Implement algorithm FindMax(L) pre-cond: L is an array of n values. post-cond: Returns an index with maximum value.
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Implement
pre-cond: L is an array of n values.
post-cond: Returns an index with maximum value.
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- In an array-based implementation of a List, why does the add operation take O(n) time in the average case? Select one: a. The time it takes to shift entries over to make room in the array depends on the number of entries in the List b. The time to copy the current entries into a newly allocated, larger array depends on the number of entries in the List c. The time to access the position in the array to add the new entry depends on the number of entries in the List d. The time to access the position in the array to add the new entry depends on the capacity of the arrayFunction PrintArray(integer array(?) dataList) returns nothing integer i for i = 0; i < dataList.size; i = i + 1 dataList[i] = Get next input Put dataList to output Put "_" to output // Your solution goes here. Modify as needed i = 0 Complete the PrintArray function to iterate over each element in dataList. Each iteration should put the element to output. Then, put "_" to output. Ex: If dataList's elements are 2 4 7, then output is: 2_4_7_ Function Main() returns nothing integer array(3) userNums integer i for i = 0; i < userNums.size; i = i + 1 userNums[i] = Get next input PrintArray(userNums)Implement a range function for a dynamic array which returns a new dynamic array that is a subset of the original. input parameters: array - (the array and any related parameters) start - index of the first elementend - index of the last elementInterval - An integer number specifying the incrementation of index This function returns a new dynamic array containing the elements from the start thru the end indices of the original array.All array indexing must be done using pointer arithmetic. For example, given the array: 49 96 99 47 76 29 22 16 30 22 If the start and end positions were 5 and 9 with step 2, return a new dynamic array: 29 16 22 Please use following main to test your function. int main(){int *p = new int[10]{49,96,99,47,76,29,22,16,30,22}; int *q = range(p,10,5,9,2);for(int i=0;i<3;i++) cout<<q[i]<<" "; // print 29 16 22 cout<<endl;delete [] q;q = range(p,10,1,8,3); for(int i=0;i<3;i++)cout<<q[i]<<" "; // print 96 76 16 cout<<endl;…
- Write Java Program to Sort an array of element by getting values from user Remove duplicate elements from a sorted array Reverse the contents inside an array Search for an element inside the array using linear searchALGO2(A)// A is an integer array, index starting at 11:for i=1 to n-1 do2: for j=n to i+1 do3: if (A[j] < A[j-1]) then4: exchange A[j] with A[j-1] Which function best describes the worst-case running time behaviour of ALGO2 on an array of size n? Option 1) a*(n^2) + b*n +c, where a,b,c are constantsOption 2) a*n + b , where a and b are constantsOption 3) a * n * log(n) + b, where a,b are constantsOption 4) a*(n^3) + b*(n^2) + c*n + d, where a,b,c,d are constantsSelect the for-loop which iterates through all even index values of an array.A. for(int idx = 0; idx < length; idx++)B. for(int idx = 0; idx < length; idx%2)C. for(int idx = 0; idx < length; idx+2)D. for(int idx = 0; idx < length; idx=idx+2)
- function Sum(A,left,right) if left > right: return 0else if left = right: return A[left] mid = floor(N/2) lsum = Sum(A,left,mid) rsum = Sum(A,mid+1,right) return lsum + rsum function CreateB(A,N)B = new Array of length 1 B[0] = Sum(A,0,N-1) return B Building on the above, in a new scenario, given an array A of non-negative integers of length N, additionally a second array B is created; each element B[j] stores the value A[2*j]+A[2*j+1]. This works straightforwardly if N is even. If N is odd then the final element of B just stores A[N-1] as we can see in the figure below: (added in image) The second array B is now introducing redundancy, which allows us to detect if there has been a hardware failure: in our setup, such a failure will mean the values in the arrays are altered unintentionally. The hope is that if there is an error in A which changes the integer values then the sums in B are no longer correct and the algorithm says there has been an error; if there were an error in B…OBJECT Lab Session 03 Application of the Linear and binary search on a list of elements stored in an array. THEORY Linear Search A nonempty array DATA with N numerical values is given. This algorithm finds the location LOC and required value of elements of DATA. The variable K is used as a counter. ALGORITHM 1. Set k := 1 & loc : = 0 Repeat step 3 & 4 while loc : = 0 &k < = n If (item = data[k]) loc : = k Else K = k + 1 If loc > = 0 or If (item = data[k]) then Print “loc is the location of item” Else Print “no. not found” Exit Binary Search Suppose DATA is an array that is sorted in increasing (or decreasing) numerical order or, equivalently, alphabetically. Then there is an extremely efficient searching algorithm, called binary search, which can be used to find the location LOC of a given ITEM of information in DATA. The binary search algorithm applied to our array DATA works as follows: During each stage of our algorithm, our search for ITEM is…: In searching an element in an array, linear search can be used, even though simple to implement, but not efficient, with only O(n) time complexity. Assuming the array is already in sorted order, modify the search function below, using a better algorithm, so the average time complexity for the search function is O(log n). include <iostream> using namespace std; int search(int al), int s, int v) { 1/ Modify below codes. for (int i = 0; i <s; i++) { if (a[i] = v) return i; return -1; int main() { int intArray:10] = { 5, 7, 8, 9, 10, 12, 13, 15, 20, 34); // Search for element '12' in 10-elements integer array. cout << search(intArray, 10, 12); // '5' will be printed out. // Search for element '35' in 10-elements integer array. cout << search(intArray, 10, 35); // '-1' will be printed out. // Index '-l' means that the element is not found. return 0;
- OBJECT Lab Session 03 Application of the Linear and binary search on a list of elements stored in an array. THEORY Linear Search A nonempty array DATA with N numerical values is given. This algorithm finds the location LOC and required value of elements of DATA. The variable K is used as a counter. ALGORITHM 1. Set k := 1 & loc : = 0 Repeat step 3 & 4 while loc : = 0 &k < = n If (item = data[k]) loc : = k Else K = k + 1 If loc > = 0 or If (item = data[k]) then Print “loc is the location of item” Else Print “no. not found” Exit Binary Search Suppose DATA is an array that is sorted in increasing (or decreasing) numerical order or, equivalently, alphabetically. Then there is an extremely efficient searching algorithm, called binary search, which can be used to find the location LOC of a given ITEM of information in DATA. The binary search algorithm applied to our array DATA works as follows: During each stage of our algorithm, our search for ITEM is…9.). integer array nums and an integer k, return the length of the shortest non-empty subarray of nums with a sum of at least k. If there is no such subarray, return -1. A subarray is a contiguous part of an array. Example 1: Input: nums = [1], k = 1 Output: 1 Example 2: Input: nums = [1,2], k = 4 Output: -1 Example 3: Input: nums = [2,-1,2], k = 3 Output: 3.JAVA Programming Write a function that returns true if you can partition an array into one element and the rest, such that this element is equal to the product of all other elements excluding itself. Examples canPartition ([2, 8, 4, 1]) → true // 8 = 2 x 4 x 1 canPartition ([-1, -10, 1, -2, 20]) → false canPartition ([-1, -20, 5, -1, -2, 2]) true