Let's say that B is an array with a size of n > 6 that has integers from 1 to n-5, inclusive, in it. There are exactly five repeats in B. Explain how to find the five numbers in B that are the same.
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Let's say that B is an array with a size of n > 6 that has integers from 1 to n-5, inclusive, in it. There are exactly five repeats in B. Explain how to find the five numbers in B that are the same.
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- Let B be an array of size n ≥ 6 containing integers from 1 to n−5 inclusive, five of which are repeated. Write and test a method for finding the five integers in B that are repeated. What is the running time of this algorithm? Hint: Sort the array B then scan it from front to end looking for the repeated entriesThe key observation is as follows. If n/2 < A[n/2], then if such anindex exists, it must be in A[1 . . . n/2−1]. This is because A[n/2+1] > A[n/2]+1 >n/2 + 1, A[n/2 + 2] > A[n/2 + 1] + 1 > n/2 + 2 and so on Could you explain why these are true by using actual number and example of array ?A[n/2+1] > A[n/2]+1 > n/2 + 1, A[n/2 + 2] > A[n/2 + 1] + 1 > n/2 + 2 and so onLet B be an array that is of size n >= 6 containing integers from 1 to n-5, inclusive, with exactly five repeated. Describe an O(n) algorithm for finding the five integers in B that are repeated.
- Imagine that we want to keep track of friendships between n people. We can do this with an array of size nXn. Each row of the array represents the friends of an individual, with the columns indicating who has that individual as a friend. For example, if person j is a friend of person i, then we place a mark in column j of row i in the array. Likewise, we should also place a mark in column i of row j if we assume that friendship works both ways. What will be the space complexity of this problemYou are given an array in which every number from 1 to N appears precisely once with the exception of one. How is the missing number to be located in O(N) time and 0(1) space? What if two numbers were absent?Write a program that reads an array A of N elements containing only 0's and 1's. Your program should find the position of a 0 and replace it with a 1 to get the longest continuous sequence of 1's. Let this position of 0 be called P. Print P if such a 0 exists and print -1 if the original array A contains only 1's. Assume the array indexing starts from 0.
- Let A be an array of numbers. In the maximum sub-array problem, your goal is to determine the sub-array A[x . . . y] of consecutive terms for which the sum of the entries is as large as possible. For example, if A = [−2, −3, 4, −1, −2, 1, 5, −3], the maximum sub-array is [4, −1, 2, 1, 5], and the largest possible sum is S = 4 − 1 − 2 + 1 + 5 = 7. Suppose A = [1, 2, −4, 8, 16, −32, 64, 128, −256, 512, 1024, −2048]. Determine S, the largest possible sum of a sub-array of A.We need to merge 4 ordered arrays with sizes a, b, c, and d, respectively. We needto get an ordered array containing all these elements for the worst case. It is knownthat 0 < a < b < c < d and a + b < d. What is the best solution for the minimumnumber of comparisons you can get? no hand writtenWrite a program (in mars ) maxarray that computes the double of maximum of values of any given array of length L=7. The computation of the maximum is done in a procedure. For example : if the values in the array are : 14, 5, 85, 17, 9, 1, 47. The result is 170.
- Given an integer n and an array a of length n, your task is to apply the following mutation to a: Array a mutates into a new array b of length n. For each i from 0 to n - 1, b[i] = a[i - 1] + a[i] + a[i + 1]. If some element in the sum a[i - 1] + a[i] + a[i + 1] does not exist, it should be set to 0. For example, b[0] should be equal to 0 + a[0] + a[1].Write a program that initialize square matrix of size N x N and check if that array is symmetric or not. The array is symmetric if array[row][column] == array[column][row].Let M(n) be the minimum number of comparisons needed to sort an array A with exactly n ele- ments. For example, M(1) = 0, M(2) = 1, and M(4) = 4. If n is an even number, clearly explain why M(n) = 2M(n/2) + n/2.