Code Analysis () If we change the first line of code in the function to be “if (low >= high) return -1;”, and we have an array defined as “int nums[] = {2, 3, 5, 7, 8, 10};”, answer the questions below if we called this function with “int index = binarySearch(nums, 0, nums.length - 1, 3);” For each iteration of binarySearch: 1) What are the values of low, high, mid, and target variables?

Code Analysis () If we change the first line of code in the function to be “if (low >= high) return -1;”, and we have an array defined as “int nums[] = {2, 3, 5, 7, 8, 10};”, answer the questions below if we called this function with “int index = binarySearch(nums, 0, nums.length - 1, 3);” For each iteration of binarySearch: 1) What are the values of low, high, mid, and target variables?

New Perspectives on HTML5, CSS3, and JavaScript

6th Edition

ISBN:9781305503922

Author:Patrick M. Carey

Publisher:Patrick M. Carey

Chapter14: Exploring Object-based Programming: Designing An Online Poker

Section14.1: Visual Overview: Custom Objects, Properties, And Methods

Problem 7QC

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Below is a recursive version of binarySearch:int binarySearch(int nums[], int low, int high, int target){ if (low > high) return -1; int mid = (low + high)/2;  if (target == nums[mid]) return mid;  else if (target < nums[mid]) return binarySearch(nums, low, mid - 1, target); else return binarySearch(nums, mid + 1, high, target);} Code Analysis ()If we change the first line of code in the function to be “if (low >= high) return -1;”, and we have an array defined as “int nums[] = {2, 3, 5, 7, 8, 10};”, answer the questions below if we called this function with “int index = binarySearch(nums, 0, nums.length - 1, 3);”For each iteration of binarySearch:2) Will a base case be true? If so, what value is being returned
Below is a recursive version of binarySearch:int binarySearch(int nums[], int low, int high, int target){ if (low > high) return -1; int mid = (low + high)/2;  if (target == nums[mid]) return mid;  else if (target < nums[mid]) return binarySearch(nums, low, mid - 1, target); else return binarySearch(nums, mid + 1, high, target);}For the statement below, indicate whether you think it is True or False. If you like, you can provide a description of your answer for partial credit in case you are incorrect.4) The code “else return binarySearch(nums, mid + 1, high, target);” is a base case for this function
Below is a recursive version of binarySearch:int binarySearch(int nums[], int low, int high, int target){ if (low > high) return -1; int mid = (low + high)/2;  if (target == nums[mid]) return mid;  else if (target < nums[mid]) return binarySearch(nums, low, mid - 1, target); else return binarySearch(nums, mid + 1, high, target);}For the statement below, indicate whether you think it is True or False. If you like, you can provide a description of your answer for partial credit in case you are incorrect. 1) The code “if (low > high) return -1;” is a base case for this function
Consider the following function: void fun_with_recursion(int x) { printf("%i\n", x); fun_with_recursion(x + 1); } What will happen when this function is called by passing it the value 0?
int functionC (int n) { int i, j, sumC = 0; for (i=n; i > 0; i=i-5) for (j=1; j 0) { if (functionC(n) % 2 == 0) { for (i=m; i > 0; i=i/3) sumE++; } else (10) Asymptotic runtime of functionE { for (i=m; i > 0; i=i-3) sumE--; } n--; } return sumE;
Below is a recursive version of binarySearch:int binarySearch(int nums[], int low, int high, int target){ if (low > high) return -1; int mid = (low + high)/2; if (target == nums[mid]) return mid; else if (target < nums[mid]) return binarySearch(nums, low, mid - 1, target); else return binarySearch(nums, mid + 1, high, target);} Code Analysis ()If we change the first line of code in the function to be “if (low >= high) return -1;”, and we have an array defined as “int nums[] = {2, 3, 5, 7, 8, 10};”, answer the questions below if we called this function with “int index = binarySearch(nums, 0, nums.length - 1, 3);”For each iteration of binarySearch:3) If the base case(s) are false, which recursive function call will be made?
java C++ Ackermann’s FunctionAckermann’s Function is a recursive mathematical algorithm that can be used to test how well a computer performs recursion. Write a function A(m, n) that solves Ackermann’s Function. Use the following logic in your function:If m = 0 then return n + 1If n = 0 then return A(m−1, 1) Otherwise, return A(m−1, A(m, n−1))Test your function in a driver program that displays the following values:A(0, 0) A(0, 1) A(1, 1) A(1, 2) A(1, 3) A(2, 2) A(3, 2) SAMPLE RUN #0: ./AckermannRF Hide Invisibles Highlight: Show Highlighted Only The·value·of·A(0,·0)=·1↵ The·value·of·A(0,·1)=·2↵ The·value·of·A(1,·1)=·3↵ The·value·of·A(1,·2)=·4↵ The·value·of·A(1,·3)=·5↵ The·value·of·A(2,·2)=·7↵ The·value·of·A(3,·2)=·29↵
In python Code so far: # TODO: Declare global variables here.recursions = 0comparisons = 0 def binary_search(nums, target, lower, upper): # Type your code here. target = (lower + upper) //2 global if target == nums[index]: return index if (lower == upper): comparisons +=1 if target == nums[lower]: return lower else: return -1 if (nums[index] < target): recursions += 1 return binary_search(nums,target, index, upper) if nums[index] > target: recursions += 1 return binary_search(nums, target, 0, lower) if __name__ == '__main__': # Input a list of nums from the first line of input nums = [int(n) for n in input().split()] # Input a target value target = int(input()) # Start off with default values: full range of list indices index = binary_search(nums, target, 0, len(nums) - 1) # Output the index where target was found in nums, and the # number of…
using namespace std; int main() int i, m=0,n=4; float arr[100] (55,66,88,1); for(i=0; i
Below is a recursive version of binary search:int binarySearch(int nums[], int low, int high, int target){ if (low > high) return -1; int mid = (low + high)/2; if (target == nums[mid]) return mid; else if (target < nums[mid]) return binarySearch(nums, low, mid - 1, target); else return binarySearch(nums, mid + 1, high, target);}For the 4 statements below, indicate whether you think it is True or False. If you like, you can provide a description of your answer for partial credit in case you are incorrect.1) The code “if (low > high) return -1;” is a base case for this function2) The code “int mid = (low + high)/2;” is a base case for this function3) The code “if (target == nums[mid]) return mid;” is a base case for this function4) The code “else return binarySearch(nums, mid + 1, high, target);” is a base case for this function
Write a recursive function (no auxiliary functions, for/while loops, STL containers or functions, static/global variables) bool mirrorMirrorOnTheWall(string n); Given a string, recursively determine if the string is the same forwards and backwards. mirrorMirrorOnTheWall(""); mirrorMirrorOnTheWall("a"); mirrorMirrorOnTheWall("mirror"); // returns false mirrorMirrorOnTheWall("racecar"); // returns true // returns true // returns true Edit View Insert Format Tools Table 12pt v Paragraph v BIU ...
Course : Data Structures & Algorithms in C++ Question : Consider the following code: int myFunc(int A[], 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[j] && 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]; return max; }a) What is the time complexity of the algorithm?