# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 ` `

Output:

```3515
```

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