[LeetCode] 202. Happy Number(C#)

LeetCode - Problems - Algorithms - 202. Happy Number

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My Solution (C#)

        static public bool IsHappy(int n)
        {
           List<int> tracks = new List<int>();
            while (!tracks.Contains(n))
            {
                tracks.Add(n);
                int remainder = 0;
                int quotient = n;
                int sum = 0;
                
                do {                
                    sum += remainder * remainder;  
                    remainder = n % 10;
                    quotient = n / 10;  
                    n = quotient;
                } while (quotient >= 10);   
                                            
                n = sum + quotient * quotient + remainder * remainder; 
            }
            if (n == 1) return true; 

            return false;
        }

 

My Explanation

// 19> 82 >68 >100 > 1: it will repeat 1. so, happy number
    // 2> 4 >8 >16 >37 >58 >89 >145 >42 >20 > 4: it will repeat and never make result 1. so, it is unhappy
    //the list to record the result of adding squares of n in
 //if the list contains n, it means it terminates this loop. so while the statement to repeat while the track does not contain n
// before starting the set of while statements we need to add n into the list, tracks to remember it for later to check if it passed or not
//an integer variable remainder by dividing 10.
//quotient that is dropped the remainder. but this will also need for checking if the quotient is a two-digit number later.
//also an integer variable sum to add each digit square.
// those three variables need to be initialized every time of result.
// square the remainder obtained from the previous loop and add it.  
//define remainder n's last digit
//define the digit number except for the last digit 
//define n as a quotient that is used to check whether the loop needs to be repeated until n has only one digit 
//this do-while statement is for repeating to add the last digit square to the total sum. we are using a do-while statement because this should be done before checking the condition of the loop. otherwise, we will miss the last step of addition.
//the condition means there is only one digit. without this condition, this will be an infinite loop.
// we had been adding the numbers from back to front. so now, the quotient is the first digit of the n
    }
//if n is contained in the list, (of course when n is 1 because when n is 1 the next n will be 1) we can escape the first while loop when the list contains n. when the number reaches 1, n is a happy number. returns true!
//otherwise false because the track is repeated infinitely between n to n.
}

Time complexity: O(1): when it repeats digit by digit the number of n's digits is reduced.

Space complexity O(1): this solution is been doing all of the calculation in a limited space

This script is written for the algorithm study group meeting.

Not confirmed professional just me who is still in junior level.