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

Problem 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

Solve Leetcode Problems and Get Offers From Your Dream Companies | by Nil Madhab | LeetCode Simplified | Medium

Nil Madhab ・ Feb 6, 2021 ・

Medium

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.

	public class LC202 {
	static HashSet<Integer> hashSet;
	public boolean isHappy(int n) {
	//Dry run of example 1
	// 19 -> 1+81 = 82 -> 64+ 4 = 68 -> 36 + 64 = 100 -> 1 . As we reach 1, we need to return 1.
	//Dry run of example 2
	// 2 -> 4 -> 16 -> 1+36 = 37 -> 9 + 49 = 58 -> 25 + 64 = 89 -> 64+81 = 145 -> 1 + 16+25 = 42 -> 16+4 = 20 -> 4 .
	// So here we can see that we are looping through same numbers in this case . Therefore we should jump out of this loop.
	hashSet = new HashSet<>();
	return isHappyUtil(n);
	}
	private boolean isHappyUtil(int n){
	if(n==1)
	return true;
	int sum = 0;
	while(n>0){
	int x = n%10;
	sum = sum + (int) Math.pow(x,2);
	n = n/10;
	}
	if(hashSet.contains(sum))
	return false;
	hashSet.add(sum);
	return isHappyUtil(sum);
	}
	}

view raw LC202.java hosted with ❤ by GitHub

The C++ code is given below.

	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;
	}
	};

view raw LC202.cpp hosted with ❤ by GitHub

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

webtutsplus / LeetCode