How Can We Optimize a Duplicate Removal Algorithm for Large Arrays Without Using Built-in Set Functions?

Optimizing Duplicate Removal Algorithm from an Array

The provided code aims to remove duplicate values from an array without utilizing built-in tools like Set or iterators. However, it faces performance bottlenecks when dealing with a large number of elements. This issue stems from the nested loop structure, where each element is compared to all subsequent elements.

To enhance the efficiency of the algorithm, consider the following optimization strategy:

Utilizing a HashSet:

Although the task explicitly prohibits using Set or HashSet, it's worth noting that a HashSet provides an efficient solution for eliminating duplicates. Its implementation employs a hash table to track the existence of each element, allowing for constant-time lookup and insertion.

Set<Integer> uniqueValues = new HashSet<>();

for (int num : arr) {
    uniqueValues.add(num);
}

The resulting uniqueValues Set will contain only distinct elements.

Preserving Element Order:

If preserving the original order of elements is crucial, a modified version of the provided algorithm can be employed:

// Create a boolean array to track duplicates
boolean[] duplicates = new boolean[arr.length];

// Find and mark duplicates in the first iteration
for (int i = 0; i < arr.length; i++) {
    for (int j = i + 1; j < arr.length; j++) {
        if (arr[i] == arr[j]) {
            duplicates[j] = true;
        }
    }
}

// Create a new array to store unique values
int[] uniqueArr = new int[arr.length - duplicates.length];
int uniqueIndex = 0;

// Copy unique values into the new array
for (int i = 0; i < arr.length; i++) {
    if (!duplicates[i]) {
        uniqueArr[uniqueIndex++] = arr[i];
    }
}

return uniqueArr;

This algorithm achieves O(n²) time complexity while preserving the original element ordering.

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