http://www.practice.geeksforgeeks.org/problem-page.php?pid=493 Sorting Elements of an Array by Frequency Given an array of integers, sort the array according to frequency of elements. For example, if the input array is {2, 3, 2, 4, 5, 12, 2, 3, 3, 3, 12}, then modify the a...
http://www.practice.geeksforgeeks.org/problem-page.php?pid=380 Largest Number formed from an Array Given a list of non negative integers, arrange them in such a manner that they form the largest number possible. The result is going to be very large, hence return the result in the form ...
Count the number of occurrences in a sorted array Given a sorted array arr[] and a number x, write a function that counts the occurrences of x in arr[]. Expected time complexity is O(Logn) Examples: Input: arr[] = {1, 1, 2, 2, 2, 2, 3,}, x = 2 Output: 4 // x (or ...
packagecom.javacodegeeks.snippets.core; importjava.util.zip.CRC32; importjava.util.zip.Checksum; publicclassCalculateCRC32ChecksumForByteArray { publicstaticvoidmain(String[] args) { String input ="Java Code Geeks - Java Examples"; // get bytes from string bytebytes[] = input.getBytes();...
The Map object is perfect for maps, or associative arrays as we call them in the PHP biz. But what about true, indexed arrays? Well actually, JavaScript has always had a great way to handle these - it's not new! It's the Array object. Well, the Array object isn't new, but it ...
He rotates the array clockwise i.e. after rotation the array A = {6,1,2,3,4,5} and delete the last element that is {5} so A = {6,1,2,3,4}. Again he rotates the array for the second time and deletes the second last element that is {2} so A = {4,6,1,3}, doing ...
Input: arr[] = {1, 1, 2, 2, 2, 2, 3,}, x = 4 Output: -1 // 4 doesn't occur in arr[] Method 1 (Linear Search) Linearly search for x, count the occurrences of x and return the count. Time Complexity: O(n) Method 2 (Use Binary Search) ...
able to find that the minimum length unsorted subarray lies between the indices 3 and 7. Solution 1. O(n * logn) runtime, O(n) space 1. make a copy of the given array. 2. sort the copy. 3. find the first and last indices where the given array and its sorted copy don't match...
1publicvoidrotateByOne(int[] arr) {2if(arr ==null|| arr.length == 0){3return;4}5inttemp = arr[arr.length - 1];6for(inti = arr.length - 1; i > 0; i--){7arr[i] = arr[i - 1];8}9arr[0] =temp;10} Follow up question: instead of rotating the given array by one po...
1. get the middle element and create root node N. 2. recursively create a balanced BST from the left half of the array and set its root as N's left node. 3. recursively create a balanced BST from the right half of the array and set its root as N's right node. ...