# Sum of the digits of square of the given number which has only 1’s as its digits

Given a number represented as a string str consisting of the digit 1 only i.e. 1, 11, 111, …. The task is to find the sum of digits of the square of the given number.

Examples:

Input: str = 11
Output: 4
112 = 121
1 + 2 + 1 = 4

Input: str = 1111
Output: 16

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach: Find the square of the given number and then find the sum of its digits.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the sum  ` `// of the digits of num ^ 2 ` `int` `squareDigitSum(string number) ` `{ ` `    ``int` `summ = 0; ` `    ``int` `num = stoi(number); ` `     `  `    ``// Store the square of num ` `    ``int` `squareNum = num * num; ` ` `  `    ``// Find the sum of its digits ` `    ``while``(squareNum > 0) ` `    ``{ ` `        ``summ = summ + (squareNum % 10); ` `        ``squareNum = squareNum / 10; ` `    ``} ` `    ``return` `summ; ` `}  ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string N = ``"1111"``; ` ` `  `    ``cout << squareDigitSum(N); ` ` `  `    ``return` `0; ` `} ` ` `  `// This code is contributed by Princi Singh `

## Java

 `// Java implementation of the approach  ` `// Java implementation of the approach ` `class` `GFG  ` `{ ` ` `  `// Function to return the sum  ` `// of the digits of num ^ 2 ` `static` `int` `squareDigitSum(String number) ` `{ ` `    ``int` `summ = ``0``; ` `    ``int` `num = Integer.parseInt(number); ` `     `  `    ``// Store the square of num ` `    ``int` `squareNum = num * num; ` ` `  `    ``// Find the sum of its digits ` `    ``while``(squareNum > ``0``) ` `    ``{ ` `        ``summ = summ + (squareNum % ``10``); ` `        ``squareNum = squareNum / ``10``; ` `    ``} ` `    ``return` `summ; ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main (String[] args) ` `{  ` `    ``String N = ``"1111"``;  ` ` `  `    ``System.out.println(squareDigitSum(N));  ` `}  ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the sum  ` `# of the digits of num ^ 2 ` `def` `squareDigitSum(num): ` ` `  `    ``summ ``=` `0` `    ``num ``=` `int``(num) ` `     `  `    ``# Store the square of num ` `    ``squareNum ``=` `num ``*` `num ` ` `  `    ``# Find the sum of its digits ` `    ``while` `squareNum > ``0``: ` `        ``summ ``=` `summ ``+` `(squareNum ``%` `10``) ` `        ``squareNum ``=` `squareNum``/``/``10` ` `  `    ``return` `summ ` `     `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``N ``=` `"1111"` `    ``print``(squareDigitSum(N)) `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Function to return the sum  ` `    ``// of the digits of num ^ 2 ` `    ``static` `int` `squareDigitSum(String number) ` `    ``{ ` `        ``int` `summ = 0; ` `        ``int` `num = ``int``.Parse(number); ` ` `  `        ``// Store the square of num ` `        ``int` `squareNum = num * num; ` ` `  `        ``// Find the sum of its digits ` `        ``while``(squareNum > 0) ` `        ``{ ` `            ``summ = summ + (squareNum % 10); ` `            ``squareNum = squareNum / 10; ` `        ``} ` `        ``return` `summ; ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main (String[] args) ` `    ``{  ` `        ``String s = ``"1111"``;  ` `     `  `        ``Console.WriteLine(squareDigitSum(s));  ` `    ``}  ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:

```16
```

Efficient approach: It can be observed that in the square of the given number, the sequence [1, 2, 3, 4, 5, 6, 7, 9, 0] repeats in the left part and the sequence [0, 9, 8, 7, 6, 5, 4, 3, 2, 1] repeats in the right part. Both of these sequences appear floor(length(str) / 9) times and the sum of both of these sequences is 81 and the square of the number adds an extra 1 in the end.

So, the sum of all these would be [floor(length(str) / 9)] * 81 + 1.

And the middle digits have a sequence such as if length(str) % 9 = a then middle sequence is [1, 2, 3….a, a – 1, a – 2, … 2]. Now, it can be observed that sum of this part [1, 2, 3….a] is equal to (a * (a + 1)) / 2 and sum of the other part [a – 1, a – 2, … 2] is ((a * (a – 1)) / 2) – 1.
Total sum = floor(length(str) / 9) * 81 + 1 + (length(str) % 9)2 – 1 = floor(length(str) / 9) * 81 + (length(str) % 9)2.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `#define lli long long int ` ` `  `// Function to return the sum ` `// of the digits of num^2 ` `lli squareDigitSum(string s) ` `{ ` `    ``// To store the number of 1's ` `    ``lli lengthN = s.length(); ` ` `  `    ``// Find the sum of the digits of num^2 ` `    ``lli result = (lengthN / 9) * 81 ` `                 ``+ ``pow``((lengthN % 9), 2); ` ` `  `    ``return` `result; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string s = ``"1111"``; ` ` `  `    ``cout << squareDigitSum(s); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG  ` `{ ` `     `  `    ``// Function to return the sum  ` `    ``// of the digits of num^2  ` `    ``static` `long` `squareDigitSum(String s)  ` `    ``{  ` `        ``// To store the number of 1's  ` `        ``long` `lengthN = s.length();  ` `     `  `        ``// Find the sum of the digits of num^2  ` `        ``long` `result = (lengthN / ``9``) * ``81` `+  ` `                      ``(``long``)Math.pow((lengthN % ``9``), ``2``);  ` `     `  `        ``return` `result;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``String s = ``"1111"``;  ` `     `  `        ``System.out.println(squareDigitSum(s));  ` ` `  `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the sum  ` `# of the digits of num ^ 2 ` `def` `squareDigitSum(num): ` ` `  `    ``# To store the number of 1's ` `    ``lengthN ``=` `len``(num) ` ` `  `    ``# Find the sum of the digits of num ^ 2 ` `    ``result ``=` `(lengthN``/``/``9``)``*``81` `+` `(lengthN ``%` `9``)``*``*``2` ` `  `    ``return` `result ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``N ``=` `"1111"` `    ``print``(squareDigitSum(N)) `

## C#

 `// C# implementation of the approach  ` `using` `System; ` `                     `  `class` `GFG  ` `{ ` `     `  `// Function to return the sum  ` `// of the digits of num^2  ` `static` `long` `squareDigitSum(String s)  ` `{  ` `    ``// To store the number of 1's  ` `    ``long` `lengthN = s.Length;  ` ` `  `    ``// Find the sum of the digits of num^2  ` `    ``long` `result = (lengthN / 9) * 81 +  ` `                  ``(``long``)Math.Pow((lengthN % 9), 2);  ` ` `  `    ``return` `result;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main (String[] args) ` `{  ` `    ``String s = ``"1111"``;  ` ` `  `    ``Console.WriteLine(squareDigitSum(s));  ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```16
```

Time Complexity O(1)

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