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# Distinct Prime Elements of a given quantity N

Given a quantity N, the duty is to search out the distinct Prime Elements of N.

Examples:

Enter: N = 12
Output: 2 3
Clarification: The elements of 12 are 1, 2, 3, 4, 6, 12.
Amongst these the distinct prime elements are 2 and three.

Enter: N = 39
Output: 3 13

Strategy: The method is to make use of a map to verify whether or not a given issue of the quantity has occurred earlier or not. Now comply with the beneath steps to resolve this drawback:

1. Create a map visited to maintain monitor of all earlier prime elements.
2. Create a variable C, and initialise it with 2.
3. Whereas N is divisible by C, print C if C will not be current within the map. Now divide N by C. Additionally increment C by 1.

Under is the implementation of the above method:

## C++

 `#embrace ` `utilizing` `namespace` `std;`   `void` `distinctPrimeFactors(``int` `N)` `{` `    ``if` `(N < 2) {` `        ``cout << -1;` `    ``}`   `    ``int` `c = 2;` `    ``unordered_map<``int``, ``bool``> visited;`   `    ``whereas` `(N > 1) {` `        ``if` `(N % c == 0) {` `            ``if` `(!visited) {` `                ``cout << c << ``" "``;` `                ``visited = 1;` `            ``}` `            ``N /= c;` `        ``}` `        ``else` `            ``c++;` `    ``}` `}`   `int` `most important()` `{` `    ``int` `N = 39;` `    ``distinctPrimeFactors(N);` `    ``return` `0;` `}`

## Javascript

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Time Complexity: O(N)
Auxiliary Area: O(N1/2)

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