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++

// 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;
}

Java

// 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));
    }
}

Python3

# 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

C#

// 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));
    }
}

PHP

<?php
// PHP implementation of the approach

$MOD = 1000000007;

// Function to return the required sum
function rangeSum($l, $r)
{
    global $MOD;
    $a = 1; $b = 9; $res = 0;
    for ($i = 1; $i <= 10; $i++)
    {
        $L = max($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 = $a * 10;
        $b = $b * 10 + 9;
    }
    return $res;
}

// Driver code
$l = 98; $r = 102;
echo rangeSum($l, $r);

// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>
// Javascript implementation of the approach
MOD=1000000007

// Function to return the required sum
function rangeSum(l, r)
{

    var a = 1, b = 9, res = 0;
    for (var i = 1; i <= 10; i++) {
        var L = Math.max(l, a);
        var R = Math.min(r, b);

        // If range is valid
        if (L <= R) {

            // Sum of AP
            var 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
var l = 98, r = 102;
document.write(rangeSum(l, r));

// This code is contributed by noob2000.
</script>

Output:

3515

Time Complexity: O(10), as we are using a loop to loop 10 times.

Auxiliary Space: O(1), as we are not using extra space.