Your program is running in O(n²) time, since you have two nested loops that both depend on the size of the input. So once you go from 10,000 to 1,000,000 elements, your program will take 100² = ten thousand times longer to complete. Furthermore, it can be that your dataset fit in the processor’s cache … Read more
Step 1: Combine all the arrays Step 2: Sort arrays using Arrays.sort(array); Step 3: Remove the duplicates. int[] a = {1,2,3,4,5}; int[] b = {1,2,3,4,5,6}; int[] c = {1,3,7}; int[] d = {2,3,4,8,9,10}; int[] resultArray1 = new int[a.length+b.length+c.length+d.length]; int arrayIndex = 0; for (int i=0; i< a.length ; i++, arrayIndex++ ) { resultArray1[arrayIndex] = a[i]; … Read more
int size=sizeof(sequence); You probably meant int size=sequence.size(); which actually returns the number of elements in the std::vector container. vector <int> s1; for(i=0;i<size;i++) { s1[i]=stoi(sequence[i]); } Here the problem is that the vector s1 is initialized as empty, so that you cannot change the i-th value (because it does not exist). Give the std::vector directly the … Read more
Your code is not right: it does not do a stable partition of the array. Imagine your array is 100, 33, 333, 22, 222, 11 After the stable partitioning around the first element, it should become 33, 22, 11, 100, 333, 222 // 33, 22, and 11 in the same order as in the original … Read more
Simplest will be bubble sort. item* sort(item *start){ item *node1,*node2; int temp; for(node1 = start; node1!=NULL;node1=node1->next){ for(node2 = start; node2!=NULL;node2=node2->next){ if(node2->draw_number > node1->draw_number){ temp = node1->draw_number; node1->draw_number = node2->draw_number; node2->draw_number = temp; } } } return start; }
Yes, instead of having an array of primitive types (like ints) or Strings, define your own class and then create an array of that. Then you can have whatever information you want in there. Alternatively, if the array must be an array of primitives or Strings because some other method requires it to be in … Read more
First of all a, e and g are variables and not necessarily strings. What you mean is probably “a”, “e” and “g”. Your question is essentially: How do I sort the values in a String[] so that numbers are sorted numerically and prioritized before letters (or words?) which are sorted alphabetically? Your question seems incomplete, … Read more
I could not post it to comment section. Please see below: List<Integer> integers = Arrays.asList(1, 2, 3, 4); int i = 0; StringBuilder sb = new StringBuilder(“[“); for (int value : integers) { sb.append((i == 0) ? “” : (i % 2 == 0 ? “,” : “:”)).append(value); i++; } sb.append(“]”); System.out.println(sb.toString()); Output is: [1:2,3:4]
array_count_values will output the count as like Array ( [keyword1] => 10 [keyword2] => 3 [keyword3] => 6 [keyword4] => 5 [keyword5] => 1 ) But For your desired output you need to use foreach Demo $arra = Array ( 0 => “keyword1”, 1 => “keyword1”, 2 => “keyword1”, 3 => “keyword1”, 4 => “keyword1”, … Read more