Breaking a number such that first part is integral division of second by a power of 10

Given a very large number N, we need to count the total ways such that if we divide the number into two parts a and b, the first part a can be obtained by integral division of second b by some power p of 10 and p>=0. 1 <= No of digits in N <= . 

Examples:

Input : 220
Output : 1
220 can be divided as a = 2 and b = 20
such that for p = 1, b/10 = a.


Input : 1111
Output : 2
We get answer 2 because we need to consider
integral division.
Let's consider the first partition a = 1,
b = 111. for p = 2, b/pow(10,p) = a thus
this is a valid partition.
now a = 11, b = 11. for p = 0, b/pow(10,p) 
= a thus this too is a valid combination.

Input : 2202200
Output : 2
for a = 2 b = 202200, p = 5 and
a = 220, b = 2200, p = 1

Since the number can be very large to be contained even in a long long int we will store it as a string. According to the conditions mentioned in the problem, division is the floor function. A simple and inefficient approach will be to divide the string into two substrings then convert those to integer and perform division. An efficient method to do it will be to use the string compare function to match the most significant digits of the two strings and ignore the rest( floor function). 

Below is the implementation of this idea : 

C++

#include <bits/stdc++.h>
using namespace std;

// c++ function to count ways to divide a
// string in two parts a and b such that
// b/pow(10, p) == a
int calculate(string N)
{
    int len = N.length();
    int l = (len) / 2;
    int count = 0;

    for (int i = 1; i <= l; i++) {

        // substring representing int a
        string s = N.substr(0, i);

        // no of digits in a
        int l1 = s.length();

        // consider only most significant
        // l1 characters of remaining string
        // for int b
        string t = N.substr(i, l1);

        // if any of a or b contains leading 0s
        // discard this combination
        if (s[0] == '0' || t[0] == '0')
            continue;

        // if both are equal
        if (s.compare(t) == 0)
            count++;
    }
    return count;
}

// driver function to test above function
int main()
{
    string N = "2202200";
    cout << calculate(N);
    return 0;
}

Java

// Java program to count ways to divide a
// String in two parts a and b such that
// b/pow(10, p) == a
import java.util.*;

class GFG
{
static int calculate(String N)
{
    int len = N.length();
    int l = (len) / 2;
    int count = 0;

    for (int i = 1; i <= l; i++)
    {

        // subString representing int a
        String s = N.substring(0, i);

        // no of digits in a
        int l1 = s.length();

        // consider only most significant
        // l1 characters of remaining String
        // for int b
        String t = N.substring(i, l1 + i);

        // if any of a or b contains leading 0s
        // discard this combination
        if (s.charAt(0) == '0' || t.charAt(0) == '0')
            continue;

        // if both are equal
        if (s.compareTo(t) == 0)
            count++;
    }
    return count;
}

// Driver Code
public static void main(String[] args)
{
    String N = "2202200";
    System.out.print(calculate(N));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program to count ways to divide
# a string in two parts a and b such that
# b/pow(10, p) == a

def calculate( N ):
    length = len(N)
    l = int((length) / 2)
    count = 0

    for i in range(l + 1):
        print(i)

        # substring representing int a
        s = N[0: 0 + i]

        # no of digits in a
        l1 = len(s)
        print(s,l1)

        # consider only most significant
        # l1 characters of remaining
        # string for int b
        t = N[i: l1 + i]

        # if any of a or b contains
        # leading 0s discard this
        try:
            if s[0] == '0' or t[0] == '0':
                continue
        except:
            continue

        # if both are equal
        if s == t:
            count+=1
        print(i,N[i],count)
    return count

# driver code to test above function
N = str("2202200")
print(calculate(N))

# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to count ways to divide a
// String in two parts a and b such that
// b/pow(10, p) == a
using System;

class GFG
{
static int calculate(String N)
{
    int len = N.Length;
    int l = (len) / 2;
    int count = 0;

    for (int i = 1; i <= l; i++)
    {

        // subString representing int a
        String s = N.Substring(0, i);

        // no of digits in a
        int l1 = s.Length;

        // consider only most significant
        // l1 characters of remaining String
        // for int b
        String t = N.Substring(i, l1);

        // if any of a or b contains leading 0s
        // discard this combination
        if (s[0] == '0' || t[0] == '0')
            continue;

        // if both are equal
        if (s.CompareTo(t) == 0)
            count++;
    }
    return count;
}

// Driver Code
public static void Main(String[] args)
{
    String N = "2202200";
    Console.Write(calculate(N));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// JavaScript program to count ways to divide
// a string in two parts a and b such that
// b/pow(10, p) == a

function calculate( N ){
    let len = N.length
    let l = Math.floor((len) / 2)
    let count = 0

    for(let i = 1; i < l + 1; i++)
    {

        // substring representing int a
        let s = N.substr(0, i)

        // no of digits in a
        let l1 = s.length

        // consider only most significant
        // l1 characters of remaining
        // string for int b
        let t = N.substr(i,l1)


        // if any of a or b contains
        // leading 0s discard this

        if (s[0] == '0' || t[0] == '0'){
            continue
        }

        // if both are equal
        if(s === t)
            count++
    }
    return count
}

// driver code to test above function
let N = "2202200"
console.log(calculate(N))

// This code is contributed by shinjanpatra

</script>

Output

2

Time Complexity: O(n²)
Auxiliary Space: O(n)