The following code is supposed to return n!, for positive n. int factorial(int n){ if (n == 0) return 1; else return (n + factorial(n - 1)); } An analysis of the code using our "Three Question" approach reveals that: Group of answer choices it fails the base-case question. it fails the smaller-caller question. it fails the general-case question. it passes on all three questions and is a valid algorithm. None of these is correct.
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30The following code is supposed to return n!, for positive n.
int factorial(int n){
if (n == 0)
return 1;
else
return (n + factorial(n - 1));
}
An analysis of the code using our "Three Question" approach reveals that:
Group of answer choices
it fails the base-case question.
it fails the smaller-caller question.
it fails the general-case question.
it passes on all three questions and is a valid
None of these is correct.
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- Use two ways to count the number of r-combinations of[n]={1,2,…,n}that contains 1 or 2 or 3. First, separate cases with Case 1 counting all r-combinations of [n] that contains 1 , Case 2 counting all r-combination of [n] that contains 2 but not 1 , and Case 3 counting allrcombination of [n] that contains 3 but not 1 or 2 . Second, count all r-combinations of [n] not containing any of1,2,3and use the subtraction rule.Correct answer will be upvoted else Multiple Downvoted. Computer science. pick a non-void adjacent substring of s that contains an equivalent number of 0's and 1's; flip all characters in the substring, that is, supplant all 0's with 1's, as well as the other way around; turn around the substring. For instance, think about s = 00111011, and the accompanying activity: Pick the initial six characters as the substring to follow up on: 00111011. Note that the number of 0's and 1's are equivalent, so this is a lawful decision. Picking substrings 0, 110, or the whole string would not be imaginable. Flip all characters in the substring: 11000111. Invert the substring: 10001111. Find the lexicographically littlest string that can be gotten from s after nothing or more tasks. Input The primary line contains a solitary integer T (1≤T≤5⋅105) — the number of experiments. Every one of the accompanying T lines contains a solitary non-void string — the input string s for…int const MULTIPLIER = 5; is a valid way to declare a constant integer variable. Group of answer choices True False ------- Given the following code segment, what is output to the screen? int num1 = 6; int num2 = 4 * num1++; cout << "num1=" << num1 << " num2=" << num2; Group of answer choices num1=7 num2=24 num1=6 num2=24 num1=6 num2=28 Nothing, because the code does not compile.
- Correct answer will be upvoted else downvoted. you can choose any substring of a containing exactly k characters 1 (and arbitrary number of characters 0) and reverse it. Formally, if a=a1a2…an, you can choose any integers l and r (1≤l≤r≤n) such that there are exactly k ones among characters al,al+1,…,ar, and set a to a1a2…al−1arar−1…alar+1ar+2…an. Find a way to make a equal to b using at most 4n reversals of the above kind, or determine that such a way doesn't exist. The number of reversals doesn't have to be minimized. Input Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤2000). Description of the test cases follows. Each test case consists of three lines. The first line of each test case contains two integers n and k (1≤n≤2000; 0≤k≤n). The second line contains string a of length n. The third line contains string b of the same length. Both strings consist of characters 0 and 1. It is guaranteed that the sum of n over all…Consider the following code segment: int a[10]; We want to use the special for (range based for) that we studied in class to print all the elements of a. Which of these will work? a.) for(int i:a) cout<< a[i]; b.)More than one correct choices listed c.)No correct answer d.)for (int i:a[i])cout<< i; e.) for(int i:a) cout<< i;What is the run-time (T(n)) and complexity of the following code segment? int y = 0; for ( int a= 0; a< n; ++a ) { for ( int b= 0; b < a; ++b) y += j;}
- mplement a Java program that applies the Newton-Raphson's method xn+1 = xn – f(xn) / f '(xn) to search the roots for this polynomial function ax6 – bx5 + cx4 – dx3+ ex2 – fx + g = 0. Fill out a, b, c, d, e, f, and g using the first 7 digits of your ID, respectively. For example, if ID is 4759284, the polynomial function would be 4x6 – 7x5 + 5x4 – 9x3+ 2x2 – 8x + 4 = 0. The program terminates when the difference between the new solution and the previous one is smaller than 0.00001 within 2000 iterations. Otherwise, it shows Not Found as the final solution.Make an analysis of the following code using different values of n and find out the best and worst running time. func( n) { if (n>10) { cout<<n<<endl; } else { i=0; while(i<n) { cout<<n<<endl; i++; } } }Language: C Pascal’s triangle is a triangular array, useful for calculating the binomial coefficients, n k , that are used in expanding binomials raised to powers, combinatorics and probability theory. 0 0 1 0 1 1 2 0 2 1 2 2 3 0 3 1 3 2 3 3 4 0 4 1 4 2 4 3 4 4 Evaluating the values of the binomial coefficients, you get the following pattern, 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 The number of the entries in each row is increased by one, as we move down. Each number in the triangle, is constructed by adding the number above it and to the left, with the number above it and to the right. The blank entries as treated as 0. Using the recursion, implement the function that computes the Pascal’s triangle. Pr
- Generate test cases for the following code by using logic coverage criteria Descision coverage. void count(char s[],int &caps ,int &low) { caps=low =0; for(int i=0;s[i] !='\0';i++) if(s[i]>='A'&& s[i]<='Z') caps++; else if(s[i]>='a'&&s[i]<='z') low++; else printf(“Not a letter!\n”);What is wrong with the following code? int *p;. //Line 1int *q;. //Line 2 p = new int [5];. //Line 3 *p = 2; //Line 4 for (int i = 1; i < 5; i++). //Line 5p[i] = p[i-1] + i;. //Line 6 q = p;. //Line 7 delete [] p; //Line 8 for (int j = 0; j < 5; j++) //Line 9 cout << q[j] << " "; //Line 10 cout << endl; //Line 11Hold * downvote for incorrect as well because code is in such same form.