# 202. Happy Number

## Description

Write an algorithm to determine if a number n is happy.

A happy number is a number defined by the following process:

• Starting with any positive integer, replace the number by the sum of the squares of its digits.
• Repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1.
• Those numbers for which this process ends in 1 are happy.

Return true if n is a happy number, and false if not.

Example 1:

Input: n = 19
Output: true
Explanation:
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1


Example 2:

Input: n = 2
Output: false


Constraints:

• 1 <= n <= 231 - 1

## Solutions

• class Solution {
public boolean isHappy(int n) {
int slow = n, fast = next(n);
while (slow != fast) {
slow = next(slow);
fast = next(next(fast));
}
return slow == 1;
}

private int next(int x) {
int y = 0;
for (; x > 0; x /= 10) {
y += (x % 10) * (x % 10);
}
return y;
}
}

• class Solution {
public:
bool isHappy(int n) {
auto next = [](int x) {
int y = 0;
for (; x; x /= 10) {
y += pow(x % 10, 2);
}
return y;
};
int slow = n, fast = next(n);
while (slow != fast) {
slow = next(slow);
fast = next(next(fast));
}
return slow == 1;
}
};

• class Solution:
def isHappy(self, n: int) -> bool:
def next(x):
y = 0
while x:
x, v = divmod(x, 10)
y += v * v
return y

slow, fast = n, next(n)
while slow != fast:
slow, fast = next(slow), next(next(fast))
return slow == 1


• func isHappy(n int) bool {
next := func(x int) (y int) {
for ; x > 0; x /= 10 {
y += (x % 10) * (x % 10)
}
return
}
slow, fast := n, next(n)
for slow != fast {
slow = next(slow)
fast = next(next(fast))
}
return slow == 1
}

• function isHappy(n: number): boolean {
const getNext = (n: number) => {
let res = 0;
while (n !== 0) {
res += (n % 10) ** 2;
n = Math.floor(n / 10);
}
return res;
};

let slow = n;
let fast = getNext(n);
while (slow !== fast) {
slow = getNext(slow);
fast = getNext(getNext(fast));
}
return fast === 1;
}


• impl Solution {
pub fn is_happy(n: i32) -> bool {
let get_next = |mut n: i32| {
let mut res = 0;
while n != 0 {
res += (n % 10).pow(2);
n /= 10;
}
res
};
let mut slow = n;
let mut fast = get_next(n);
while slow != fast {
slow = get_next(slow);
fast = get_next(get_next(fast));
}
slow == 1
}
}