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  1. Prepare
  2. Algorithms
  3. Strings
  4. Separate the Numbers
  5. Discussions

Separate the Numbers

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682 Discussions

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  • phanvantienpvt01
     + 0 comments

    My python sollution :

    def separateNumbers(s):
    for i in range(len(s)//2):
        n = s[:(i+1)]
        l = 0
        while l < len(s):
            t = str(int(n)+1)
            if s.find(t,l) != -1 :
                if s.find(n,l) + len(n) != s.find(t,l) :
                    break
                else :
                    l += len(n)
                    n = t
            else:
                l += len(n)
                break
        if l == len(s):
            print(f'YES {s[:(i+1)]}')
            return
    else :
        print('NO')

  • geovanni_luna
     + 0 comments

    Here is a O(n^2) time and and O(n) space:

    O(n^2) time because in the worse case scenario the outer loop will run through the entires string by 1/2 and the while loop will run through all charecters in the string. O(n) space because the string created will be as long as 's'

    def separateNumbers(s):
    start = ""
    
    for i in range(1, (len(s) // 2 + 1)):
        start = s[:i]
        num = int(start)
        res = start
        
        while len(res) < len(s):
            num += 1
            res += str(num)
            
            if s[:len(res)] != res:
                break
        
        if s == res:
            print(f"YES {start}")
            return
    
    print("NO")

  • yashdeora98294
     + 0 comments

    Here is problem solution in python, java c++ c and javascript- https://programmingoneonone.com/hackerrank-separate-the-numbers-solution.html

  • yashdeora98294
     + 0 comments

    thanks

  • shiaramoon
     + 0 comments
    #!/bin/python3

from collections import deque

def separateNumbers(s):
    d = 1
    n = len(s)
    
    while d < n:
        dq = deque(list(s[d:]))
        first = s[:d]
        a = s[:d]
        b = ""
        while dq:
            if not a or a[0] == '0':
                break
                
            if all(n == '9' for n in a):
                d += 1
            
            for _ in range(d):
                if dq:
                    b += dq.popleft()
            
            if not b or b[0] == '0' or int(b) - int(a) != 1:
                break
                
            if not dq:
                print(f"YES {first}")
                return
            else:
                a, b = b + '', ''
                
        d = len(first) + 1
    
    print("NO")
    return