Get count from array using recursion

Question

I need to get the count of numbers less than the first integer in an array using recursion. I am given a function definition as

public static int countGreaterThanFirst(int[]
numbers, int startIndex, int endIndex, int firstNumber){}

I am not supposed to use a loop or global/static variable. How can I convert my implementation below to satisfy the two conditions above. I have recently asked another similar question but this is a bit different due to the need to keep track of the count variable. If someone can help, I will really appreciate. Below is my implementation with a loop.

public static int countGreaterThanFirst(int[] numbers, int startIndex, int endIndex, int firstNumber) {
    int greater_than_first = 0;
    for (int count = startIndex; count <= endIndex; count++) {
        if (numbers[count] > firstNumber) {
            greater_than_first++;
        }
    }
    return greater_than_first;
}

It works, but that's not the expected implementation.

mungaih pk
– mungaih pk

2016-03-29 14:10:10 +00:00
Commented Mar 29, 2016 at 14:10 — mungaih pk
– mungaih pk, Commented Mar 29, 2016 at 14:10

Konstantin Yovkov · Accepted Answer · 2016-03-29 14:12:07Z

3

Probably you don't need that much parameters:

public static int countGreaterThanFirst(int[] numbers, int currentIndex) {
    if (currentIndex == numbers.length) return 0;
    else {
        if (numbers[currentIndex] > numbers[0]) {
            return 1 + countGreaterThanFirst(numbers, currentIndex + 1);
        } else {
            return countGreaterThanFirst(numbers, currentIndex + 1);
        }
    }
}

and you should invoke it with (for example):

 countGreaterThanFirst(someArray, 1);

If you meant to find "all the numbers between numbers[startIndex] and numbers[endIndex] that are greater than firstNumber, then the implementation should be pretty similar to the above one:

public static int countGreaterThanFirst(int[] numbers, int startIndex, int endIndex, int firstNumber) {
    if (startIndex > endIndex) return 0;
    else {
        if (numbers[startIndex] > firstNumber) {
            return 1 + countGreaterThanFirst(numbers, startIndex + 1, endIndex, firstNumber);
        } else {
            return countGreaterThanFirst(numbers, startIndex + 1, endIndex, firstNumber);
        }
    }
}

edited Mar 29, 2016 at 14:12

answered Mar 29, 2016 at 14:05

Konstantin Yovkov

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6 Comments

mungaih pk Over a year ago

The parameters is a condition.

mungaih pk Over a year ago

Yes, and I have to implement the function as per the definition given

mungaih pk Over a year ago

unfortunately, the count is always 0.

Konstantin Yovkov Over a year ago

How do you invoke it? Why don't you try to debug it at the least? It works fine on my environment.

mungaih pk Over a year ago

@ Konstantin thanks for your help. It is working. I was invoking it wrongly.

|

arezooalipanah · Accepted Answer · 2023-06-10 11:00:19Z

0

public static int doCount(int[] list) {
    if (list.length == 0) {
        return 0;
    }
    return 1 + doCount(Arrays.copyOfRange(list, 1, list.length));
}

answered Jun 10, 2023 at 11:00

arezooalipanah

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