Search⌘ K
AI Features
Home
Courses
Algorithms for Coding Interviews in Java

Solution: Count Element Occurrence

Explore different algorithmic solutions for counting the occurrences of an element in an array. Learn the brute force approach with linear search, a binary search method that counts contiguous duplicates, and a modified binary search to find first and last occurrence indices. Understand the time complexities and trade-offs of each solution in the context of sorting and searching techniques using Java.

We'll cover the following...

Solution #1: Brute force with linear search

Java
class Count {
 public static int calcFreq(int arr[], int key) {
  int count = 0;
  //traversing the array
  for (int i = 0; i < arr.length; i++) {
   //checking if current index value matches key
   if (arr[i] == key)
    //incrementing count if it does
    count++;
  }
  return count;
 }
 public static void main(String args[]) {
  int arr[] = {-5,-3,0,1,3,3,3,3,4,5};
  int key = 3;
  System.out.println("The key \"" + key + "\" occurs " + calcFreq(arr, key) + " times in the Array.");
 }
}

This is an extremely simple way to solve this problem. We simply initialize a variable, count, to keep count and set it to 0 (line 3). We then iterate over the array (line 5), increasing count by one every time the key (value to search) is encountered (lines 7-9).

Time complexity

The time complexity of this algorithm is in O(n)O(n) since the array is iterated over once.

Solution #2: Using binary search

Java
class Count
{
  public static int calcFreq(int[]arr, int key) 
  { 
    int arrSize = arr.length;
    Helper obj = new Helper();
    int index = obj.binarySearch(key, arr, arrSize);
    
    if (index == -1)   // If element is not present 
      return 0; 
    int count = 1; // Initialzing a count
    
    int start = index - 1; // Counting the ones present on the left
    while (start >= 0 && arr[start] == key) {
      count++;
      start--; // Decrement the start index if found
    }
    
    int end = index + 1;  // Counting the ones present on the right
    while (end < arrSize && arr[end] == key) {
      count++;
      end++; // Increment the end index if found
    }
    return count; 
  }
  public static void main(String args[]) 
  {
    int arr[] = {-5,-3,0,1,3,3,3,3,4,5};
    int key = 3;
    System.out.println("The key \""+ key + "\" occurs " + calcFreq(arr, key) + " times in the Array.");
  }
}

This is also a very simple solution. It draws upon the fact that if an element exists in a sorted array, all of its ...