In this Separate the Numbers HackerRank solution, A numeric string, , is beautiful if it can be split into a sequence of two or more positive integers, , satisfying the following conditions:
- for any (i.e., each element in the sequence is more than the previous element).
- No contains a leading zero. For example, we can split into the sequence , but it is not beautiful because and have leading zeroes.
- 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:
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.
Constraints
Separate the Numbers HackerRank Solution
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Problem Solution in Python
def separateNumbers(s):
for i in range(len(s)//2):
starting_digit = int(s[:i+1])
new_s = s[:i+1]
if len(new_s)*2 <= len(s):
digit = starting_digit
while len(new_s) < len(s):
digit += 1
new_s += str(digit)
if (new_s == s):
print("YES", str(starting_digit))
return
print("NO")Problem Solution in JavaScript
function separateNumbers(s) {
// Write your code here
for (let i = 1; i < s.length; i++) {
// use BigInt for test cases with large numbers
let curr = BigInt(s.slice(0, i));
let temp = curr.toString();
while (temp.length < s.length) {
curr++;
temp += curr.toString();
}
if (temp === s) {
return console.log(`YES ${ s.slice(0, i) }`);
}
}
return console.log("NO");
}