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Check if a numeric string can be split into substrings having difference between consecutive numbers equal to K

  • Difficulty Level : Hard
  • Last Updated : 07 Sep, 2021

Given a numeric string S consisting of N digits and a positive integer K, the task is to check if the given string can be split into more than one substrings with difference between consecutive substrings equalt to K.

Examples:

Input: S = “8642”, K = 2
Output: Yes
Explanation: Split the given string as {“8”, “6”, “4”, “2”}. Now, the difference between the consecutive substrings is K(= 2).

Input: S = “1009896”, K = 0
Output: No

Approach: The given problem can be solved by generating all possible substring of the given string and check if the concatenation of any subset of the generated substring is equal to the given string S and the consecutive difference of the number as a substring is K, then print Yes. Otherwise, print No. Follow the below steps to solve the problem:



Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if a numeric string
// can be split into substrings such that
// the difference between the consecutive
// substrings is K
void isPossible(string s, int K)
{
    bool valid = false;
    long firstx = -1;
 
    // Stores the size of the string
    int n = s.length();
 
    // Iterate over the range [1, N] and
    // try each possible starting number
    for (int i = 1; i <= n / 2; ++i) {
 
        long x = stol(s.substr(0, i));
        firstx = x;
 
        // Convert the number to string
        string test = to_string(x);
 
        // Build up a sequence
        // starting with the number
        while (test.length() < s.length()) {
            x -= K;
            test += to_string(x);
        }
 
        // Compare it with the
        // original string s
        if (test == s) {
            valid = true;
            break;
        }
    }
 
    // Print the result
    cout << ((valid == true) ? "Yes " : "No");
}
 
// Driver Code
int main()
{
    string S = "8642";
    int K = 2;
    isPossible(S, K);
 
    return 0;
}

Python3




# python 3 program for the above approach
 
# Function to check if a numeric string
# can be split into substrings such that
# the difference between the consecutive
# substrings is K
def isPossible(s,K):
    valid = False
    firstx = -1
 
    # Stores the size of the string
    n = len(s)
 
    # Iterate over the range [1, N] and
    # try each possible starting number
    for i in range(1,n//2+1,1):
        x = int(s[0:i])
        firstx = x
 
        # Convert the number to string
        test = str(x)
 
        # Build up a sequence
        # starting with the number
        while (len(test) < len(s)):
            x -= K
            test += str(x)
 
        # Compare it with the
        # original string s
        if (test == s):
            valid = True
            break
 
    # Print the result
    print("Yes") if valid == True else print("No")
 
# Driver Code
if __name__ == '__main__':
    S = "8642"
    K = 2
    isPossible(S, K)
     
    # This code is contributed by ipg2016107.

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to check if a numeric string
// can be split into substrings such that
// the difference between the consecutive
// substrings is K
function isPossible(s, K)
{
    let valid = false;
    let firstx = -1;
 
    // Stores the size of the string
    let n = s.length;
 
    // Iterate over the range [1, N] and
    // try each possible starting number
    for (let i = 1; i <= n / 2; ++i) {
 
        let x = (s.substr(0, i));
        firstx = x;
 
        // Convert the number to string
        let test = x.toString();
 
        // Build up a sequence
        // starting with the number
        while (test.length < s.length) {
            x -= K;
            test += x.toString();
        }
 
        // Compare it with the
        // original string s
        if (test == s) {
            valid = true;
            break;
        }
    }
 
    // Prlet the result
    document.write((valid == true) ? "Yes " : "No");
}
// Driver Code
 
    let S = "8642";
    let K = 2;
    isPossible(S, K);
     
</script>
Output: 
Yes

 

Time Complexity: O(N2)
Auxiliary Space: O(N)

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