How to Find Duplicates in an Array with Optimal Time and Space Complexity?

Finding Duplicates with Optimal Time and Space Complexity

Introduction:
The task of detecting duplicate elements in an array is a prevalent problem in programming. While solutions using auxiliary data structures like hash tables offer simplicity, they compromise space efficiency. This article presents an ingenious algorithm that identifies duplicates in linear time, O(n), while maintaining constant space consumption, O(1).

Problem Statement:
Given an array of n elements ranging from 0 to n-1, where some numbers may appear multiple times, find the duplicate elements in the array.

Optimal Algorithm:

// Iterate through the array
for i = 0 to n - 1:
    // Swap elements until A[A[i]] = A[i]
    while A[A[i]] != A[i]:
        swap(A[i], A[A[i]])
    end while

// Identify duplicates based on A[i] != i
for i = 0 to n - 1:
    if A[i] != i:
        print A[i]
    end if
end for

Insight:
The algorithm leverages the fact that each element in the array should ideally reside at its index. It uses the array's elements to create a permutation, ensuring that elements that exist will map to their correct indices. The first loop iterates through the array and ensures that each element reaches its intended index through swaps.

Explanation:
The first loop permutes the array so that for any element x, if it appears at least once, one instance will be at index A[x]. In each swap, an entry corresponding to an incorrect element is corrected, guaranteeing that the total number of swaps is bounded by the number of elements, N-1.

The second loop identifies duplicates by printing the index of each element that does not match its index, indicating its absence in the original array.

Conclusion:
This algorithm efficiently identifies duplicate elements in an array by making clever use of the input array itself. Its O(n) time and O(1) space complexity make it suitable for a wide range of practical applications.

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