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

19 Mar, 2022
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 = 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++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to return the sum// of the digits of num ^ 2int 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 codeint 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 approachclass GFG{ // Function to return the sum// of the digits of num ^ 2static 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 codepublic 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 ^ 2def 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 codeif __name__ == "__main__":     N = "1111"    print(squareDigitSum(N))
// C# implementation of the approachusing 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 ^ 2function 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 codevar N = "1111"; document.write(squareDigitSum(N)); // This code is contributed by todaysgaurav </script>
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++
Java
Python3
C#
Javascript
// 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^2lli 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 codeint main(){    string s = "1111";     cout << squareDigitSum(s);     return 0;}
// Java implementation of the approachclass 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 ^ 2def 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 codeif __name__ == "__main__" :     N = "1111"    print(squareDigitSum(N))
// C# implementation of the approachusing System;                     class GFG{     // Function to return the sum// of the digits of num^2static 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 codepublic 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^2function 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) 
ankthon
29AjayKumar
Rajput-Ji
princi singh
todaysgaurav
rishavmahato348
subhamkumarm348
number-digits
Mathematical
Pattern Searching
Mathematical
Pattern Searching
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