Sam and substrings

Problem Statement :

Samantha and Sam are playing a numbers game. Given a number as a string, no leading zeros, determine the sum of all integer values of substrings of the string.

Given an integer as a string, sum all of its substrings cast as integers. As the number may become large, return the value modulo 10^9 +7.

n = '42'
Here n is a string that has 3 integer substrings: 4, 2, and 42. Their sum is 48, and 48 modulo{10^9 +7)=48.

Function Description

Complete the substrings function in the editor below.

substrings has the following parameter(s):

string n: the string representation of an integer

int: the sum of the integer values of all substrings in n, modulo 10^9 + 7
Input Format

A single line containing an integer as a string, without leading zeros.


1 <= ncastasaninteger <= 2 x 10^5

Solution :


                            Solution in C :

In C++ :

#include <iostream>
#include <string.h>
using namespace std;

#define MOD 1000000007

int main() {
	char num[200005];
	int len = strlen(num);
	long long factor = 1, ans = 0;

	for (int i = len-1; i >= 0; --i) {
		long long tmp = (num[i]-'0') * (i+1) * factor % MOD;
		ans += tmp;
		ans %= MOD;
		factor = (factor * 10 + 1) % MOD;
	printf("%d\n", ans);

	return 0;

In Java :

import java.util.Scanner;

public class Solution {

	static long mod = 1000000007;
	public static void main(String[] args) {
		Scanner cin = new Scanner(;
		char s[] =;
		long ten[] = new long[s.length + 2];
		ten[0] = 1;
		for (int i=1; i<ten.length; i++) {
			ten[i] = ten[i - 1] * 10 + 1;
			ten[i] %= mod;
		long ans = 0;
		for (int i=0; i<s.length; i++) {
			ans += ((s[i] - '0') * ten[s.length - i - 1]) % mod * (i + 1) % mod;
			ans %= mod;


In C :

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAXLEN 200001
#define MODULUS 1000000007LL

int main() {
	long long total, multiplier, partialsum;
	char c;
	total = partialsum = 0;
	multiplier = 1;
	while (1) {
		if (c<'0' || c>'9') break;
		c -= '0';
		partialsum += multiplier * c;
		total = (total * 10 + partialsum) % MODULUS;
	return 0;

In Python3 :

import operator

MOD = 1000000007

s = input()
n = len(s)
a = []
p10 = 0
for i in range(n, 0, -1):
	p10 = (p10 * 10 + 1) % MOD
	a.append(p10 * i % MOD)

b = [ord(c) - ord('0') for c in s]
print(sum(map(operator.mul, reversed(a), b)) % MOD)

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