How to identify it?

• Querying item presence in A against B.
• Prioritizing linear time complexity (O(n)) over quadratic (O(n²)), avoiding nested loops.

TO ACCOMPLISH THIS OBJECTIVE...

• Utilize individual O(n) loops instead of nested loops for efficiency.
• Employ object mapping to facilitate element comparison.
• Establish keys as elements and values as their respective frequencies.

---Pseudocode
            A) Iterate through the array 
            B) Grab the value of each item in the array and set it to variable key 
            C) If key exists as a key in myObject... 
            D) ... increment it by 1 
            E) If not (key doesn't exist in myObject)... 
            F) ... create key in myObject and set it equal to 1

FREQUENCY COUNTER IN ACTION

Problem: Write a function frequencyCounter, which accepts two arrays. The function should return true if every value in the array has its corresponding value squared in the second array. The frequency values must be the same.

Code to solve this problem:

Mapping each list's elements to their frequencies in separate dictionaries enables efficient comparison, ensuring a streamlined process. This technique sidesteps nested loops, resulting in a time complexity of O(n) and superior efficiency compared to alternative methods.

With two individual loops, each element in the arrays is processed separately, allowing for direct comparison and frequency counting. Nested loops, on the other hand, would lead to redundant comparisons between every pair of elements, resulting in unnecessary computations.

ADDITIONAL EXERCISES:

• Valid Anagram
• Are There Duplicates?
• Same Frequency