Sum of i * countDigits(i)^countDigits(i) for all i in range [L, R]
Last Updated :
24 Nov, 2021
Given two numbers L, R which signify a range [L, R], the task is to find the sum i * countDigits(i)countDigits(i) for all i ? [L, R] where countDigits(i) is the count of digits in i.
Examples:
Input: L = 8, R = 11
Output: 101
Explanation:
For the given numbers, we need to find the values for all the numbers in the range [L, R].
=> 8 * 11 + 9 * 11 + 10 * 22 + 11 * 22
=> 8 + 9 + 40 + 44
=> 101
Input: L = 98, R = 102
Output: 8969
Explanation:
For the given numbers, we need to find the values for all the numbers in the range [L, R].
=> 98 * 22 + 99 * 22 + 100 * 33 + 101 * 33 + 102 * 33
=> 8969
Approach: The idea is to break the range into the segments according to the number of digits. That is, every segment contains numbers with same number of digits:
[1 - 9], [10 - 99], [100 - 999], [1000 - 9999], [10000 - 99999] ...
From the above segments, clearly, both [L, R] of each segment have the same lengths. Therefore, the required sum for the segment is:
countDigits(L)countDigits(L) * (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
22 * (10 + 11 + 12 + 13 + 14)
=> totalDigitstotalDigits * (Sum of AP)
- Sum of AP = (no of terms / 2) * (first term + last term) i.e. (R – L + 1) * (L + R) / 2.
- Therefore, the required sum is:
countDigits(L)countDigits(L) * (L + R) * (R - L + 1) / 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 += pow (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
import java.util.*;
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 += Math.pow(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 + = pow (i, i) * ( sum % MOD)
res % = MOD
a = a * 10
b = b * 10 + 9
return res
if __name__ = = "__main__" :
l = 98
r = 102
print (rangeSum(l, r))
|
C#
using System;
class GFG{
static readonly 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 += ( int )Math.Pow(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;
Console.Write(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 += Math.pow(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)
Auxiliary Space: O(1)
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