Happy Number (original) (raw)

Last Updated : 2 Sep, 2024

A number is called happy if it leads to 1 after a sequence of steps wherein each step number is replaced by the sum of squares of its digit that is if we start with Happy Number and keep replacing it with digits square sum, we reach 1.

**Examples :

Input: n = 19
Output: True
19 is Happy Number,
1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1
As we reached to 1, 19 is a Happy Number.

Input: n = 20
Output: False

A number will not be a Happy Number when it makes a loop in its sequence that is it touches a number in sequence which already been touched. So to check whether a number is happy or not, we can keep a set, if the same number occurs again we flag result as not happy. A simple function on the above approach can be written as below –

Try It Yourself

C++ `

// method return true if n is Happy Number int numSquareSum(int n) { int num = 0; while (n != 0) { int digit = n % 10; num += digit * digit; n /= 10; } return num; }

int isHappyNumber(int n) { set st; while (1) { n = numSquareSum(n); if (n == 1) return true; if (st.find(n) != st.end()) return false; st.insert(n); } }

Java

// method return true if n is Happy Number public static int numSquareSum(int n) { int num = 0; while (n != 0) { int digit = n % 10; num += digit * digit; n /= 10; } return num; }

static boolean isHappyNumber(int n) { HashSet st = new HashSet<>(); while (true) { n = numSquareSum(n); if (n == 1) return true; if (st.contains(n)) return false; st.add(n); } }

// This code is contributed by Princi Singh

Python

method return true if n is Happy Number

def numSquareSum(n): num = 0 while(n): digit = n % 10 num = num + digit*digit n = n // 10 return num

def isHappyNumber(n): st = set() while (1): n = numSquareSum(n) if (n == 1): return True if n not in st: return False st.insert(n)

C#

// Method return true if n is Happy Number

static int numSquareSum(int n) { int num = 0; while (n != 0) { int digit = n % 10; num += digit * digit; n /= 10; } return num; }

static int isHappyNumber(int n) { HashSet st = new HashSet<>(); while (1) { n = numSquareSum(n); if (n == 1) return true; if (st.Contains(n)) return false;

    st.Add(n);
}

}

// This code is contributed by 29AjayKumar

JavaScript

`

**Complexity Analysis:

**Time Complexity: O(n*log(n)).
**Auxiliary Space: O(n) since using extra set for storage

We can solve this problem without using extra space and that technique can be used in some other similar problems also. If we treat every number as a node and replacement by square sum digit as a link, then this problem is same as finding a loop in a linklist :

So as a proposed solution from the above link, we will keep two numbers slow and fast both initialize from a given number, slow is replaced one step at a time and fast is replaced two steps at a time. If they meet at 1, then the given number is Happy Number otherwise not.

C++ `

// C++ program to check a number is a Happy number or not #include <bits/stdc++.h> using namespace std;

// Utility method to return sum of square of digit of n int numSquareSum(int n) { int squareSum = 0; while (n) { squareSum += (n % 10) * (n % 10); n /= 10; } return squareSum; }

// method return true if n is Happy number bool isHappynumber(int n) { int slow, fast; // initialize slow and fast by n slow = fast = n; do { // move slow number by one iteration slow = numSquareSum(slow); // move fast number by two iteration fast = numSquareSum(numSquareSum(fast)); } while (slow != fast); // if both number meet at 1, then return true return (slow == 1); }

// Driver code to test above methods int main() { int n = 13; if (isHappynumber(n)) cout << n << " is a Happy number\n"; else cout << n << " is not a Happy number\n"; }

// This code is contributed by divyeshrabadiya07

C

// C program to check a number is a Happy number or not #include <stdbool.h> #include <stdio.h>

// Utility method to return sum of square of digit of n int numSquareSum(int n) { int squareSum = 0; while (n) { squareSum += (n % 10) * (n % 10); n /= 10; } return squareSum; }

// method return true if n is Happy number bool isHappynumber(int n) { int slow, fast; // initialize slow and fast by n slow = fast = n; do { // move slow number by one iteration slow = numSquareSum(slow); // move fast number by two iteration fast = numSquareSum(numSquareSum(fast)); } while (slow != fast); // if both number meet at 1, then return true return (slow == 1); }

// Driver code to test above methods int main() { int n = 13; if (isHappynumber(n)) printf("%d is a Happy number\n", n); else printf("%d is not a Happy number\n", n); }

// This code is contributed by Sania Kumari Gupta // (kriSania804)

Java

// Java program to check a number is a Happy // number or not

class GFG {

// Utility method to return sum of square of // digit of n static int numSquareSum(int n) { int squareSum = 0; while (n!= 0) { squareSum += (n % 10) * (n % 10); n /= 10; } return squareSum; }

// method return true if n is Happy number static boolean isHappynumber(int n) { int slow, fast;

//  initialize slow and fast by n
slow = fast = n;
do
{
    //  move slow number
    // by one iteration
    slow = numSquareSum(slow);

    //  move fast number
    // by two iteration
    fast = numSquareSum(numSquareSum(fast));

}
while (slow != fast);

//  if both number meet at 1,
// then return true
return (slow == 1);

}

// Driver code to test above methods public static void main(String[] args) { int n = 13; if (isHappynumber(n)) System.out.println(n + " is a Happy number"); else System.out.println(n + " is not a Happy number"); } }

Python

Python3 program to check if a number is a Happy number or not

Utility method to return the sum of squares of digits of n

def num_square_sum(n): square_sum = 0 while n: square_sum += (n % 10) ** 2 n //= 10 return square_sum

Method returns True if n is a Happy number

def is_happy_number(n): # Initialize slow and fast pointers slow = n fast = n

while True:
    # Move slow pointer by one iteration
    slow = num_square_sum(slow)
    
    # Move fast pointer by two iterations
    fast = num_square_sum(num_square_sum(fast))
    
    if slow != fast:
        continue
    else:
        break

# If both pointers meet at 1, then return True
return slow == 1

Driver Code

n = 13 if is_happy_number(n): print(n, "is a Happy number") else: print(n, "is not a Happy number")

C#

// C# program to check a number // is a Happy number or not using System; class GFG {

// Utility method to return // sum of square of digit of n static int numSquareSum(int n) { int squareSum = 0; while (n!= 0) { squareSum += (n % 10) * (n % 10); n /= 10; } return squareSum; }

// method return true if // n is Happy number static bool isHappynumber(int n) { int slow, fast;

// initialize slow and
// fast by n
slow = fast = n;
do
{
    
    // move slow number
    // by one iteration
    slow = numSquareSum(slow);

    // move fast number
    // by two iteration
    fast = numSquareSum(numSquareSum(fast));

}
while (slow != fast);

// if both number meet at 1,
// then return true
return (slow == 1);

}

// Driver code public static void Main() { int n = 13; if (isHappynumber(n)) Console.WriteLine(n + " is a Happy number"); else Console.WriteLine(n + " is not a Happy number"); } }

// This code is contributed by anuj_67.

JavaScript

PHP

`

**Output :

13 is a Happy Number

**Complexity Analysis:

**Time Complexity: O(n*log(n)).
**Auxiliary Space: O(1).

**Another approach for solving this problem using no extra space.
A number cannot be a happy number **if, at any step, the sum of the square of digits obtained is a single-digit number except 1 or 7. This is because 1 and 7 are the only single-digit happy numbers. Using this information, we can develop an approach as shown in the code below -

C++ `

// C++ program to check if a number is a Happy number or // not. #include <bits/stdc++.h> using namespace std;

// Method - returns true if the input is a happy number else // returns false bool isHappynumber(int n) {

int sum = n, x = n;
// This loop executes till the sum of square of digits
// obtained is not a single digit number
while (sum > 9) {
    sum = 0;
    // This loop finds the sum of square of digits
    while (x > 0) {
        int d = x % 10;
        sum += d * d;
        x /= 10;
    }
    x = sum;
}
if (sum == 7 || sum == 1)
    return true;
return false;

}

int main() { int n = 13;

if (isHappynumber(n))
    cout << n << " is a Happy number";
else
    cout << n << " is not a Happy number";
return 0;

}

// This code is contributed by Sania Kumari Gupta

C

// C program to check if a number is a Happy number or // not. #include <stdbool.h> #include <stdio.h>

// Method - returns true if the input is a happy number else // returns false bool isHappynumber(int n) { int sum = n, x = n;

// This loop executes till the sum of square of digits
// obtained is not a single digit number
while (sum > 9) {
    sum = 0;
    // This loop finds the sum of square of digits
    while (x > 0) {
        int d = x % 10;
        sum += d * d;
        x /= 10;
    }
    x = sum;
}
if (sum == 7 || sum == 1)
    return true;

return false;

}

int main() { int n = 13; if (isHappynumber(n)) printf("%d is a Happy number", n); else printf("%d is not a Happy number", n); return 0; }

// This code is contributed by Sania Kumari Gupta

Java

// This code is contributed by Vansh Sodhi. // Java program to check if a number is a Happy number or // not.

class GFG {

// method - returns true if the input is a happy
//  number else returns false
static boolean isHappynumber(int n)
{
    int sum = n, x = n;

    // this loop executes till the sum of square of
    // digits obtained is not a single digit number
    while (sum > 9) {
        sum = 0;

        // this loop finds the sum of square of digits
        while (x > 0) {
            int d = x % 10;
            sum += d * d;
            x /= 10;
        }
        x = sum;
    }
    if (sum == 1 || sum == 7)
        return true;
    return false;
}

// Driver code
public static void main(String[] args)
{
    int n = 13;
    if (isHappynumber(n))
        System.out.println(n + " is a Happy number");
    else
        System.out.println(n
                           + " is not a Happy number");
}

}

Python

Python3 program to check if a number is a Happy number or not.

Method - returns true if the input is

a happy number else returns false

def isHappynumber(n): Sum, x = n, n

# This loop executes till the sum
# of square of digits obtained is
# not a single digit number
while Sum > 9:
    Sum = 0

    # This loop finds the sum of
    # square of digits
    while x > 0:
        d = x % 10
        Sum += d * d
        x = int(x / 10)

    x = Sum

if Sum == 1 or Sum == 7:
    return True

return False

n = 13

if isHappynumber(n): print(n, "is a Happy number") else: print(n, "is not a Happy number")

This code is contributed by mukesh07.

C#

// C# program to check if a number // is a Happy number or not. using System;

class GFG {

// Method - returns true if the input is
// a happy number else returns false
static bool isHappynumber(int n)
{
    int sum = n, x = n;

    // This loop executes till the sum
    // of square of digits obtained is
    // not a single digit number
    while (sum > 9) {
        sum = 0;

        // This loop finds the sum of
        // square of digits
        while (x > 0) {
            int d = x % 10;
            sum += d * d;
            x /= 10;
        }

        x = sum;
    }
    if (sum == 1 || sum == 7)
        return true;

    return false;
}

// Driver code
public static void Main(String[] args)
{
    int n = 13;

    if (isHappynumber(n))
        Console.WriteLine(n + " is a Happy number");
    else
        Console.WriteLine(n + " is not a Happy number");
}

}

// This code is contributed by 29AjayKumar

JavaScript

`

Output

13 is a Happy number

**Complexity Analysis:

**Time Complexity: O(n*log(n)).
**Auxiliary Space: O(1).

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