Sum of i * countDigits(i)^2 for all i in range [L, R]
Last Updated :
23 May, 2022
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++
#include <bits/stdc++.h>
using namespace std;
#define MOD 1000000007
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 (L <= R) {
int sum = (L + R) * (R - L + 1) / 2;
res += (i * i) * (sum % MOD);
res %= MOD;
}
a = a * 10;
b = b * 10 + 9;
}
return res;
}
int main()
{
int l = 98, r = 102;
cout << rangeSum(l, r);
return 0;
}
|
Java
class GFG {
static final int MOD = 1000000007 ;
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 (L <= R) {
int sum = (L + R) * (R - L + 1 ) / 2 ;
res += (i * i) * (sum % MOD);
res %= MOD;
}
a = a * 10 ;
b = b * 10 + 9 ;
}
return res;
}
public static void main(String args[])
{
int l = 98 , r = 102 ;
System.out.print(rangeSum(l, r));
}
}
|
Python3
MOD = 1000000007 ;
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 (L < = R) :
sum = (L + R) * (R - L + 1 ) / / 2 ;
res + = (i * i) * ( sum % MOD);
res % = MOD;
a * = 10 ;
b = b * 10 + 9 ;
return res;
if __name__ = = "__main__" :
l = 98 ; r = 102 ;
print (rangeSum(l, r));
|
C#
using System;
class GFG {
const int MOD = 1000000007;
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 (L <= R) {
int sum = (L + R) * (R - L + 1) / 2;
res += (i * i) * (sum % MOD);
res %= MOD;
}
a = a * 10;
b = b * 10 + 9;
}
return res;
}
public static void Main()
{
int l = 98, r = 102;
Console.WriteLine(rangeSum(l, r));
}
}
|
PHP
<?php
$MOD = 1000000007;
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 ( $L <= $R )
{
$sum = ( $L + $R ) * ( $R - $L + 1) / 2;
$res += ( $i * $i ) * ( $sum % $MOD );
$res %= $MOD ;
}
$a = $a * 10;
$b = $b * 10 + 9;
}
return $res ;
}
$l = 98; $r = 102;
echo rangeSum( $l , $r );
?>
|
Javascript
<script>
MOD=1000000007
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 (L <= R) {
var sum = (L + R) * (R - L + 1) / 2;
res += (i * i) * (sum % MOD);
res %= MOD;
}
a = a * 10;
b = b * 10 + 9;
}
return res;
}
var l = 98, r = 102;
document.write(rangeSum(l, r));
</script>
|
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.
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