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 –
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<int> st;
while (1)
{
n = numSquareSum(n);
if (n == 1)
return true;
if (st.find(n) != st.end())
return false;
st.insert(n);
}
}
// 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<Integer> 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
# 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)
// 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<int> 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
<script>
// method return true if n is Happy Number
function numSquareSum(n) {
let num = 0;
while (n !== 0) {
let digit = n % 10;
num += digit * digit;
n = Math.floor(n / 10);
}
return num;
}
let st = new Set();
while (1)
{
n = numSquareSum(n);
if (n == 1)
return true;
if (st.has(n))
return false;
st.add(n);
}
}
//This code is contributed by Mayank Tyagi
</script>
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 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 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");
}
}
# 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# 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.
<script>
// Javascript program to check a number is a Happy
// number or not
// Utility method to return sum of square of
// digit of n
function numSquareSum(n)
{
var squareSum = 0;
while (n!= 0)
{
squareSum += (n % 10) * (n % 10);
n = parseInt(n/10);
}
return squareSum;
}
// method return true if n is Happy number
function isHappynumber(n)
{
var 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
var n = 13;
if (isHappynumber(n))
document.write(n +
" is a Happy number");
else
document.write(n +
" is not a Happy number");
// This code contributed by Princi Singh
</script>
<?php
// PHP program to check a number
// is a Happy number or not
// Utility method to return
// sum of square of digit of n
function numSquareSum( $n)
{
$squareSum = 0;
while ($n)
{
$squareSum += ($n % 10) *
($n % 10);
$n /= 10;
}
return $squareSum;
}
// method return true if
// n is Happy number
function isHappynumber( $n)
{
$slow; $fast;
// initialize slow
// and fast by n
$slow = $n;
$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
$n = 13;
if (isHappynumber($n))
echo $n , " is a Happy number\n";
else
echo n , " is not a Happy number\n";
// This code is contributed by anuj_67.
?>
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 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
// 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");
}
}
# 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# 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
<script>
// This code is contributed by Vansh Sodhi.
// javascript 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
function isHappynumber(n)
{
var 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)
{
var d = x % 10;
sum += d * d;
x /= 10;
}
x = sum;
}
if(sum == 1 || sum == 7)
return true;
return false;
}
// Driver code
var n = 13;
if (isHappynumber(n))
document.write(n +
" is a Happy number");
else
document.write(n +
" is not a Happy number");
// This code is contributed by 29AjayKumar
</script>
Output13 is a Happy number
Complexity Analysis:
Time Complexity: O(n*log(n)).
Auxiliary Space: O(1).
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