int sum (int n) {//assume n is non-negative if (0 == n) return 0; else n+ sum (n-1); |
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- int n = 1; int k - 2; int r = n; if (k < n) { r - ka) FindMinIterative public int FindMin(int[] arr) { int x = arr[0]; for(int i = 1; i < arr.Length; i++) { if(arr[i]< x) x = arr[i]; } return x; } b) FindMinRecursive public int FindMin(int[] arr, int length) { if(length == 1) return arr[0]; return Math.Min(arr[length - 1], Find(arr, length - 1)); } What is the Big-O for this functions. Could you explain the recurisive more in details ?// 9. Is It Prime? function isPrime(n) { if (n < 2 || n % 1 != 0) { return false; } for (let i if (n % i } return true; 2; i < n; ++i) { 0) return false; == }
- int funcB(int); int funcA(int n) { if (n 4) { return n* funcA(n - 5); } else { return n- funcB(n - 1); int main() { cout << funcA(13); return 0; What is the output of this program? Please show our work.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. PrST(1/2) - +n T(n) = Answer: Display response n> 1 n<1
- Given the following pseudocode: function fun2(n) { var outer_count = 0; var inner_count %3D for (i=0; i 0, as an expression of normal arithmetic and n, what is the value of outer_count that counts the number of times the outer loop executes? Enter an expression Fun.2.inner: Assuming n E N and n > 0, as an expression of normal arithmetic and n, what is the value of inner_count that counts the number of times the inner loop executes? Enter an expressioninhinclude Rsing namespace std; int maxResult( ) int maxVal for (int i { for (int j = 0; j <= n 0; i <= n; i += a) i; j += b) %3D %3D float z = (float)(n (i + j)) / (float)(c); if (floor(z) { int x = int y ceil(z)) i / a; j/ b; maxVal = max(maxVal, x + y (int)z); return maxVal; } int main() { max cout << maxResult( ); return 0; 1 } Input Compilation failed due to fellowing ons main.cpp:7:23: error: 'n' was not declared in this scope 7| for (int i = 0; i <= n; i a) main.cpp:7:31: error: 'a was not declared in this scope for (int i = 0; i <= n; i t= a) %3D main.cpp:9:36: error: b' was not declared in this scope | 6 for (int j = 0; j <= n - i; j - b) %3D main.cpp:11:45: error: 'c was not declared in this scopevoid changeAll(int x[]) { x[0] = x[0] + 5; return; The first element in an array would be increased by 5 The array would be the same because it was passed by value x would be increased by 5 All the elements in the passed array would be incremented by 5
- b) int [] num = new int [6]; for (int i = 1; i 2) { num [i] = 2 * num[i]-num[i-1]; System.out.println( "num [" + i + ") =" + num [i]);* How many elements in the array A are * also in the array B? Assume B is sorted. 01: int overlap (int* A, int* B, int N) 02: { 03: int count = 0; 04: for (int i = 0; i < N; ++i) 05: { %3D 06: int x = A[i]; 07: int pos = lower_bound (B, B+N, x) - B: %3D if (pos <0 && B[pos] == x) 09: 10: 08: ++count; 11: 12: } 13: return count; 14:} According to the copy-and-paste technique, how would you annotate line 8 of this function by the time you had determined the complexity of lines 8-11?#include | int main() { int n=5; for (int i=1;iSEE MORE QUESTIONS