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- 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 symmetricExecute the following using Array Implementation. Use diagrams in showing your final answer A ENQUEUE(A) ENQUEUE(B) ENQUEUE(C) DEQUEUE() ENQUEUE(A) ENQUEUE(B) ENQUEUE(C) DEQUEUE() DEQUEUE() ENQUEUE(A) DEQUEUE() ENQUEUE(B) ENQUEUE(C) B PUSH (x, S) PUSH (y, S) PUSH (z, S) POP (S) PUSH (x, S) PUSH (y, S) POP (S) POP (S) C INSERT(Alice, 1, ) INSERT(Bob, 2, L) INSERT(Karen, 3, L) INSERT(Ram, 4, L) DELETE( 2, L) What will happen to your array?Using recursion, write a Python function def before(k,A) which takes an integer k and an array A of integers as inputs and returns a new array consisting of all the integers in A which come before the last occurrence of k in A, in the same order they are in A. For example, if A is [1,2,3,6,7,2,3,4] then before(3,A) will return [1,2,3,6,7,2]. If k does not occur in A, the function should return None.
- Using recursion, write a Python function def countOdds(A) which which takes an array of integers A as input and returns the number of elements in the array that are odd. For example, if A is [1,2,5,3,6,5,3,5,5,4] the the function should return 7. If A is empty then it should return 3. Hint: To check if n is odd it suffices to check if dividing it by 2 gives remainder 1 (in Python, this remainder is calculated by n%2)./** * Write a Java function to determine if the elements of an array can be split into two groups such that the sum of one group is a non-zero multiple of 10 and the sum of the other group is odd. Your solution must use recursive backtracking. <code> * Example: * Given: {1, 2, 6, 3, 1} * Result: true because {1,2,6,1}, {3} is one possibility. * * Example: * Given: {4, 3, 5, 2} * Result: false * </code> */ public static boolean splitOdd10(int[] num) { }Write a recursive function that, given a sequence of comparable values, returns the count of elements where the current element is less than the following ( next ) element in the given sequence. See the examples given below. def count_ordered ( seq ) : """ Input : A sequence of comparable elements Output : The number of elements that are less than the following element in the sequence Example : >>> count_ordered ( [ 1 , 2 , 3 , 4 , 5 , 6 ] ) 5 >>> count_ordered ( ( 1 , 12, 7.3 , -2,4 ) ) 2 >>> count_ordered ( 'Python' ) 2 >>> count_ordered ( [ 6 ] ) 0 >>> count_ordered ( [ ] ) 0 """ In the first example above , count_ordered ( [ 1,2,3,4,5,6 ] )the returned answer is 5 because for all the first 5 numbers the current number is less than the next number. In the second example above, count_ordered ( ( 1,12,7.3 , -2,4 ) )the…
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- Write C code that does the following: Using the experimental data below, write a computer program to determine the coefficients a and b for the best line that represents the data. Note that the equation for a line is y = ax = b. x=1.01, 2.8, 4.6, 6.79, 9.3, 10.94, 13.03, 15.21, 16.89, 18.94, 21.2, 23, 24.99 y=2.05, 5.96, 9.85, 13.5, 19.3, 21.54, 25.88, 29.13, 33.45, 38.19, 43.01, 45.89, 49.34 then a set of 500 float numbers is given. I can show the numbers but not via image. I would like to know where I copy and paste them in the code. . Write a C program to show how many out of 500 are between 60.0 and 100. Form an array B containing the elements from a) Calculate average, variance, and standard deviation of the array B and print those on the screen.The goal is to rewrite the function, below, such that passes in a different list of parameters, particularly eliminating the need to pass low and high for each recursive call to binary_search. defbinary_search(nums,low,high,item): mid=(low+high)//2iflow>high:returnFalse #The item doesn't exist in the list!elifnums[mid]==item:returnTrue# The item exists in the list!elifitem<nums[mid]:returnbinary_search(nums,low,mid-1,item)else:returnbinary_search(nums,mid+1,high,item) The new function should be prototyped below. The number of changes between the given version, and the one requested is not significant. defbinary_search(nums,item):pass# Remove this and fill in with your code Tip: If you consider that high and low are used to create a smaller version of our problem to be processed recursively, the version requested will do the same thing, just through a different, more Pythonic technique.The following function f uses recursion: def f(n): if n <= 1 return n else return f(n-1) + f(n-2) Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion? a) def f(n): a <- 0 b <- 1 if n = 0 return a elsif n = 1 return b else for i in 1..n c <- a + b a <- b b <- c return b b) def f(n): a <- 0 i <- n while i > 0 a <- a + i + (i-1) return a c) def f(n): arr[0] <- 0 arr[1] <- 1 if n <= 1 return arr[n] else for i in 2..n arr[i] <- arr[i-1] + arr[i-2] return arr[n] d) def f(n): arr[0..n] <- [0, ..., n] if n <= 1 return arr[n] else a <- 0 for i in 0..n a <- a + arr[i] return a