Java Program #11 - Find Armstrong Number in Java

Java Program to find Armstrong Number in java
In this video by Programming for Beginners we will learn to write Java Program for Armstrong Number, using Java Tutorial videos.
This Java program is very important for your Java interview questions or if you are learning Java Programming language as a student.
An Armstrong number is a positive m-digit number that is equal to the sum of the mth powers of their digits.
Examples:
1: 1^1 = 1
2: 2^1 = 2
3: 3^1 = 3
153: 1^3 + 5^3 + 3^3 = 1 + 125+ 27 = 153
125: 1^3 + 2^3 + 5^3 = 1 + 8 + 125 = 134 (Not an Armstrong Number)
1634: 1^4 + 6^4 + 3^4 + 4^4 = 1 + 1296 + 81 + 256 = 1643
1741725 is an Armstrong Number.
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We can learn Java Programming language to make web applications or mobile applications for android phones and several other applications for windows, Mac OS and Unix operating systems. Also we can make android applications using Java programming. The concepts covered will be related to basic java and core java that will help you in your next interview questions. You can use any version of Java but all concepts will be same for all java versions. I will be using Java SE 8.
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Thanks for watching :)

[@programmingforbeginners7392]

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[@shashank125sanji]

bro really thank for teaching it properly it was asked in aptitude test but i was blank in the test i will start learning properly with ur vedios

[@programmingforbeginners7392]

Welcome bro. All the best ☺️

[@abhijit6350]

bhai you earned a subscriber, hats off bro, may god bless you

[@programmingforbeginners7392]

Thank you.. keep watching

[@diptimayeepatro6097]

I think this one is the best way to solve and nice explanation....it would be help to crack the interview... thank you so much sir for understanding nicely 👌🥰🙏

[@programmingforbeginners7392]

Welcome. Glad it was helpful. Subscribe and Share with your friends 😀

[@pavankumar7864]

public static void main(String[] args) { int n=407; int orginal_number=n; int sum=0; int numOfdigits=String.valueOf(n).length(); while(n>0){ int digit= n%10; n=n/10; sum= (int) (sum+Math.pow(digit,numOfdigits)); } if (sum==orginal_number){ System.out.println(orginal_number+" is the Armstrongnumber"); }else { System.out.println(orginal_number+" is not a Armstrongnumber"); } }

[@programmingforbeginners7392]

good. keep practicing

[@shiwangikar5422]

Complicated😅

[@Giriagroandfashion]

Good but Very Confusing

[@programmingforbeginners7392]

Please practice example programs to be more clear

[@nagamanim3390]

Can you please tell what is the time complexity and space complexity for this solution

[@thenameis-jatin]

thank you bro for clearing my doubts ❤

[@programmingforbeginners7392]

Welcome. Subscribe and share with your friends 😄

[@HimaBindu-ki5js]

Nice explanation

[@programmingforbeginners7392]

Thanks 👍

[@anandbalmiki8346]

Nice explanation and very clear code

[@programmingforbeginners7392]

Thank you so much 👍

[@abdulrahim7763]

tried without while loop..this is working fine for me (give any number within the limit of integer data type) Scanner sc=new Scanner(System.in); System.out.println("Enter any number to find out if it is a Armstrong number or not"); int input=sc.nextInt(); String s=""; String count=s.valueOf(input); int input_size = count.length(); int digit; double Armstrong=0; double result=0; int actual=input; if(input>0) { for(int i=1;i

[@crazzyraghu8726]

Bro, I m getting error that " Sum cannot resolved to be variable'

[@programmingforbeginners7392]

Define sum variable after: int digits, temp, sum = 0; This should solve your issue

[@crazzyraghu8726]

@@programmingforbeginners7392 Now i got it bro, Thanks for the Response.

[@crazzyraghu8726]

@@programmingforbeginners7392 int number; Scanner sc = new Scanner(System.in); System.out.println("Enter Number : "); number = sc.nextInt(); System.out.println("Is amstrong num : " + isAmstrong(number)); } static boolean isAmstrong(int n) { int digits = 0, temp; temp = n; while(temp > 0) { temp = temp/10; digits++; } System.out.println("No of digits: " + digits); temp = n; int sum = 0; while(temp> 0) { int lastDigit = temp %10; sum = (int)(sum +Math.pow(lastDigit, digits)); temp =temp/10; } System.out.println("sum is : " + sum); if(sum == n) return true; return false; } }

[@Faith-b9w]

Can we compile this program in cmd?

[@abhijit6350]

ig only for eclipse , vs code, etc. not for cmd prompt

[@Faith-b9w]

@@abhijit6350 ok ...thank u

[@programmingforbeginners7392]

Java programs can be executed from cmd

[@crazzyraghu8726]

public static void main(String[] args) { int number; Scanner sc = new Scanner(System.in); System.out.println("Enter Number : "); number = sc.nextInt(); System.out.println("Is amstrong num ; " + isAmstrong(number)); } static boolean isAmstrong(int n) { int digits, temp; temp = n; while(temp > 0) { temp = temp/10; digits++; } System.out.println("No of digits: " + digits); temp = n; while(temp> 0) { int lastDigit = temp %10; sum = (int)(sum +Math.pow(lastDigit, digits)); temp =temp/10; } System.out.println("sum is ; " + sum); if(sum == n) return true; return false; } } Receiving error that sum cannot resolved to a Variable

[@programmingforbeginners7392]

Define sum variable after: int digits, temp, sum = 0; This should solve your issue

[@reddybasha3255]

Nice video bro Please simplify the code

[@programmingforbeginners7392]

Thank you

[@nagamanim3390]

Can you tell me what is time complexity and space complexity for this solution?