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How to select N unique random indices from an existing array in C#

1 Answer

0 votes
using System;
using System.Collections.Generic;
using System.Linq;

/*
    Select N unique random indices from an existing array in C#.
    Return the indices and print both the index and the corresponding value.

    Approach:
    - Build a list of indices: 0, 1, 2, ..., size-1.
    - Shuffle the list using Random + OrderBy.
    - Take the first N shuffled indices — guaranteed unique.
    - Return those indices to the caller.
*/

class UniqueRandomIndices
{
    // Return N unique random indices
    static List<int> PickUniqueIndices(int arraySize, int count)
    {
        if (count > arraySize)
            throw new ArgumentException("Cannot pick more unique indices than array size.");

        // Build index list
        List<int> indices = new List<int>(arraySize);
        for (int i = 0; i < arraySize; i++)
            indices.Add(i);

        // Shuffle indices
        Random rng = new Random();
        indices = indices.OrderBy(x => rng.Next()).ToList();

        // Return first N indices
        return indices.Take(count).ToList();
    }

    static void Main()
    {
        // Example array
        int[] data = {5, 12, 5, 19, 5, 33, 47, 5, 58, 61, 17, 3, 5, 74, 83, 90, 6};

        int N = 6; // number of unique indices to pick

        // Get unique random indices
        List<int> indices = PickUniqueIndices(data.Length, N);

        // Print results
        Console.WriteLine("Random unique indices and their values:");
        foreach (int idx in indices)
        {
            Console.WriteLine("index " + idx + " -> value " + data[idx]);
        }
    }
}


/*
run:

Random unique indices and their values:
index 7 -> value 5
index 2 -> value 5
index 11 -> value 3
index 5 -> value 33
index 4 -> value 5
index 14 -> value 83

*/

 



answered Sep 12 by avibootz
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