To detect the duplicate, you will have to write a nested loop that compares each element with all the previous elements.
To actually delete the duplicate, once you find it, you have to move over all the elements after the duplicate. If the order of the elements doesn't matter, it is faster to just move the LAST array element, overwriting the duplicate element. Use a variable to keep track how many elements of the array are "usable". For example, if your array had 10 elements, and you delete 1, the array size will still be 10... but (after moving the elements over) only 9 of those elements have useful information.
The only way to delete objects in an object oriented programming language (unless they were created in heap memory) is for the object to go out of scope. If the object is declared in the heap, in c++ you would use delete[] ptr; or delete ptr; where ptr is a pointer to your object.
n-1 times
binary search system
algorithm on multiple queues in a single dimensional array
O(N) where N is the number of elements in the array you are searching.So it has linear complexity.
// Assuming you dynamically allocated this array using "new"... delete array[arraysize - 1]; arraysize--;
There is no such condition. The algorithm to locate the second largest number in an array of numbers is: Sort the array in descending order. Remove the duplicate values from the array. If there are two or more elements remaining, return the second number. If there is less than two elements remaining, return the NaN value (not a number).
You cannot delete elements from an array. But you can move the elements: if (del_index < no_of_elements-1) { memmove (&array [del_index], &array [del_index+1], sizeof (array [0]) * (no_of_elements - del_index - 1)); } --no_of_elements;
The only way to delete objects in an object oriented programming language (unless they were created in heap memory) is for the object to go out of scope. If the object is declared in the heap, in c++ you would use delete[] ptr; or delete ptr; where ptr is a pointer to your object.
The inplace quicksort algorithm efficiently sorts elements in an array by recursively dividing the array into smaller subarrays based on a chosen pivot element. It then rearranges the elements so that all elements smaller than the pivot are on one side, and all elements larger are on the other. This process is repeated until the entire array is sorted. The algorithm's efficiency comes from its ability to sort elements in place without requiring additional memory allocation for new arrays.
The best search algorithm to use for an unsorted array is linear search. It involves checking each element in the array one by one until the desired element is found. This algorithm has a time complexity of O(n), where n is the number of elements in the array.
To implement the quicksort algorithm with a 3-way partition in Java, you can modify the partitioning step to divide the array into three parts instead of two. This involves selecting a pivot element and rearranging the elements so that all elements less than the pivot are on the left, all elements equal to the pivot are in the middle, and all elements greater than the pivot are on the right. This approach can help improve the efficiency of the quicksort algorithm for arrays with many duplicate elements.
The runtime complexity of the mergesort algorithm is O(n log n), where n is the number of elements in the input array.
In a binary search algorithm, typically log(n) comparisons are made when searching for a specific element in a sorted array, where n is the number of elements in the array.
In a binary search algorithm, typically log(n) comparisons are required to find a specific element in a sorted array, where n is the number of elements in the array.
The maximum number of comparisons required in a binary search algorithm to find a specific element in a sorted array is log(n), where n is the number of elements in the array.
The running time of the bubble sort algorithm is O(n2), where n is the number of elements in the array being sorted.