Given a number "n", The task is to find if it is Disarium or not. A number is called Disarium if sum of its digits powered with their respective positions is equal to the number itself.
Examples:
Input: n = 135
Output: Yes
Explanation: 1^1 + 3^2 + 5^3 = 135
Therefore, 135 is a Disarium number
Input: n = 89
Output: Yes
Explanation: 8^1+9^2 = 89
Therefore, 89 is a Disarium number
Input: n = 80
Output: No
Explanation: 8^1 + 0^2 = 8
Approach:
The idea is to first count digits in given numbers. Once we have count, we traverse all digits from right most (using % operator), raise its power to count and decrement the count.
C++
// C++ program to check whether a number is Disarium
// or not
#include <iostream>
#include <cmath>
using namespace std;
// Finds count of digits in n
int countDigits(int n) {
int count = 0;
// Count number of digits in n
int x = n;
while (x) {
x = x / 10;
// Count the no. of digits
count++;
}
return count;
}
// Function to check whether a number is disarium or not
bool isDisarium(int n) {
// Count digits in n.
int count = countDigits(n);
// Compute sum of terms like digit multiplied by
// power of position
// Initialize sum of terms
int sum = 0;
int x = n;
while (x) {
// Get the rightmost digit
int r = x % 10;
// Sum the digits by powering according to
// the positions
sum = sum + pow(r, count--);
x = x / 10;
}
// If sum is same as number, then number is
return (sum == n);
}
int main() {
int n = 135;
if (isDisarium(n))
cout << "True";
else
cout << "False";
return 0;
}
// Java program to check whether a number is disarium
// or not
// Java program to check whether a number is Disarium or not
class GfG {
// Function to count the number of digits in n
static int countDigits(int n) {
int count = 0;
int x = n;
// Count the number of digits in n
while (x > 0) {
x = x / 10;
count++;
}
return count;
}
// Function to check whether a number is Disarium or not
static boolean isDisarium(int n) {
// Count digits in n
int count = countDigits(n);
// Compute sum of digits raised to the power of their positions
int sum = 0; // Initialize sum of terms
int x = n;
while (x > 0) {
// Get the rightmost digit
int r = x % 10;
// Sum the digits by raising them to the power of their positions
sum += Math.pow(r, count--);
x = x / 10;
}
// If sum is same as the number, then it is a Disarium number
return (sum == n);
}
public static void main(String[] args) {
int n = 135;
if (isDisarium(n))
System.out.println("True");
else
System.out.println("False");
}
}
# Python program to check whether a number is Disarium
# or not
# Finds count of digits in n
def countDigits(n):
count = 0
# Count number of digits in n
x = n
while x:
x = x // 10 # Integer division
# Count the no. of digits
count += 1
return count
# Function to check whether a number is disarium or not
def isDisarium(n):
# Count digits in n.
count = countDigits(n)
# Compute sum of terms like digit multiplied by
# power of position
sum = 0
x = n
while x:
# Get the rightmost digit
r = x % 10
# Sum the digits by powering according to
# the positions
sum = sum + pow(r, count)
count -= 1
# Integer division
x = x // 10
# If sum is same as number, then number is
return (sum == n)
if __name__ == "__main__":
n = 135
if isDisarium(n):
print("True")
else:
print("False")
// C# program to check whether a number is Disarium
// or not
using System;
class GfG
{
// Finds count of digits in n
static int countDigits(int n)
{
int count = 0;
// Count number of digits in n
int x = n;
while (x > 0)
{
x = x / 10;
// Count the no. of digits
count++;
}
return count;
}
// Function to check whether a number is disarium or not
static bool isDisarium(int n)
{
// Count digits in n.
int count = countDigits(n);
// Compute sum of terms like digit multiplied by
// power of position
int sum = 0;
int x = n;
while (x > 0)
{
// Get the rightmost digit
int r = x % 10;
// Sum the digits by powering according to
// the positions
sum = sum + (int)Math.Pow(r, count--);
x = x / 10;
}
// If sum is same as number, then number is
return (sum == n);
}
static void Main()
{
int n = 135;
if (isDisarium(n))
Console.WriteLine("True");
else
Console.WriteLine("False");
}
}
// Finds count of digits in n
function countDigits(n) {
let count = 0;
// Count number of digits in n
let x = n;
while (x > 0) {
x = Math.floor(x / 10);
// Count the no. of digits
count++;
}
return count;
}
// Function to check whether a number is disarium or not
function isDisarium(n) {
// Count digits in n.
let count = countDigits(n);
// Compute sum of terms like digit multiplied by
// power of position
let sum = 0;
let x = n;
while (x > 0) {
// Get the rightmost digit
let r = x % 10;
// Sum the digits by powering according to
// the positions
sum = sum + Math.pow(r, count--);
x = Math.floor(x / 10); // Integer division
}
// If sum is same as number, then number is
return sum === n;
}
// Driver code
(function () {
let n = 135;
if (isDisarium(n))
console.log("True");
else
console.log("False");
})();
Time complexity : O(logn)
Auxiliary Space : O(1)
Python Program to Check a Number is Disarium Number
Explore
DSA Fundamentals
Data Structures
Algorithms
Advanced
Interview Preparation
Practice Problem