Software Developer

Store elements and their counts in a hash table. After storing counts, we traverse input array again and print those elements whose counts are more than once. To make sure that every output element is printed only once, we set count as 0 after printing the element.

Traverse the complete matrix and search for the target element. If the target element is found, return true. Otherwise, return false.

Use any sorting algorithm such as selection sort, bubble sort, insertion sort, or merge sort or quick sort.

The idea is to use a hash map to store the count of all values. Then iterate through the map and return the value with count 1.

Start from the leftmost element and keep track of the index of smaller (or equal) elements as i . While traversing, if we find a smaller element, we swap the current element with arr[i]. Otherwise, we ignore the current element.

I used binary search for solving this question.

Sort the given array in descending order and return the element at the index K - 1 (zero-based indexing).

Traverse linked list using two pointers.
Move one pointer(slow) by one step ahead and another pointer(fast) by two steps ahead.
If these pointers meet at the same node then there is a loop. If pointers do not meet then the linked list doesn’t have a loop.