How to Optimize Your Algorithm for Finding the k-th Smallest Element in the Union of Two Sorted Arrays

Efficiently locating the k-th smallest element in the union of two sorted arrays is a common challenge in algorithm design. The problem is often encountered in scenarios such as finding percentiles, statistics, or data analysis. This article is designed to guide you through an optimization problem and provide a refined solution that ensures correct functionality and maintains optimal performance.

The Original Code

Below is an initial approach to solving the problem, which contains an error:

public int find(int a[], int b[], int k) {
    int i  0, j  0, count  0, smallest  -1;
    if (a.length   b.length  k) {
        return smallest;
    }
    while (count  k  i  a.length  j  b.length) {
        if (a[i]  b[j]) {
            smallest  a[i];
            i  ;
        } else {
            smallest  b[j];
            j  ;
        }
        count  ;
    }
    if (count  k) {
        return smallest;
    }
    if (i  a.length) {
        return b[k - count - 1];
    }
    if (j  b.length) {
        return a[k - count - 1];
    }
    return smallest;
}

Identifying and Fixing the Issue

The code provided above has several issues that need correction. Notably, the condition `a.length b.length k` is incorrect and may lead to premature termination. Additionally, the return statement `return b[k - count - 1]` is flawed as it returns the (k-1)th element. Correcting these issues is critical for the overall functionality of the algorithm.

Improved Solution

Here is the corrected and optimized version of the code:

private static int find(int[] A, int[] B, int k) {
    int a  0, b  0;
    if (A.length   B.length  k) {
        throw new IllegalArgumentException("k is out of bounds for the combined array");
    }
    while (a  A.length  b  B.length) {
        if (A[a]  B[b]) {
            if (a   b - 1  k) {
                return A[a];
            }
            a  ;
        } else {
            if (a   b - 1  k) {
                return B[b];
            }
            b  ;
        }
    }
    if (a  A.length) {
        return B[k - a];
    } else {
        return A[k - b];
    }
}

Explanation

The improved version maintains a balance between simplicity and correctness. Here are the key points:

Initialization: The `a b - 1 k` condition ensures the correct index match. Corner Case Handling: If one of the arrays is fully traversed, the function returns the corresponding k-th element from the other array.

Compact Version

For a more concise and elegant solution, the algorithm can be rewritten as:

private static int find(int[] A, int[] B, int k) {
    if (A.length   B.length  k) {
        throw new IllegalArgumentException();
    }
    for (int a  0, b  0; a  A.length  b  B.length;   a,   b) {
        int currValue  -1;
        if (a  A.length  b  B.length) {
            currValue  A[a]  B[b] ? A[a] : B[b];
        } else {
            currValue  a  A.length ? B[b] : A[a];
        }
        if (a   b - 1  k) {
            return currValue;
        }
    }
    throw new IllegalStateException();
}

Advantages of the Compact Version

Clarity: The compact version is more straightforward and easier to understand. Efficiency: It maintains the same operational complexity while reducing redundancy.

Conclusion

Optimizing the algorithm for finding the k-th smallest element in the union of two sorted arrays not only improves the correctness of the solution but also enhances its efficiency. The corrected and compact versions provided in this article demonstrate key improvements and considerations for handling edge cases and ensuring robust functionality.