Sum of the digits of square of the given number which has only 1’s as its digits
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++
// 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 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
# 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> |
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++
// 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 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
# 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> |
16
Time Complexity O(1)
Auxiliary Space: O(1)
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