Let f be a two-dimensional NumPy array, such that the element f[0, 0] is equal to 0. Which of the following pieces of code would not overwrite f to make this element equal to 1? O a. h = f h[0, 0] = 1 O b. f.reshape((2, 3, 2))[0, 0, 0] = 1 c. f.copy() [0, 0] = 1 O d. f[0, 0] = 1 O e. f[, ][0] = 1 O f. f.view() [0, 0] = 1
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- Write a function that partitions the array into two subarrays: one with all even integers, and the other with all odd integers. Return your result in the following format: [[evens], [odds]] Example: evenoddpartition ([2, 4, 8, 9]) → [[2, 4, 8], [9]] WRITE IN PYTHON PLEASED3 no. D4. Coding. TheTwo arrays are provided to you: one longer and one shorter (with all distinguishable items). Search the longer array for the smallest subarray that includes every member in the shorter array. Any arrangement of the components is possible.EXAMPLEInput: {1, 5, 9}{7, 5, 9, 0, 2, 1, 3, 5. 7, 9. 1, 1, 5, 8, 8, 9, 7}Result: [7, 10] (the underlined portion above)
- Write a program / Pseudo code / Algorithm that inserts the following numbers into two Separate arrays. L1 50 30 25 75 82 28 77 L2 50 40 25 75 80 21 37 30 After Insertion median should be calculated for each adjacent element from both array structures, after computing median print / Output that array M. if M is the median them M0 = (L1[0]+L2[0]) /2M1 = (L1[1]+L2[1]) /2An array is special if every even index contains an even number and every odd index contains an odd number. Create a function that returns true if an array is special, and false otherwise. Examples isSpecialArray([2, 7, 4, 9, 6, 1, 6, 3]) → true // Even indices: [2, 4, 6, 6]; Odd indices: [7, 9, 1, 3] isSpecialArray([2, 7, 9, 1, 6, 1, 6, 3]) → false // Index 2 has an odd number 9. isSpecialArray ([2, 7, 8, 8, 6, 1, 6, 3]) → false // Index 3 has an even number 8.Given an 8-element array: A = {x1, X2, X3, X4, X5, X6, 7, x8}, we would like to find its 3rd smallest element. It is known that x₁ and 8 are the two extreme elements: max and min, but we do not know which one is the max and which one is the min. It is also known that x4 and x5 are the two medians: left-median and right-median. What is the minimum number of comparisons you need to find the 3rd smallest element of the original array? Ans:
- Select 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)Number of students should be 279, not 278 what did I do wrong?Return array of odd rows and even columns from below NumPy array. Expected Output: Printing Input Array [[ 3 6 9 12] [15 18 21 24] [27 30 33 36] [39 42 45 48] [51 54 57 60]] Printing array of odd rows and even columns [[ 6 12] [30 36] [54 60]]
- solve in java Integer numValues is read from input, representing the number of integers to be read next. Then, the remaining integers are read and stored into array numbersArray. For each element in numbersArray that is divisible by 5: Output the element, followed by ": adjusted to a number not divisible by 5" and a newline. Assign the element with the element's current value plus 1. Ex: If the input is: 3 60 97 95 then the output is: Original numbers: 60 97 95 60: adjusted to a number not divisible by 5 95: adjusted to a number not divisible by 5 New numbers: 61 97 96 Note: (x % 5 == 0) returns true if x is divisible by 5.Function 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)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…