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: 4
112 = 121
1 + 2 + 1 = 4Input: 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++ 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 implementation of the approach // Java implementation of the approach import java.io.*;
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 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# 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 |
<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> |
16
Time complexity: O(log10n), where n is no of digits in the given number
Auxiliary Space: O(1)
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++ 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 implementation of the approach import java.io.*;
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 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# 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 |
<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> |
16
Time Complexity O(1)
Auxiliary Space: O(1)