Given an array of integers, find the first missing positive integer in linear time and constant space. In other words, find the lowest positive integer that does not exist in the array. The array can contain duplicates and negative numbers as well. For example, the input [3, 4, -1, 1] should give 2. The input [1, 2, 0] should give 3 . You can modify the input array in- place.
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- reversing the elements of an array involves swapping the corespodning elements of the array: the first with the last, the second with the next to the last, and so on, all the way to the middle of the array. given an array a, an int variable n containing the number of elements in a, and two other int variables, k and temp, write a loop that reverses the elements of the array. do not use any other variables besides a, n, k, and temp. in c++You 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 aprogram that, given an array of N distinct int values in ascending order, determineswhether a given integer is in the array. You may use only additions and subtractionsand a constant amount of extra memory. The running time of your program should beproportional to log N in the worst case.
- Write an c progra, that takes an array of N integers (N will be specified by the user at run time). It then finds how many numbers in the array are divisible by 3 and how many numbers are divisible by 7. the it copies, from the input array above, the numbers that are divisible by 3 into a new array and those divisible by 7 into a different new array. Once copied, the algorithm should compute the average of each of these two new arrays and print the averages on the screen with suitable message to let the user know which average is being printed Note: an integer number is divisible by 3 (respectively 7) if the remaining of its division by 3 (respectively 7) is 0. You are allowed to use the modulus operator %.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].You are given an array with all the numbers from 1 to N appearing exactly once,except for one number that is missing. How can you find the missing number in O(N) time and 0( 1) space? What if there were two numbers missing?
- 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.Write a function that gets an array of ints of length n, and a number k, and returns the longest subsequence of consecutive k’s in the array. For example - On input ([1, 1, 3, 3, 3, 1, 5, 1], 1), the function returns 3 - On input ([1, 3, 3, 2, 3, 3, 3, 3], 3), the function returns 4 - On input ([3, 5, 2, 5, 2, 5, 5], 3), the function returns 2 - On input ([1, 2, 3, 3, 3, 3], 5), the function returns 0 // returns the length of the longest subsequence // of consecutive k’s in the array int longest_seq(const int* ar, int n, int k); test for the function: void test_q1() { inta1[] = {1, 1, 3, 3, 3, 1, 5, 1}; if (longest_seq(a1, 8, 1) == 2) printf("Q1-1 ok\n"); else printf("Q1-1 ERROR\n"); inta2[] = {1,3,3,2,3,3,3,3}; if (longest_seq(a2, 8, 3) == 4) printf("Q1-2 ok\n"); else printf("Q1-2 ERROR\n"); }Given an ArrayList of integers, return an ArrayList where all the odds are before the evens in the order that they occur. public ArrayList<Integer> solution(ArrayList<Integer> nums) {ArrayList<Integer> finalList = new ArrayList<>();int index = 0;for (int i = 0; i < nums.size(); i++) {if (nums.get(i) % 2 == 1) {finalList.add(index, nums.get(i));index++;} elsefinalList.add(nums.get(i));}return finalList;} This is my failing code. for [1.-3.3] where [1,3,-3] is being returned
- Write a program that randomly fills in 0s and 1sinto an n-by-n matrix, prints the matrix, and finds the rows and columns with themost 1s. Here is a sample run of the program: Enter the array size n: 4 ↵EnterThe random array is0011001111011010The largest row index: 2The largest column index: 2, 3Write a program that reads an integer array a[] from the console and writes out the subsequence of the array that sums to the largest value. If there are more than one such subsequence, then the program should write out the subsequence that starts at the smallest position in the array and, if there are more than one subsequence of largest sum starting at the same smallest position, it should write out the shortest such subsequence.Given an array, find the total number of inversions of it. If (i < j) and (A[i] > A[j]), then pair (i, j) is called an inversion of an array A. We need to count all such pairs in the array. For example, Input: A[] = [1, 9, 6, 4, 5] Output: The inversion count is 5 There are 5 inversions in the array: (9, 6), (9, 4), (9, 5), (6, 4), (6, 5)