SOLVED: What will be the output of the code given below? public class Tester { publ ...

What will be the output of the code given below?

public class Tester {
	public static void main(String[] args) {
		System.out.println(demo(5, 1));
	}

	public static int demo(int x, int y) {
		if (x == 0)
			return 1;
		else if (y > x)
			return 0;
		else
			return (y + demo(x - 1, y + 1));
	}
}

6

0

15

18

Verified Answer

Correct Option - a

Anush said

The output of the code will be 6. Here's the explanation:
1. The function demo(x, y) is a recursive method that follows these conditions:

If x == 0, it returns 1.
If y > x, it returns 0.
Otherwise, it returns y + demo(x - 1, y + 1).

2. Initially, demo(5, 1) is called, and the recursion continues until x == 0. 3. Here are the steps of the recursion:

demo(5, 1): returns 1 + demo(4, 2)
demo(4, 2): returns 2 + demo(3, 3)
demo(3, 3): returns 3 + demo(2, 4)
demo(2, 4): returns 0 because y > x

4. The final calculation is: 1 + 2 + 3 = 6.

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What will be the output of the code given below?

`public class Tester { public static void main(String[] args) { System.out.println(demo(5, 1)); } public static int demo(int x, int y) { if (x == 0) return 1; else if (y > x) return 0; else return (y + demo(x - 1, y + 1)); } }`

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What will be the output of the code given below? public class Tester { public static void main(String[] args) { System.out.println(demo(5, 1)); } public static int demo(int x, int y) { if (x == 0) return 1; else if (y > x) return 0; else return (y + demo(x - 1, y + 1)); } }

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What will be the output of the code given below?

`public class Tester { public static void main(String[] args) { System.out.println(demo(5, 1)); } public static int demo(int x, int y) { if (x == 0) return 1; else if (y > x) return 0; else return (y + demo(x - 1, y + 1)); } }`