r': 1, 'm': 1}
Collaborative software or groupware.
//=indicating you to do the programming part on your own relating to the description provided against. This done because different programming languages require different coding for that.
n=integer value
n1=dummy storage for n
r=variable used to do the function
{
int n,r,n1,rev=0;
//do the coding here for storing the integer in the variable n
n1=n;
while(n>0){
r=n%10;
rev=(rev*10)+r;
n=n/10;
}
//now add a command for displaying the value of rev
}
this is just a logic i used for java
done.
The correct answer for the question that is being presented above is this one: "monopolistic." Suppose barriers to entry exist in the telecommunications industry. This best describes a monopolistic market. In a <span>monopolistic market, that specific source of service or good, is being handled by a single company.</span>
