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 t...
/*if x is present in arr[] then returns the count of occurrences of x, otherwise returns -1.*/intcount(intarr[],intx,intn) {inti;//index of first occurrence of x in arr[0..n-1]intj;//index of last occurrence of x in arr[0..n-1]/*get the index of first occurrence of x...
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 2) occurs 4 times in arr[] Input: arr[] = {...
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 ...
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 ...
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...
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 ...
3. copy temp to the first element. O(n) runtime, O(1) space 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...
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. ...