Problem statement

A disarium number is a number where the sum of its digits, each raised to the power of its 1-based position, equals the number itself.

Given a positive integer n, return true if n is a disarium number.

Example 1

Input: n = 89
Output: true
Explanation: 8¹ + 9² = 8 + 81 = 89.

Example 2

Input: n = 135
Output: true
Explanation: 1¹ + 3² + 5³ = 1 + 9 + 125 = 135.

Example 3

Input: n = 76
Output: false
Explanation: 7¹ + 6² = 43, not 76.

Constraints

• 1 <= n <= 1000000

Hints for Disarium Number

1. Hint 1

   Positions run left to right starting at 1, so the leftmost digit gets exponent 1.

2. Hint 2

   That makes the string form convenient: index i carries exponent i + 1.

3. Hint 3

   Contrast this with an Armstrong number, where every digit shares the same exponent.

How to solve Disarium Number

Positions count from the left starting at 1, so index i of the decimal string carries exponent i + 1. Sum digit^(i+1) and compare.

Approach

1. Write n as a digit string.
2. For each index i, add digit[i] ^ (i + 1) to a total.
3. Return total == n.

Why it works

Unlike an Armstrong number, where every digit shares the same exponent, a disarium number weights each digit by its position; iterating the string left to right lines the exponent up with the position directly.

Complexity

• Time — O(d)
• Space — O(d)

Pitfalls

• Peeling digits from the right gives positions in reverse; either count the digits first or work on the string.
• 89 = 8¹ + 9² and 175 = 1 + 49 + 125 are handy test values.

Reference solution

Python

def isDisarium(n: int) -> bool:
    s = str(n)
    total = 0
    for i, ch in enumerate(s):
        total += int(ch) ** (i + 1)
    return total == n

Java

public static boolean isDisarium(int n) {
    String s = String.valueOf(n);
    long total = 0;
    for (int i = 0; i < s.length(); i++) {
        long d = s.charAt(i) - '0';
        long p = 1;
        for (int e = 0; e <= i; e++) p *= d;
        total += p;
    }
    return total == n;
}

C++

bool isDisarium(int n) {
    string s = to_string(n);
    long long total = 0;
    for (int i = 0; i < (int) s.size(); i++) {
        long long d = s[i] - '0';
        long long p = 1;
        for (int e = 0; e <= i; e++) p *= d;
        total += p;
    }
    return total == (long long) n;
}

JavaScript

var isDisarium = function(n) {
    const s = String(n);
    let total = 0;
    for (let i = 0; i < s.length; i++) {
        total += Math.pow(s.charCodeAt(i) - 48, i + 1);
    }
    return total === n;
};

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