Austin Ian

#include class listNode int value; listNode next; listNode(int x) value = x; next = null; class remove_linked_list_elements { public static void print(listNode head) { if(head { listNode head = new listNode(1); head.next = new listNode(2); head.next.next = new listNode(2); head.next.next.next = new listNode(3); head.next.next.next.next = new listNode(4); int val = 2; print(removeElements(head , val)); }1 3 4 O(N), #include class listNode int value; listNode next; listNode(int x) value = x; next = null; class remove_linked_list_elements { public static void print(listNode head) { if(head ; return dummy.next; public static void main(String args[]) { listNode head = new listNode(1); head.next = new listNode(2); head.next.next = new listNode(2); head.next.next.next = new listNode(3); head.next.next.next.next = new listNode(4); int val = 2; print(removeElements(head , val)); 1 3 4 O(N), as we iterate the whole list once.

[] queries = {{1 , 0} , {-3 , 1} , {-4 , 0} , {2 , 3}}; int [][] queries) { int evenSum = 0; for(int x : A) { if(x % 2 == 0) { evenSum += x; int n 2 == 0) evenSum += A[idx]; ans[i] = evenSum; return ans; }8 6 2 4 O(N + Q), where, N = size of the given array. Queries for GCD of all numbers of an array except…Queries for Number of Distinct Elements in a SubarrayQueries on Probability of Even or Odd Number in given RangesQueries on XOR of greatest odd divisor of the rangeQueries for counts of array elements with values in…Queries for Decimal Values of Subarrays of a Binary ArrayFind Maximum Sum Possible Equal Sum of Three StacksFind

We call a string balanced if it has the same number of ‘R’s and ‘L’s. The goal is to find the maximum possible number of balanced split strings. Now, we need to maximize this number of splits to get max number of partitions. ; class balanced_splits { public static int balancedStringSplit(String s) { int balance = 0 , splits = 0; for(int i = 0 ; i < s.length() ; i++) { char c = s.charAt(i); balance +=

["KthLargest", "add", "add", "add", "add", "add"] [[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]4 5 5 8 8 And since we bound the heap size to be equal to k after every query, this top element would be the Kth largest in the overall *; class comp implements Comparator4 5 5 8 8 O(N + Q), where N = size of the initial array (while calling the constructor). when modification to BST…Find Top K (or Most Frequent) Numbers in a StreamFind Largest d in Array such that a + b + c = dLargest subarray with equal number of 0s and 1sLength of the largest subarray with contiguous elements