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

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++

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// 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

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Java

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// 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

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Python3

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# 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))

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C#

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// 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

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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++

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// 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;
}

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Java

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// 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

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Python3

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# 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))

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C#

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// 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

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Output:

16

Time Complexity O(1)

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