Find All Prime Numbers in Array
Medium
C++
Arrays
42 views
Problem Statement
Identify and display all prime numbers present in given array. This combines prime checking logic with array traversal.
Logic: For each number, check if it's prime using division method
Logic: For each number, check if it's prime using division method
Input Format
An integer n (size of array)
Then n integers.
Then n integers.
Output Format
All prime numbers present in the array (space-separated).
If no prime exists, print nothing (or empty output).
If no prime exists, print nothing (or empty output).
Example
6 2 4 5 6 7 9
2 5 7
Constraints
• n ≥ 1
• Array elements ≥ 1
• Array elements ≥ 1
Concept Explanation
A prime number is a number greater than 1 that has only two divisors: 1 and itself.
Each array element is checked separately.
Each array element is checked separately.
Step-by-Step Explanation
Prime Check Logic (for one number)
1.If number ≤ 1 → not prime.
2.Check divisibility from 2 to √number.
3.If divisible by any value → not prime.
4.If no divisor found → prime.
Array Traversal Logic
1.Traverse the array from index 0 to n-1.
2.For each element, apply the prime check logic.
3.If the number is prime:
• Print it.
4.Continue until the end of the array.
1.If number ≤ 1 → not prime.
2.Check divisibility from 2 to √number.
3.If divisible by any value → not prime.
4.If no divisor found → prime.
Array Traversal Logic
1.Traverse the array from index 0 to n-1.
2.For each element, apply the prime check logic.
3.If the number is prime:
• Print it.
4.Continue until the end of the array.
Concept Explanation
A prime number is a number greater than 1 that has only two divisors: 1 and itself.
Each array element is checked separately.
Each array element is checked separately.
Step-by-Step Explanation
Prime Check Logic (for one number)
1.If number ≤ 1 → not prime.
2.Check divisibility from 2 to √number.
3.If divisible by any value → not prime.
4.If no divisor found → prime.
Array Traversal Logic
1.Traverse the array from index 0 to n-1.
2.For each element, apply the prime check logic.
3.If the number is prime:
• Print it.
4.Continue until the end of the array.
1.If number ≤ 1 → not prime.
2.Check divisibility from 2 to √number.
3.If divisible by any value → not prime.
4.If no divisor found → prime.
Array Traversal Logic
1.Traverse the array from index 0 to n-1.
2.For each element, apply the prime check logic.
3.If the number is prime:
• Print it.
4.Continue until the end of the array.
Input / Output Format
Input Format
An integer n (size of array)
Then n integers.
Then n integers.
Output Format
All prime numbers present in the array (space-separated).
If no prime exists, print nothing (or empty output).
If no prime exists, print nothing (or empty output).
Constraints
• n ≥ 1
• Array elements ≥ 1
• Array elements ≥ 1
Examples
Input:
6
2 4 5 6 7 9
Output:
2 5 7
Example Solution (Public)
void question10_prime_in_array() {
int arr[] = {10, 13, 15, 17, 20, 23};
int size = 6;
cout << "Prime numbers: ";
for(int i = 0; i < size; i++) {
bool isPrime = true;
if(arr[i] < 2) isPrime = false;
for(int j = 2; j * j <= arr[i]; j++) {
if(arr[i] % j == 0) {
isPrime = false;
break;
}
}
if(isPrime) {
cout << arr[i] << " ";
}
}
cout << endl;
}
Official Solution Code
void question10_prime_in_array() {
int arr[] = {10, 13, 15, 17, 20, 23};
int size = 6;
cout << "Prime numbers: ";
for(int i = 0; i < size; i++) {
bool isPrime = true;
if(arr[i] < 2) isPrime = false;
for(int j = 2; j * j <= arr[i]; j++) {
if(arr[i] % j == 0) {
isPrime = false;
break;
}
}
if(isPrime) {
cout << arr[i] << " ";
}
}
cout << endl;
}
Please login to submit solutions.