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Prime Numbers between 1 and 100
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A number is called prime number if the number is only divisible by 1 and itself. The prime numbers has exactly two factors- 1 and the number itself.
For example, if the number is 5. 5 is only divisible by 1 and 5. The number of factors of 5 is 2. So, 5 is a prime number.
If the number is 4, it is divisible by 1, 2 and 4. So, instead of 1 and 4, it has another factor 2. The number of factors of 4 is 3. So, it is not a prime number.
So, the main logic of the program is to take the number as input and the number of factors are counted. If the number of factor is 2 then it is a prime number otherwise it is not a prime number.
The program is written here to find the prime numbers between 1 and 100. For each number it is checked for prime, if it is prime then it will be printed.
Algorithm
OUTPUT: Prime numbers between 1 and 100.
PROCESS:
Step 1: [printing the prime numbers between 1 and 100]
For j=1 to 100 repeat
Set c<-0
[counting the number of factors]
For i=1 to j repeat
If j mod i=0 then
Set c<-c+1
[End of ‘if’]
[End of ‘for’ loop]
If c=2 then
Print j
[End of ‘if’]
Step 2: Stop.
Code
TIME COMPLEXITY:
for(j=1;j<=100;j++)----------------------------------O(100)
{
c=0;
//counting the number of factors
for(i=1;i<=j;i++)---------------------------O(j)
{
if(j%i==0)------------------------O(1)
c++;
}
if(c==2)-------------------------------------O(1)
printf("%d ",j);
}
The time complexity of finding prime numbers here is O(n*j) where n is the number of elements to be checked for prime and ‘j’ is each number to be checked. Here, the value of n is 100. For the simplicity of the program here the loop to check prime is executed from 1 to j.
But for more optimization we can check from 2 to because a number ‘j’ can have maximum number of factors. So, we can say most optimized complexity is O().
SPACE COMPLEXITY:
The space complexity of this program is O(1) as it requires a constant number of memory spaces to find the prime numbers between 1 and 100.
