Sum of i * countDigits(i)^2 for all i in range [L, R]

Given a range [L, R], the task is to find the sum i * countDigits(i)2 for all i ∈ [L, R] where countDigits(i) is the count of digits in i.

That is, find:

L * countDigits(L)2 + (L + 1) * countDigits(L + 1)2 + ….. + R * countDigits(R)2.

Examples:

Input: L = 8, R = 11
Output: 101
8 * 12 + 9 * 12 + 10 * 22 + 11 * 22 = 8 + 9 + 40 + 44 = 101

Input: L = 98, R = 102
Output: 2
98 * 22 + 99 * 22 + 100 * 32 + 101 * 32 + 102 * 32 = 3515



Approach: We break the segment [L, R] into several segments of the numbers with the same number of digits.
[1 – 9], [10 – 99], [100 – 999], [1000 – 9999], [10000 – 99999], [100000 – 999999], [10000000 – 99999999] and so on.
When L and R are of the same length then the required sum will be countDigits(L)2 * (L + R) * (R – L + 1) / 2

Proof:

Let [L, R] = [10, 14] where L and R are of the same length i.e. 2.
Therefore, the sum for the segment [L, R] will be 10 * 22 + 11 * 22 + 12 * 22 + 13 * 22 + 14 * 22.
Take 22 common, 22 * (10 + 11 + 12 + 13 + 14) = totalDigits2 * (Sum of AP)
Sum of AP = (no of terms / 2) * (first term + last term) i.e. (R – L + 1) * (L + R) / 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 MOD 1000000007
  
// Function to return the required sum
int rangeSum(int l, int r)
{
  
    int a = 1, b = 9, res = 0;
    for (int i = 1; i <= 10; i++) {
        int L = max(l, a);
        int R = min(r, b);
  
        // If range is valid
        if (L <= R) {
  
            // Sum of AP
            int sum = (L + R) * (R - L + 1) / 2;
            res += (i * i) * (sum % MOD);
            res %= MOD;
        }
        a = a * 10;
        b = b * 10 + 9;
    }
    return res;
}
  
// Driver code
int main()
{
    int l = 98, r = 102;
    cout << rangeSum(l, r);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG {
  
    static final int MOD = 1000000007;
  
    // Function to return the required sum
    static int rangeSum(int l, int r)
    {
  
        int a = 1, b = 9, res = 0;
        for (int i = 1; i <= 10; i++) {
            int L = Math.max(l, a);
            int R = Math.min(r, b);
  
            // If range is valid
            if (L <= R) {
  
                // Sum of AP
                int sum = (L + R) * (R - L + 1) / 2;
                res += (i * i) * (sum % MOD);
                res %= MOD;
            }
            a = a * 10;
            b = b * 10 + 9;
        }
        return res;
    }
  
    // Driver code
    public static void main(String args[])
    {
  
        int l = 98, r = 102;
        System.out.print(rangeSum(l, r));
    }
}

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Python3

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# Python3 implementation of the approach 
  
MOD = 1000000007;
  
# Function to return the required sum 
def rangeSum(l, r) :
  
    a = 1; b = 9; res = 0;
    for i in range(1, 11) :
        L = max(l, a);
        R = min(r, b); 
          
        # If range is valid
        if (L <= R) :
              
            # Sum of AP
            sum = (L + R) * (R - L + 1) // 2;
            res += (i * i) * (sum % MOD); 
            res %= MOD; 
          
        a *= 10;
        b = b * 10 + 9;
      
    return res; 
  
# Driver code 
if __name__ == "__main__" :
  
    l = 98 ; r = 102
      
    print(rangeSum(l, r)); 
  
# This code is contributed by Ryuga

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C#

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// C# implementation of the approach
using System;
class GFG {
  
    const int MOD = 1000000007;
  
    // Function to return the required sum
    static int rangeSum(int l, int r)
    {
  
        int a = 1, b = 9, res = 0;
        for (int i = 1; i <= 10; i++) {
            int L = Math.Max(l, a);
            int R = Math.Min(r, b);
  
            // If range is valid
            if (L <= R) {
  
                // Sum of AP
                int sum = (L + R) * (R - L + 1) / 2;
                res += (i * i) * (sum % MOD);
                res %= MOD;
            }
            a = a * 10;
            b = b * 10 + 9;
        }
        return res;
    }
  
    // Driver code
    public static void Main()
    {
        int l = 98, r = 102;
        Console.WriteLine(rangeSum(l, r));
    }
}

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PHP

Output:

3515


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