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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

Solution 1

Thinking

Repeatedly replacing a number by the sum of squares of its digits follows the definition, but the sequence may cycle instead of reaching \(1\). The values quickly fall into a bounded range, so a hash set can record numbers already seen.

A repeat means a cycle (not happy); reaching \(1\) means it is happy. Simulation and cycle checks run together.

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class Solution:
    def isHappy(self, n: int) -> bool:
        vis = set()
        while n != 1 and n not in vis:
            vis.add(n)
            x = 0
            while n:
                n, v = divmod(n, 10)
                x += v * v
            n = x
        return n == 1

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class Solution {
    public boolean isHappy(int n) {
        Set<Integer> vis = new HashSet<>();
        while (n != 1 && !vis.contains(n)) {
            vis.add(n);
            int x = 0;
            while (n != 0) {
                x += (n % 10) * (n % 10);
                n /= 10;
            }
            n = x;
        }
        return n == 1;
    }
}

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class Solution {
public:
    bool isHappy(int n) {
        unordered_set<int> vis;
        while (n != 1 && !vis.count(n)) {
            vis.insert(n);
            int x = 0;
            for (; n; n /= 10) {
                x += (n % 10) * (n % 10);
            }
            n = x;
        }
        return n == 1;
    }
};

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func isHappy(n int) bool {
    vis := map[int]bool{}
    for n != 1 && !vis[n] {
        vis[n] = true
        x := 0
        for ; n > 0; n /= 10 {
            x += (n % 10) * (n % 10)
        }
        n = x
    }
    return n == 1
}

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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;
    };
    const set = new Set();
    while (n !== 1) {
        const next = getNext(n);
        if (set.has(next)) {
            return false;
        }
        set.add(next);
        n = next;
    }
    return true;
}

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use std::collections::HashSet;
impl Solution {
    fn get_next(mut n: i32) -> i32 {
        let mut res = 0;
        while n != 0 {
            res += (n % 10).pow(2);
            n /= 10;
        }
        res
    }

    pub fn is_happy(mut n: i32) -> bool {
        let mut set = HashSet::new();
        while n != 1 {
            let next = Self::get_next(n);
            if set.contains(&next) {
                return false;
            }
            set.insert(next);
            n = next;
        }
        true
    }
}

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int getNext(int n) {
    int res = 0;
    while (n) {
        res += (n % 10) * (n % 10);
        n /= 10;
    }
    return res;
}

bool isHappy(int n) {
    int slow = n;
    int fast = getNext(n);
    while (slow != fast) {
        slow = getNext(slow);
        fast = getNext(getNext(fast));
    }
    return fast == 1;
}

Solution 2

Thinking

The hash set is correct but uses extra memory. The square-sum map is a deterministic iteration, so a cycle can be found without storing the full history.

Floyd's pointers suffice: the slow pointer moves once and the fast pointer twice; if they meet at \(1\), the number is happy, and space becomes constant.

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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
    }
}

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