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How to sort an array with a single loop in Kotlin

1 Answer

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/**
 * Swaps two elements in an IntArray at the specified indices.
 */
private fun IntArray.swap(i: Int, j: Int) {
    val temp = this[i]
    this[i] = this[j]
    this[j] = temp
}

/**
 * Sorts an IntArray in-place in non-decreasing order using Gnome Sort.
 *
 * Algorithm Logic (Single Loop):
 * - Advances through the array using a single while loop index.
 * - Moves forward when adjacent elements are in correct relative order.
 * - When an out-of-order adjacent pair is encountered, swaps the elements
 *   and steps backward one index to verify order against preceding items.
 * - Time Complexity: O(N) best case (already sorted), O(N^2) worst case.
 * - Space Complexity: O(1) auxiliary space.
 *
 * @receiver IntArray to be sorted in place.
 */
fun IntArray.singleLoopSort() {
    var pos = 0
    val len = this.size

    while (pos < len) {
        // Move forward if at index 0 or if adjacent pair is in correct order
        if (pos == 0 || this[pos] >= this[pos - 1]) {
            pos++
        } else {
            // Swap out-of-order elements in place and step backward
            swap(pos, pos - 1)
            pos--
        }
    }
}

/**
 * Main application entry point
 */
fun main() {
    val numbers = intArrayOf(42, -5, 12, 0, 89, -18, 33, 7)

    println("Original array:")
    println(numbers.joinToString(" "))

    numbers.singleLoopSort()

    println("\nSorted array:")
    println(numbers.joinToString(" "))
}


/*
run:

Original array:
42 -5 12 0 89 -18 33 7

Sorted array:
-18 -5 0 7 12 33 42 89

*/

 



answered 18 hours ago by avibootz
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