Check if a string can be split into substrings starting with N followed by N characters

Given a string str, the task is to check if it can be split into substrings such that each substring starts with a numeric value followed by a number of characters represented by that numeric integer. Examples:

Input: str = “4g12y6hunter” Output: Yes Explanation: Substrings “4g12y” and “6hunter” satisfy the given condition Input: str = “31ba2a” Output: No Explanation: The entire string cannot be split into substrings of desired types

Approach:

• Check for the conditions when a split is not possible:
• If the given string does not start with a number.
• If the integer, in the beginning of a substring, is greater than the total number of succeeding characters in the remaining substring.
• If the above two condition are not satisfied, an answer is definitely possible. Hence find the substrings recursively.

Below is the implementation of the above approach:

C++

// C++ Program to implement the
// above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if the given
// can be split into desired
// substrings
bool helper(string& s, int pos)
{
    // Length of the string
    int len = s.size();

    if (pos >= len)
        return true;

    if (!isdigit(s[pos]))
        return false;

    int num = 0;

    // Traverse the string
    for (int i = pos; i < len; i++) {

        // Extract the digit
        num = num * 10 + s[pos] - '0';

        // Check if the extracted number
        // does not exceed the remaining
        // length
        if (i + 1 + num > len)
            return false;

        // Check for the remaining
        // string
        if (helper(s, i + 1 + num))
            return true;
    }

    // If generating desired
    // substrings is not possible
    return false;
}

// Driver Code
int main()
{
    string s = "123abc4db1c";
    if (helper(s, 0))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java program to implement the
// above approach
import java.util.*;

class GFG{

// Function to check if the given
// can be split into desired
// substrings
public static boolean helper(String s, int pos)
{

    // Length of the string
    int len = s.length();

    if (pos >= len)
        return true;
    if (!Character.isDigit(s.charAt(pos)))
        return false;

    int num = 0;

    // Traverse the string
    for(int i = pos; i < len; i++)
    {

    // Extract the digit
    num = num * 10 + s.charAt(pos) - '0';

    // Check if the extracted number
    // does not exceed the remaining
    // length
    if (i + 1 + num > len)
        return false;

    // Check for the remaining
    // string
    if (helper(s, i + 1 + num))
        return true;
    }

    // If generating desired
    // substrings is not possible
    return false;
}

// Driver code
public static void main (String[] args)
{
    String s = "123abc4db1c";

    if (helper(s, 0))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to implement the
# above approach

# Function to check if the given
# can be split into desired
# substrings
def helper(s, pos):

    # Length of the string
    size = len(s)

    if(pos >= size):
        return True

    if(s[pos].isdigit() == False):
        return False

    num = 0

    # Traverse the string
    for i in range(pos, size):

        # Extract the digit
        num = num * 10 + ord(s[pos]) - 48

        # Check if the extracted number
        # does not exceed the remaining
        # length
        if (i + 1 + num > size):
            return False

        # Check for the remaining
        # string
        if (helper(s, i + 1 + num)):
            return True

    # If generating desired
    # substrings is not possible
    return False

# Driver Code
s = "123abc4db1c";

if (helper(s, 0)):
    print("Yes")
else:
    print("No")

# This code is contributed by Sanjit_Prasad

C#

// C# program to implement the
// above approach
using System;

class GFG{

// Function to check if the given
// can be split into desired
// substrings
public static bool helper(String s, int pos)
{

    // Length of the string
    int len = s.Length;

    if (pos >= len)
        return true;
    if (!char.IsDigit(s[pos]))
        return false;

    int num = 0;

    // Traverse the string
    for(int i = pos; i < len; i++)
    {

    // Extract the digit
    num = num * 10 + s[pos] - '0';

    // Check if the extracted number
    // does not exceed the remaining
    // length
    if (i + 1 + num > len)
        return false;

    // Check for the remaining
    // string
    if (helper(s, i + 1 + num))
        return true;
    }

    // If generating desired
    // substrings is not possible
    return false;
}

// Driver code
public static void Main(String[] args)
{
    String s = "123abc4db1c";

    if (helper(s, 0))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}

// This code is contributed by amal kumar choubey

Javascript

// JavaScript program to implement the above approach

// Function to check if the given
// can be split into desired
// substrings
function helper(s, pos) {
    // Length of the string
    let size = s.length;
    if (pos >= size) {
        return true;
    }

    if (!/^\d+$/.test(s[pos])) {
        return false;
    }

    let num = 0;

    // Traverse the string
    for (let i = pos; i < size; i++) {
        // Extract the digit
        num = num * 10 + parseInt(s[i]);

        // Check if the extracted number
        // does not exceed the remaining
        // length
        if (i + 1 + num > size) {
            return false;
        }

        // Check for the remaining
        // string
        if (helper(s, i + 1 + num)) {
            return true;
        }
    }

    // If generating desired
    // substrings is not possible
    return false;
}

// Driver Code
let s = "123abc4db1c";

if (helper(s, 0)) {
    console.log("Yes");
} else {
    console.log("No");
}

//contributed by adityasha4x71

Output

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)