Question

A number is given as input. Find the odd digits in the number, add them and find if the sum is odd or not.if even return -1, if odd return 1 input:52315 logic:5+3+1 +5=14(even) output:-1 input1112 logic: 1+1+1=3(odd) output: 1​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

A number is given as input. Find the odd digits in the number, add them and find if the sum is odd or not. If even return -1, if odd return 1

input:52315

logic:5+3+1+5=14(even)

output:-1

input:1112

logic:1+1+1=3(odd)

output:1

Explanation:

Given:

• First, calculate the reverse of the given number.
• To the reverse number we apply modulus operator and extract its last digit which is actually the first digit of a number so it is odd positioned digit.
• The next digit will be even positioned digit, and we can take the sum in alternating turns.

Find: Java program for the above scenario of the problem

Program:

// Java implementation of the approach

import java.util.*;

class SORE {

// Function to return the reverse of a number

static int reverse(int n)

{

int rev = 0;

while (n != 0) {

rev = (rev * 10) + (n % 10);

n /= 10;

}

return rev;

}

// Function to find the sum of the odd

// and even positioned digits in a number

static void getSum(int n)

{

n = reverse(n);

int sumOdd = 0, sumEven = 0, c = 1;

while (n != 0) {

// If c is even number then it means

// digit extracted is at even place

if (c % 2 == 0)

sumEven += n % 10;

else

sumOdd += n % 10;

n /= 10;

c++;

}

System.out.println("Sum odd = " + sumOdd);

System.out.println("Sum even = " + sumEven);

}

// Driver code

public static void main(String args[])

{

int n = 457892;

getSum(n);

}

}