CareerCup

The idea is to use Merge function of Merge sort.

Create an array arr3[] of size n1 + n2.
Simultaneously traverse arr1[] and arr2[].
Pick smaller of current elements in arr1[] and arr2[], copy this smaller element to next position in arr3[] and move ahead in arr3[] and the array whose element is picked.
If there are remaining elements in arr1[] or arr2[], copy them also in arr3[].

This is how I'd solve that, considering arrays a and b to merge to c

check one element from a and b and add the min to c until one of the arrays runs out of elements

add remaining elements from a or b to c

fun main() {
    val a = arrayOf(1, 3, 5, 7, 8, 9)
    val b = arrayOf(2, 4, 6)
    print("A = ${a.asList()}")
    print("B = ${b.asList()}")
    
    val c = Array(a.size + b.size) { 0 }
    
    var i = 0
    var j = 0
    var k = 0
    // find the min of a[j],b[j] and add it at c[k]
    // until one of the arrays(a or b) is empty
    while (i < a.size && j < b.size) {
        if (a[i] < b[j]) {
            c[k] = a[i]
            i++
            k++
        } else {
            c[k] = b[j]
            j++
            k++
        }
    }
    
    // add remaining elements to c
    while (i < a.size) {
        c[k] = a[i]
        i++
        k++
    }
    while (j < b.size) {
        c[k] = b[j]
        j++
        k++
    }
    
    print("C = ${c.asList()}")
}

note: i contains position of current element in A similarly j for B and k for C