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Yes, an array that is in sorted order is considered a min-heap because the smallest item in the array is the root. Also, the rest of the items in the array will gradually get bigger from the root until the end of the array.
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Is an array that is in sorted order a min-heap?
Yes, an array that is in sorted order is considered a min-heap because the smallest item in the array is the root. Also, the rest of the items in the array will gradually get bigger from the root until the end of the array.
What is the difference between binary heap and binomial heap?
The difference between Binomial heap and binary heap is Binary heap is a single heap with max heap or min heap property and Binomial heap is a collection of binary heap structures(also called forest of trees).
How to find max and min values?
Find the minimum and maximum of what? An array?
What are the key differences between a heap and a binary search tree (BST)?
A heap is a complete binary tree where each node has a value greater than or equal to its children (max heap) or less than or equal to its children (min heap). A binary search tree is a binary tree where the left child of a node has a value less than the node and the right child has a value greater than the node. The key difference is that a heap does not have a specific order between parent and child nodes, while a binary search tree maintains a specific order for efficient searching.
Write a java programme to find minimum number in the given array?
public int min(int[] arr) { int min = Integer.MAX_VALUE; for(int e : arr) if(e<min) min=e; return min; }
What is the purpose and functionality of a median heap in data structures and algorithms?
A median heap is a data structure used to efficiently find the median value in a set of numbers. It combines the properties of a min heap and a max heap to quickly access the middle value. This is useful in algorithms that require finding the median, such as sorting algorithms and statistical analysis.
What are the key differences between a binary tree and a heap in terms of their structure and functionality?
A binary tree is a data structure where each node has at most two children, while a heap is a specialized binary tree with specific ordering properties. In a binary tree, the structure is more flexible and can be balanced or unbalanced, while a heap follows a specific order, such as a min-heap where the parent node is smaller than its children. Functionally, a heap is commonly used for priority queues and efficient sorting algorithms, while a binary tree is more versatile for general tree-based operations.
How is the bottom-up heap construction process implemented in data structures and algorithms?
In the bottom-up heap construction process, a heap is built by starting with individual elements and gradually combining them into a complete heap structure. This is done by repeatedly "heapifying" smaller sub-heaps until the entire heap is formed. The process involves comparing elements and swapping them if necessary to maintain the heap property, which ensures that the parent node is always greater (for a max heap) or smaller (for a min heap) than its children. This method is commonly used in data structures and algorithms to efficiently create and maintain heap structures.
Q1 Find the minimum and the maximum number of keys that a heap of height h can contain?
Q1. Find the minimum and the maximum number of keys that a heap of height h can contain.
Difference between insertion and selection sort?
Before we examine the differences, let us examine the implementations. The code that follows is written in C++, but can be easily adapted to other languages. Firstly, let us assume we have an array of num elements (we'll use integers here, but typically you will use pointers to maintain efficiency): int array[num]; A selection sort has the following nested loop structure: for( int outer=0; outer < num-1; ++outer ) { int min = outer; for( inner=outer+1; inner < num; ++inner ) { if( array[inner] < array[min] ) min = inner; } if( min != outer ) swap( array[min], array[outer] ); } Whereas an insertion sort has the following nested loop structure: for( index=1; index < num; ++index ) { value = array[index]; hole = index; while( hole > 0 && value < array[hole-1] ) array[hole] = array[--hole]; array[hole] = value; } Although both algorithms split an array into two subsets (a sorted subset to the left and an unsorted subset to the right) and both move one element from one set to the other on each iteration of the outer loop, the selection sort begins with an empty sorted subset while the insertion sort begins with a sorted subset of one element (any array with only one element can always be regarded as being sorted). In both cases, the outer loop keeps track of the initial index of the current unsorted subset and both perform a complete traversal of the unsorted subset. However selection sort stops when there is only one unsorted element left, which automatically becomes the largest value in the sorted set and therefore doesn't need to move, whereas insertion sort traverses the entire unsorted subset. The key difference is in the inner loops. With selection sort, the inner loop skips the first unsorted element and traverses the remainder of the unsorted subset locating the index of the lowest value (recorded by min). At the end of the inner traversal, if min is not the same as the outer loop index then the values are swapped. Thus the sorted subset gains a new element containing the next largest value on each iteration, beginning with the lowest value. With insertion sort, the outer loop extracts the first unsorted element's value and stores its index (hole). There isn't really a hole in the array, it's simply a marker to determine where the extracted value will be inserted. The inner while loop then traverses the sorted subset. On each iteration, if the hole marker is non-zero and the value is less than the value of the element to the left of the hole marker, then that element's value is copied into the hole to its right and the hole marker moves left. When the inner loop ends, the hole marker denotes where the extracted value should be placed. Thus if the extracted value is greater than or equal to the largest sorted value, then the value doesn't move (it was already sorted), otherwise it is inserted into its correct position within the sorted subset. It can be proved that although selection sort only requires one swap at most for every iteration of the outer loop, it must perform (num-1)! comparisons in total (that is, for num=6 there are 5+4+3+2+1 comparisons in total). It should be noted that every swap incurs three copy processes, however a single copy takes less time than a single comparison. By contrast, insertion sort stops comparing when the insert point is located, thus, on average, there will be fewer comparisons than (num-1)!. However, while every iteration of the outer loop incurs two copy processes even if the current element doesn't need to move, and additional copy processes for each movement of the hole marker, it's wrong to assume selection sort is always faster. On average, it is actually slower. Looking at the worst case example, let us suppose the array is in reverse order. Selection sort requires (num-1)! comparisons no matter what and will also incur n-1 swaps in total (which is 3(num-1) copies in total). By contrast, insertion sort also requires (num-1)! comparisons but requires 2(num-1)+(num-1)! copy processes. The end result is that selection sort performs slightly better, and will improve as the array size increases. Another worst case example is the already-sorted array. Again, selection sort requires (num-1)! comparisons but no swaps at all. However, insertion sort only requires num-1 comparisons and 2(num-1) copies. In this case, insertion sort performs slightly better, and will improve as the array size increases. While both algorithms are fairly efficient when dealing with small arrays (typically up to 20 elements), with insertion sort performing better on average, they are highly inefficient when dealing with larger arrays. For this you need recursive, logarithmic, divide-and-conquer algorithms, such as quicksort, which are capable of sorting num elements with only (log num) comparisons. However, these algorithms don't perform well with smaller subsets, so it pays to combine the two into a hybrid algorithm. That is, when a logarithmic algorithm reduces a subset to 20 elements or less, then that subset can be sorted far more efficiently with an insertion sort.
C plus plus prog a function min that returns the minimum of all Values in the given array int min int list int arraySize?
int min(int list[], int arraySize) { int min=arraySize?list[0]:0; for(int i=1; i<arraySize; ++i ) if(list[i]<min) m=list[i]; return(min); }
What is the c plus plus programme for merge sort without recursion?
The following template function will sort an array of any type using the merge sort algorithm, non-recursively. Note that you must call the function by passing a pointer to the array, rather than passing the array by reference as you normally would with sorting algorithms. This is because the function uses a second array (a work array) of the same size to perform the actual merge, switching from one to the other upon each iteration, so there's no guarantee the final sorted array will be the one you originally referred to (it could be the work array). You could maintain a count of the iterations and perform an extra copy of the work array back to your referenced array if the count is odd, but it's more efficient to simply return the sorted array through a pointer. The code can be difficult to follow if you're not familiar with the merge sort algorithm, so I've commented the code verbosely to explain what's going on. In essence, we're treating the original array as being several subsets of 1 element each. A subset of 1 is already sorted, so we simply merge these subsets in pairs, so the new array contains sorted subsets of 2 elements each. We repeat the process, merging each pair of subsets to create sorted subsets of 4, then 8 and so on, until there is only 1 sorted subset in the array. When we merge two subsets into one, we simply compare the first element of each subset and place the lowest into the merged subset. Note that in the final iteration, the second subset of the pair may be smaller than the first, or it may not exist at all (in which case the remaining subset may be smaller). However, the merge algorithm caters for this eventuality quite efficiently. If there's only one subset remaining (which can happen during any iteration as elements are removed from the subsets and merged into the work array), the remaining elements are simply copied sequentially from that subset since they will already be in order (as determined by the previous iteration). template<typename T> void merge_sort(T* A[], size_t size) { // Arrays of length 0 or 1 are already sorted. if( size < 2 ) return; // Instantiate a new array of the same size as A. T *B = new T[size]; // Array A initially contains subsets of width 1 element, then 2, 4, 8... for( size_t width=1; width<size; width<<=1 ) { // Dereference A (now known as C). T *C = *A; // Array B initially holds subsets of width 2 elements, then 4, 8, 16... size_t width2 = 2*width; // Iterate through each pair of subsets in C... // sub1 is the start index of the first subset of the pair in C. for( size_t sub1=0; sub1<size; sub1+=width2 ) { // sub2 is the start index of the second subset of the pair in C. size_t sub2 = min( sub1+width, size ); // next is the start of the next pair of subsets in C. size_t next = min( sub1+width2, size ); // Start with the first elements in each pair of subsets in C // (the ones with the lowest values in each subset). size_t c1 = sub1; size_t c2 = sub2; // Iterate through B's elements from index [sub1] to index [next-1]. for( size_t b=sub1; b<next; ++b ) // Copy the lowest of the two indexed values in C to B. B[b] = ( c1<sub2 && ( c2>=next C[c1]<=C[c2] )) ? C[c1++] : C[c2++]; } // Swap the roles of *A and B (note that C still points to the dereferenced A at this point). *A=B; B=C; } // Finished with B (*A is now the fully-sorted array). delete[]B; }
