Separate the Numbers Hacker Rank Solution Best & Easiest

Posted by admin October 11, 2022
Separate-the-Numbers-Problem-Solution
Separate-the-Numbers-Problem-Solution

In this post, we will solve the Separate the Numbers HackerRank Solution. This problem (Separate the Numbers) is a part of the HackerRank Problem Solving series.

A numeric string, s , is beautiful if it can be split into a sequence of two or more positive integers, a[1], a[2], …, a[n] , satisfying the following conditions:

  1.  a[i] – a[i-1] = 1 for any 1<i<=n (i.e., each element in the sequence is 1 more than the previous element).
  2. No a[i] contains a leading zero. For example, we can split  into the sequence , but it is not beautiful because  and  have leading zeroes.
  3. The contents of the sequence cannot be rearranged. For example, we can split  into the sequence , but it is not beautiful because it breaks our first constraint (i.e., ).

The diagram below depicts some beautiful strings:

image

Perform  queries where each query consists of some integer string . For each query, print whether or not the string is beautiful on a new line. If it is beautiful, print YES x, where  is the first number of the increasing sequence. If there are multiple such values of , choose the smallest. Otherwise, print NO.

Function Description

Complete the separateNumbers function in the editor below.

separateNumbers has the following parameter:

  • s: an integer value represented as a string

Prints
– string: Print a string as described above. Return nothing.

Input Format

The first line contains an integer , the number of strings to evaluate.
Each of the next  lines contains an integer string  to query.

Separate the Numbers Hacker Rank Solution

Problem solution in Python programming:

def separateNumbers(s):
    l = len(s)
    ls = []
    flag = 0
    for i in range(1,l//2+1):
        st = s[:i]
        temp = st
        count = 1
        while len(temp) < len(s):
            temp = temp + str(int(st)+count)
            count += 1
            if temp == s:
                print(f"YES {st}")
                flag = 1
                break
    if flag == 0:
        print("NO")

Problem solution in JavaScript programming:

function separateNumbers(s) {
    const len = s.length;
    let first = 0; // first number of the increasing sequence
    let nDigits = 1; // "91011", 1 = "9", 2 = "91", 3 = "910", ...
    let p = 0; // start index
    let q = 1;  // end index
    let incr = 1; // increment
    let number = 0; // string to BigInt
    let search = 0; // number to search (number + 1)
    let next = 0; // string segment for look up

    while (q < len) {
        number = BigInt(s.slice(p, q));
        search = (number + 1n).toString();
        nDigits = search.length;

        next = s.slice(q, q + nDigits);

        if (search[0] !== '0' && search === next) {
            if (p === 0) first = number;
            p = q;
            q = p + nDigits;
            if (q >= len) {
                console.log('YES ' + first);
                return 1;
            }
        } else {
            p = 0;
            q = (++incr);
        }
    }
    console.log('NO');

}

Problem solution in Java programming:

public static void separateNumbers(String s) {
    // Write your code here

 String sub = "";
        boolean isValid = false;
        for (int i = 1; i <= s.length() / 2; i++) {
            sub = s.substring(0, i);
            long num = Long.parseLong(sub);
            StringBuilder valid = new StringBuilder(sub);
            while (valid.length() < s.length()) {
                valid.append(++num);
            }
            if (s.equals(valid.toString())) {
                isValid = true;
                break;
            }
        }
        System.out.println(isValid ? "YES " + sub : "NO");
    }