int FactNotFact (int N) { if (N< 0 ) { fac=-1 } else { if (N==0 || N==1){ fac=1 } else { fac=N; K=N-1; while ( k <> 0 ) { fac = fac * k; k = k – 1; } } } Fact=fac; }.Using McCabe’s cyclomatic complexity, how many paths are there in this code

Do you have choices because im pretty sure i could get it

Public static int mystery(int a[], int x) {     int c = 0;     for(int i = 0; i < a.length; i++) {        if (a[i] == x) c++;     } return c; }the method counts how many times an integer shows up in an integer array.the c variable is declared in the method to keep track of how many times the certain value has shown up.the for-loop loops through the entire array.the line    if (a[i] == x) c++;   checks if each element of the array is equal to the inputted integer to check. for each number in the array that is equal to the inputted integer x, the c variable will be incremented by 1. otherwise it will do nothing.then c is returned at the end of the loop.for example, if we have, in the main method with the mystery method coded in the class:     int arraytest[] = {2, 23, 23, 23, 3};     int testme = 23;     int count = mystery(arraytest, testme);     system.out.println(count); then the print out will be    3 because there are three 23s in that array.

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