Java code for getting count in the sorted ArrayList for particular range

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Need to find the count of the elements in the sorted ArrayList from the given range(x,y). And the count should have count of range elements also if it is in ArrayList.

So for, I have done this by traversing the whole list and getting the count.


count = 0; 
for (i=0; i<length(list); i++)
if (list[i]>= startrange and list[i]<=endrange)
    count = count+1;

Current solution is taking more time because input array size is more than 1000000. Help me to optimize the solution.


Input array looks like this [1,4,5,8,9,12,16,19,23,26,28,29,30,31,33,35,37].

Input range: (12,30)

Output should be like 8

Use binary search to find the first element in the range and start your counting after it.

Java Program for Counting Sort, Counting sort is a sorting technique based on keys between a specific range. It works by counting the number of objects having distinct key values (kind of hashing). Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. 2) Sort arraylist of integers. Java program to sort arraylist of integers using different sorting techniques discussed above. All methods work exactly same for integers as well. //Unsorted list List<Integer> numbers = Arrays.asList(1, 3, 5, 4, 2); //1.

Queries for counts of array elements with values in given range , Queries for counts of array elements with values in given range Simple C++ program to count number of elements sort(arr, arr + n);. Java ArrayList. The ArrayList class is a resizable array, which can be found in the java.util package.. The difference between a built-in array and an ArrayList in Java, is that the size of an array cannot be modified (if you want to add or remove elements to/from an array, you have to create a new one).

int cnt=Stream.of(arr).filter(o->(o>=12&& o<=30)).count();

The java.util.ArrayList.removeRange(int fromIndex, int toIndex) method removes from this list all of the elements whose index is between fromIndex, inclusive, and toIndex, exclusive. Shifts any succeeding elements to the left and reduces their index. Declaration. Following is the declaration for java.util.ArrayList.removeRange() method

Following Java 8 Stream one-liner will work fine & return the result as expected without using cumbersome for-loop .

int[] xyz = { 1, 4, 5, 8, 9, 12, 16, 19, 23, 26, 28, 29, 30, 31, 33, 35, 37 };
long elementCountWithinRange = -> (x > 12 && x <= 31)).count();
    System.out.println(elementCountWithinRange); // will return 8

Note : Earlier similar answer given by @Gaurav Dhiman is incorrect as the expression won't compile as count() method returns a long and not an int . Also , even if you resolve that it will give below error :

The operator >= is undefined for the argument type(s) int[], int

To resolve that i have used instead of Stream.of() to create a Stream .

Count the occurrences of an element in an array in Java. We will be performing the below steps to count the occurrence. As a first step we will be creating a HashMap “countMap” to hold the element (Key) and the count as the value. For each of the element in the input array, check if it is present in the countMap, using containsKey() method.

Let us compile and run the above program, this will produce the following result − Number = 15 Number = 22 Number = 30 Number = 40 Retrieved element is = 40 java_util_arraylist.htm

In this program, instead of using a while loop, we use a for loop without any body. On each iteration, the value of num is divided by 10 and count is incremented by 1. The for loop exits when num != 0 is false, i.e. num = 0. Since, for loop doesn't have a body, you can change it to a single statement in Java as such: for(; num != 0; num/=10

The subList() method of java.util.ArrayList class is used to return a view of the portion of this list between the specified fromIndex, inclusive, and toIndex, exclusive. . (If fromIndex and toIndex are equal, the returned list is emp

  • So the array is sorted? In all cases?
  • Use binary search to find index of12, then again binary seacrh for 30. The difference 13 - 5 == 8 is the range; time complexity is O(log(n))
  • @Robby Cornelissen: depends on binary search implementation. Often, binary search either returns the index of item (>= 0) or negative value which is the 2-complement of index of an item which is smaller then required
  • @DmitryBychenko Good point! Totally forgot that Arrays.binarySearch() will return the insertion point if element not found. If not all values are unique, results might still be wrong though...
  • Hi guys, this list is sorted for sure. And there is a chance that values may get repeats and min rangevalue's presence in the list is not guaranteed.
  • I can all but guarantee that that solution will be even slower.
  • While this code may provide a solution to the question, it's better to add context as to why/how it works. This can help future users learn, and apply that knowledge to their own code. You are also likely to have positive feedback from users in the form of upvotes, when the code is explained.