Skip to content
Related Articles

Related Articles

Improve Article

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

  • Difficulty Level : Basic
  • Last Updated : 03 May, 2021

Given a number represented as 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:
112 = 121 
1 + 2 + 1 = 4

Input: str = 1111 
Output: 16 
 

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 <bits/stdc++.h>
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

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the sum
// of the digits of num ^ 2
function squareDigitSum(number)
{
    var summ = 0;
    var num = parseInt(number);
 
    // Store the square of num
    var squareNum = num * num;
 
    // Find the sum of its digits
    while (squareNum > 0)
    {
        summ = summ + (squareNum % 10);
        squareNum = parseInt(squareNum / 10);
    }
    return summ;
}
 
// Driver code
var N = "1111";
 
document.write(squareDigitSum(N));
 
// This code is contributed by todaysgaurav
 
</script>
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 <bits/stdc++.h>
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

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the sum
// of the digits of num^2
function squareDigitSum(s)
{
    // To store the number of 1's
    let lengthN = s.length;
 
    // Find the sum of the digits of num^2
    let result = parseInt(lengthN / 9) * 81
                 + Math.pow((lengthN % 9), 2);
 
    return result;
}
 
// Driver code
    let s = "1111";
 
    document.write(squareDigitSum(s));
 
</script>
Output: 
16

 

Time Complexity O(1)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :