How to find common elements in two unsorted arrays with sizes n and m avoiding double for loop?

How to find common elements in two unsorted arrays with sizes n and m avoiding double for loop?

Programmers, software engineers, coders, IT professionals, and software architects all face the common challenge of needing to find common elements in two unsorted arrays with sizes n and m. This can be a difficult task, especially if you don’t want to use a double for loop.

In this blog post, we will be discussing how to find common elements in two unsorted arrays with sizes n and m avoiding double for loop. We will be discussing various methods that can be used to solve this problem and comparing the time complexity of each method.

There are several ways that you can find common elements in two unsorted arrays with sizes n and m avoiding double for loop. One way is by using the hashing technique. With this technique, you can create a hash table for one of the arrays. Then, you can traverse through the second array and check if the element is present in the hash table or not. If the element is present in the hash table, then it is a common element. Another way that you can find common elements in two unsorted arrays with sizes n and m avoiding double for loop is by using the sorting technique. With this technique, you can sort both of the arrays first. Then, you can traverse through both of the arrays simultaneously and compare the elements. If the elements are equal, then it is a common element.

Method 1: Linear Search

The first method we will discuss is linear search. This method involves iterating through both arrays and comparing each element. If the element is found in both arrays, it is added to the result array. The time complexity of this method is O(nm), where n is the size of the first array and m is the size of the second array.

Method 2: HashMap Method

The second method we will discuss is the HashMap method. This method involves creating a HashMap of all the elements in the first array. Then, we iterate through the second array and check if the elements are present in the HashMap. If they are, we add them to the result array. The time complexity of this method is O(n+m), where n is the size of the first array and m is the size of the second array.

Method 3: Sort and Compare Method

The third method we will discuss is the Sort and Compare Method. This method involves sorting both arrays using any sorting algorithm like merge sort or quick sort. Once both arrays are sorted, we compare each element of both arrays one by one until we find a match. If a match is found, we add it to our result array. The time complexity of this method is O(nlogn+mlogm), where n is the size of the first array and m is the size of the second array.

The naïve algorithm for finding common elements in two unsorted arrays with sizes nn and mm is O(nm)O(nm), i.e. quadratic.

The algorithm for sorting an array is O(nlogn)O(nlog⁡n), and you can find common elements in two sorted arrays in O(n+m)O(n+m). In other words, for large enough arrays, it is significantly faster to first sort them, then look for the common elements, because the sorting algorithm will dominate the complexity, so your final algorithm ends up at O(nlogn)O(nlog⁡n) as well.

Conclusion:

In this blog post, we discussed how to find common elements in two unsorted arrays with sizes n and m avoiding double for loop. We discussed three different methods that can be used to solve this problem and compared their time complexities. We hope that this blog post was helpful in understanding how to solve this problem.

There are many different ways to find common elements in two unsorted arrays with sizes n and m avoiding double for loop. The most straight forward way is by using a double for loop but this approach is not very efficient. A more efficient way is by using a hash table which has a time complexity of O(n+m). This algorithm is faster because we only need to loop through one of the arrays. We can then use the values from that array to check if there are any duplicates in the second array. This approach also uses less memory because we are not creating a new list to store the common elements.