If one element in the array is larger than all of the items to its right, it is recognised as a leader. Find all the leaders from a set of elements. For example, in the array A = [1 8 6 7 3 1 5 2], the numbers 8, 7, and 5 are leaders.
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If one element in the array is larger than all of the items to its right, it is recognised as a leader. Find all the leaders from a set of elements.
For example, in the array A = [1 8 6 7 3 1 5 2], the numbers 8, 7, and 5 are leaders.
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Solved in 3 steps
- Given an array, find the next greater element for each element in the array, ifavailable. If not available, print the element itself. The next greater element y for anelement x in the array is the first element that is greater than x and occurs on its rightside. The next greater element of the right most element in an array is the elementitself.Example: Given A = [ 6 8 4 3 9] the next greater element listB = [8 9 9 9 9]An element that is "peak" in an array of integers is one that is more than or equal to the adjacent integers, while an element that is "valley" is one that is less than or equal to the surrounding numbers. For instance, in the array, the values 5, 8, 6, 2, 3, 4, 6 are peaks, whereas the values 5, 2 are valleys. Sort an integer array into an alternating pattern of peaks and valleys given the array of numbers. EXAMPLEInput: {5, 3, 1, 2, 3}Output: {5, 1, 3, 2, 3}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
- If an array is given, identify the next bigger element for each element in the array, if one exists. If the element is not accessible, print the element itself. The next bigger element y in the array for an element x is the first element that is greater than x and appears on its right side. The element itself is the next bigger member of the array's rightmost element.For instance, if A = [6 8 4 3 9], the next bigger element list B = [8 9 9 9 9].Write a program in java that randomly fills in 0s and 1s into an n-by-n matrix, prints the matrix, and finds the rows and columns with the most 1s. (Hint: Use two ArrayLists to store the row and column indices with the most 1s.)given an array of integer values , return true if 6 appears as either the first or last element in the array. firstLast6([1,2,36]) true firstLast([6,1,2,3]) true firstLast([13,6,1,2,3]) false
- : In an array of integers, a "peak" is an element which is greater than or equal tothe adjacent integers and a "valley" is an element which is less than or equal to the adjacent integers. For example, in the array {5, 8, 6, 2, 3, 4, 6}, {8, 6} are peaks and {5, 2} are valleys. Given an arrayof integers, sort the array into an alternating sequence of peaks and valleys.EXAMPLEInput: {5, 3, 1, 2, 3}Output: {5, 1, 3, 2, 3}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.For this exercise, you need to shift all the elements in the array one space to the right (increase the index of the element by 1). The last element will wrap around and become the new first element. For example, given: moveUp({"first", "second", "third", "fourth"}); You would return: {"fourth", "first", "second", "third"} in java
- A magic square is a square array of non-negative integers where the sums of the numbers on each row,each column, and both main diagonals are the same. An N × N occult square is a magic square with Nrows and N columns with additional constraints:• The integers in the square are between 0 and N, inclusive.• For all 1 ≤ i ≤ N, the number i appears at most i times in the square.• There are at least two distinct positive integers in the square.For example, the following is a 5 × 5 occult square, where the sums of the numbers on each row, eachcolumn, and both main diagonals are 7:0 0 0 3 42 4 0 0 10 0 3 4 05 0 0 0 20 3 4 0 0For a given prime number P, you are asked to construct a P × P occult square, or determine whether nosuch occult square exists.InputInput contains a prime number: P (2 ≤ P ≤ 1000) representing the number of rows and columns in theoccult square.OutputIf there is no P × P occult square, simply output -1 in a line. Otherwise, output P lines, each contains Pintegers…Removing an item from the middle of a large ArrayList (n items) is a single operation. What is the time complexity of this operation?What will be the problem with the following algorithm for finding an item in an unsorted array? How can you solve it? int location = 0; bool moreToSearch = (location > length); found = false; while (moreToSearch && !found) { if(item == info[location]) { item = info[location]; } else { location++; moreToSearch = (location > length); } }