int myFunc(int AI, int n) { int i, j, max = 0; int msis[n]; for (i = 0; i < n; i++) msis[i] = A[i); for (i = 1; i < n; i++) for (j = 0; j< i; j++) if (A[i] > A[i] && msis[i] < msis[j] + A[i]) msis[i] = msis[j] + A[i]; for (i = 0; i < n; i++) if ( max < msis[i] ) max = msis[i]; %3D %D return max;
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- Q#2 Write a recursive function zeroCount ( int a[ ], int s, int e) that receives an array of integers a [], a start index s, and an end index e. The function should return the number of zeros in that array between s and e. int zeroCount ( int a[ ], int s, int e); Trace your function given the following array and function call. Draw your steps. int a[ ] = {1, 0, 0, 5}; int zeros = zeroCount(a, 0, 3); language c++13. Implement function triangleArea(a,b,c) that takes as input the lengths of the 3 sides of a triangle and returns the area of the triangle. By Heron's formula, the area of atriangle with side lengths a, b, and c is √ s ( s−a )( s−b)( s−c ) , where s=(a+b+ c )/ 2 .>>> triangleArea(2,2,2)1.7320508075688772C++ Given a positive integer, N, the ’3N+1’ sequence starting from N is defined as follows: If N is an even number, then divide N by two to get a new value for N If N is an odd number, then multiply N by 3 and add 1 to get a new value for N. Continue to generate numbers in this way until N becomes equal to 1 For example, starting from N = 3 the complete ’3N+1’ sequence would be: 3, 10, 5, 16, 8, 4, 2, 1 Write code to ask the user to enter a positive integer (N) in the main() function. Write a function sequence() that receives the integer value N and display the ‘3N+1’ sequence starting from the integer value that was received (entered by the user). The function must also count and return the numbers that the sequenceconsists of. The returned value must be displayed from the main() function.
- Question 1 Work your way through the following code fragments. What would be printed? a) int i = -23; int * p = &i; printf("*p = %i\n", *p); b) int i; int * p = &i; printf("*p = %i\n", *p); c) int i = 48; int * p; printf("*p = %i\n", *p); d) int i = 10;int * p = &i;int j; j = ++*p; printf("j = %i\n", j);printf("i = %i\n", i); e) int i = 10, j = 20;int * p = &i;int * q = &j; *p = *q;printf("i = %i, j = %i\n", i, j);printf("*p = %i, *q = %i\n", *p, *q); i = 10; j = 20; p = q;printf("i = %i, j = %i\n", i, j);printf("*p = %i, *q = %i\n", *p, *q);iv) Write a JAVA program to merge two sorted arrays into a single sorted array. Take the sample arrays given below and print the result in the output. The time complexity of the program should be O(N+M) where N and M are the sizes of both the arrays. int []arr1 = {1, 6, 10, 14} int []arr2 = {2, 4, 5, 11, 16, 19} Print the merged array in the output.Answer the given question with a proper explanation and step-by-step solution. In Java, please. Consider the following problem. Input: an array of integers A[] Output: rearranges the array to have the following property: Suppose the first element in the original array A[], A[0] = x. In the rearranged array, suppose that x is at index i, that is A[i] = x. Then, we want the rearranged array to be such that A[j] <= x for all j < i and A[j] > x for all j > i. The rearranging should place all the values less than (or equal to) x to the "left" of x and all values larger than x to the right of x. Here's an example. Suppose the array has the elements in this initial order: 4 3 9 2 7 6 5 A[0] = 4. After rearranging elements, we get: 3 2 4 5 9 7 6 That is, A[0] = 4, is positioned in the resulting array so that all elements less than 4 (that is, 2, 3) are to its left (in no particular order), and all elements larger than 4 are to its right (again, in no particular order).
- Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for the rugged hero to tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The number of ways that a positive integer n can be represented as a sum of consecutive integers is called its politeness, and can also be computed by tallying up the number of odd divisors of that number. However, note that the linked Wikipedia de0inition…Problem Statement: Consider a permutation of numbers from 1 to N written on a paper. Let’s denote the product of its element as ‘prod’ and the sum of its elements as ‘sum’. Given a positive integer N, your task is to determine whether ‘prod’ is divisible by ‘sum’ or not. Input Format: First input will be an integer T. It depicts a number of test cases. Followed by the value for each test case. Each test case will contain an integer N (1<= N <=10^9). It is nothing but the length of the permutation. Output Format: For each test case, print “YEAH” if ‘prod’ is divisible by ‘sum’, otherwise print “NAH”..write java codes to do the following: Write a recursive function to check if an integer array is negative symmetric, for example, an array of 10, 20, 30, 90, -30, -20, -10 is considered as negative symmetric, while an array of 10, 20, 30, 90, 30, 20, 10 is not considered as negative symmetric. public static boolean checkArraySym(int [ ] A, int first, int last) that receives an array A, first index, last index and checks if the array is negative symmetric
- Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for someone to come and tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The count of how many different ways a positive integer n can be represented as a sum of consecutive integers is also called its politeness, and can be alternatively computed by counting how many odd divisors that number has. However, note that the linked…1.Given a positive integer, N, the ’3N+1’ sequence starting from N is defined as follows:If N is an even number, then divide N by two to get a new value for NIf N is an odd number, then multiply N by 3 and add 1 to get a new value for N.Continue to generate numbers in this way until N becomes equal to 1For example, starting from N = 3 the complete ’3N+1’ sequence would be:3, 10, 5, 16, 8, 4, 2, 1Write c++ code to ask the user to enter a positive integer (N) in the main() function. Write a function sequence()that receives the integer value N and display the ‘3N+1’ sequence starting from the integer value that wasreceived (entered by the user). The function must also count and return the numbers that the sequenceconsists of. The returned value must be displayed from the main() function.Do not use static variables to implement recursive methods. USING JAVA USING: // P4 public static int min(int [] a, int begin, int end) { } Implement a recursive method min that accepts an array and returns the minimum element in the array. The recursive step should divide the array into two halves and find the minimum in each half. Demonstrate the output of min on the array int [] a = { 2, 3, 5, 7, 11, 13, 17, 19, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 23, 29, 31, 37, 41, 43 }