Solve Leetcode Problems and Get Offers From Your Dream Companies | Problem 202. Happy Number(Easy)

Start writiProblem 202. Happy Number (Leetcode Easy)

In this series, I am going to solve Leetcode medium problems live with my friend, which you can see on our youtube channel, Today we will do Problem Problem 202. Happy Number.

A little bit about me, I have offers from Uber India and Amazon India in the past, and I am currently working for Booking.com in Amsterdam.

Motivation to Learn Algorithms

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Problem Statement

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

Youtube Discussion

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Solution

This is a basic recursion based problem. We need to replace a number with the sum of the square of its digits. We need to repeat this process until the number becomes 1 or it loops endlessly in a cycle. In the case of reaching 1, we should return true else false.

So to keep track of all the last encountered numbers we can use a HashSet and then we can keep checking if our current number is present in that HashSet or not. If it is present , then we immediately terminate the method and return false , else we keep on finding the sum of square of it’s digits.

The following is the Java code for the given problem.

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The C++ code is given below.

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class LC202 {
public:
    int convert(int n){
        int ans = 0;
        while(n){
            
            ans += pow(n%10, 2);
      n/=10;
        
        }
        return ans;
    }
    
    bool isHappy(int n) {
        bool map[1000];
      
        memset(map, 0, sizeof(map));
    
        n = convert(n);
        
        while(!map[n]){
      map[n] = true;
            if(n == 1)
                return true;   
      n = convert(n);
        }
        
        return false;
    }
};

The code for this problem can be found in the following repository.

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