# HackerRank Separate the Numbers Solution

#### ByBrokenprogrammers

Dec 13, 2022

Hello Programmers, In this post, you will learn how to solve HackerRank Separate the Numbers Solution. This problem is a part of the HackerRank Algorithms Series.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## HackerRank Separate the Numbers Solution

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 s = 10203 into the sequence {1, 02, 03}, but it is not beautiful because 02 and 03 have leading zeroes.
3. The contents of the sequence cannot be rearranged. For example, we can split s = 312 into the sequence {3, 1, 2}, but it is not beautiful because it breaks our first constraint (i.e.1 – 3 != 1).

The diagram below depicts some beautiful strings:

Perform q queries where each query consists of some integer string s. For each query, print whether or not the string is beautiful on a new line. If it is beautiful, print `YES x`, where x is the first number of the increasing sequence. If there are multiple such values of x, 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 q, the number of strings to evaluate.
Each of the next q lines contains an integer string s to query.

Constraints

• 1 <= q <= 10
• 1 <= |s| <= 32
• s[i] ∈ [0 – 9]

Sample Input 0

```7
1234
91011
99100
101103
010203
13
1
```

Sample Output 0

```YES 1
YES 9
YES 99
NO
NO
NO
NO
```

Explanation 0

The first three numbers are beautiful (see the diagram above). The remaining numbers are not beautiful:

• For s = 101103, all possible splits violate the first and/or second conditions.
• For s = 010203, it starts with a zero so all possible splits violate the second condition.
• For s = 13, the only possible split is {1, 3}, which violates the first condition.
• For s = 1, there are no possible splits because s only has one digit.

Sample Input 1

```4
99910001001
7891011
9899100
999100010001
```

Sample Output 1

```YES 999
YES 7
YES 98
NO```

## HackerRank Separate the Numbers Solution

### Separate the Numbers Solution in C

```#include <stdio.h>
#include <string.h>
typedef unsigned long long int Long;
char s[33];
int q;
Long x;
int isZeroLead(int i) { return s[i] == '0'; }
Long read(int i, int sz) {
char *pt = s + i;
Long ans = 0;
while(sz-- && *pt) {
ans = ans*10 + (*pt - '0');
pt++;
}
return ans;
}
int digs(Long x) {
int ll = 0;
while(x) {
ll++;
x /= 10;
}
return ll;
}
int check(Long fst, int len) {
Long last = fst, curr;
int lsz = digs(fst);
for(int i = lsz; i < len; i += lsz) {
if(digs(last + 1) != digs(last)) { lsz++; }

if(curr - last != 1) return 0;
last = curr;
}

return 1;
}
int main() {
scanf("%d",&q);
while(q--) {
scanf("%s", s);
Long x = -1, fst;
for(int i = 1, len = strlen(s); i <= (len>>1); i++) {
if(check(fst, len)) {
x = fst;
break;
}
}

if(x == -1) puts("NO");
else printf("YES %lld\n", x);
}

return 0;
}```

### Separate the Numbers Solution in Cpp

```#include <bits/stdc++.h>
#define modx 1000000009
#define ll long long
#define pb push_back
#define mp make_pair
#define PI 3.14159265359
using namespace std;
long long int gcd( long long int a , long long int b )
{
return b == 0 ? a : gcd( b , a%b );
}
#define N 34
string str ;
int main( )
{
int q ;
cin >> q ;
while( q-- ) {
cin >> str ;
if( str[ 0 ] == '0' ) {
cout << "NO" << endl ;
continue ;
}
bool flag = false ;
int n = str.length() ;
long long cur_val = 0 ;
for( int i = 0 ; i < n - 1 ; i++ ) {
cur_val = cur_val * 10 + str[ i ] - '0' ;
int j = i + 1 ;
long long exp_val = cur_val + 1 , new_val = 0 ;
while( j < n ) {
if( new_val == 0 && str[ j ] == '0' ) {
new_val = 1 ;
break ;
}

new_val = new_val * 10 + str[ j ] - '0' ;
if( new_val == exp_val ) {
exp_val++ ;
new_val = 0 ;
}
j++ ;
}
if( new_val == 0 ) {
cout << "YES" << " " << cur_val << endl ;
flag = true ;
break ;
}
}
if( flag == false ) cout << "NO" << endl ;
}
return 0 ;
}```

### Separate the Numbers Solution in Java

```import java.io.*;
import java.util.StringTokenizer;
/**
* @author Aydar Gizatullin a.k.a. lightning95, [email protected]
*         Created on 17.02.17.
*/
public class Main {
private void solve() {
int n = rw.nextInt();
main:
for (int i = 0; i < n; ++i) {
String s = rw.next();
if (s.startsWith("0") || s.length() == 1) {
rw.println("NO");
continue;
}
long x, cur;
cy:
for (int j = 1; j <= s.length() / 2; ++j) {
x = Long.parseLong(s.substring(0, j));
cur = x + 1;
int c = j;
while (c < s.length()) {
String p = String.valueOf(cur);
cur += 1;
if (s.startsWith(p, c)) {
c += p.length();
} else {
continue cy;
}
}
rw.println("YES" + " " + x);
continue main;
}
rw.println("NO");
}
}
private RW rw;
private String FILE_NAME = "file";
public static void main(String[] args) {
new Main().run();
}
private void run() {
rw = new RW(FILE_NAME + ".in", FILE_NAME + ".out");
solve();
rw.close();
}
private class RW {
private StringTokenizer st;
private PrintWriter out;
private boolean eof;
RW(String inputFile, String outputFile) {
out = new PrintWriter(new OutputStreamWriter(System.out));
File f = new File(inputFile);
try {
out = new PrintWriter(new FileWriter(outputFile));
} catch (IOException e) {
e.printStackTrace();
}
}
}
private String nextLine() {
String s = "";
try {
} catch (IOException e) {
e.printStackTrace();
}
return s;
}
private String next() {
while (st == null || !st.hasMoreTokens()) {
try {
} catch (IOException e) {
eof = true;
return "-1";
}
}
return st.nextToken();
}
private long nextLong() {
return Long.parseLong(next());
}
private int nextInt() {
return Integer.parseInt(next());
}
private void println() {
out.println();
}
private void println(Object o) {
out.println(o);
}
private void print(Object o) {
out.print(o);
}
private void close() {
try {
br.close();
} catch (IOException e) {
e.printStackTrace();
}
out.close();
}
}
}```

### Separate the Numbers Solution in Python

```def ok(s):
return s == "0" or s[0] != "0"
def can(s, x):
n = len(s)
s += 'x'*100
p = 0
while p < n:
t = str(x)
if s[p:p+len(t)] != t:
return False
x += 1
p += len(t)
return True
def solve(s):
for i in range(1, len(s)):
t = s[:i]
if ok(t) and can(s, int(t)):
print "YES", t
return
print "NO"
n = int(raw_input())
for i in range(n):
solve(raw_input())```

### Separate the Numbers Solution using JavaScript

```process.stdin.resume();
process.stdin.setEncoding('ascii');
var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;
process.stdin.on('data', function (data) {
input_stdin += data;
});
process.stdin.on('end', function () {
input_stdin_array = input_stdin.split("\n");
main();
});
return input_stdin_array[input_currentline++];
}
/////////////// ignore above this line ////////////////////
function foundRound(s, len){
var length = len;
var curr = s.substring(0, length);
for(var i=length; i < s.length; i+= length){
if(s[i] === '0'){
return false;
}

var all = true;
for(var j = 0; j < curr.length; j++){
if(parseInt(curr[j]) % 10 !== 9){
all = false;
}
}
if(all === true){
length++ ;

curr = '1';
for(var j=0; j < length-1; j++){
curr += '0';
}
}
else{
var string = curr;
var index = string.length-1;
while(index>-1 && string[index]==='9'){
string[index] = '0';
index--;
}
var newValue = parseInt(string[index]) + 1;
curr = string.substring(0,index) + newValue.toString();
while(index < string.length-1){
index++;
curr += '0';
}

}

if(i + length > s.length){
return false;
}

var next = s.substring(i, i + length);
if(curr !== next){
return false;
}
}
return true;
}
function main() {
for(var a0 = 0; a0 < q; a0++){
if(s.length < 2 || s[0]==='0'){
console.log('NO');
}
else{
var found = false;
for(var len = 1; len <= s.length/2 && !found; len++){
console.log('YES ' + s.substring(0,len));
found = true;
}
}
if(!found){
console.log('NO');
}
}
}
}```

### Separate the Numbers Solution in Scala

```import scala.io.StdIn

object Solution {
def main(args: Array[String]) {
(0 until q) foreach { i =>
val x = (1 to s.length / 2) flatMap { j => test(s, j) }
println(if (x.length > 0) s"YES \${x.min}" else "NO")
}
}

def test(s: String, i: Int): Option[Long] = {
def _test(s: String, n: Long): Boolean =
if (s.isEmpty) true
else {
val t = n.toString
if (s.startsWith(t)) _test(s.substring(t.length), n + 1)
else false
}
val n = s.substring(0, i).toLong
if (_test(s.substring(i), n + 1)) Some(n) else None
}
}```

### Separate the Numbers Solution in Pascal

```var
arr:array[1..2000] of longint;
i,j,k,c,casos,L,n:longint;
si:boolean;
res:int64;
procedure compara(i,t,a:int64);
var
x:int64;
j:longint;
begin
if i=L+1 then
si:=true
else
if i=1 then
begin
x:=0;
for i:=1 to L div 2 do
begin
compara(i+1,i,x);
if si then
begin
res:=x;
exit;
end;
exit;
end;
end
else
if t+i-1<=L then
begin
exit;
x:=0;
for j:=i to t+i-1 do
if x<=a then
begin
if t+i>L then
exit;
t:=t+1;
end;
if x=a+1 then
compara(i+t,t,x);
end;
end;
begin
for c:=1 to casos do
begin
if L=1 then
si:=false
else
si:=true;
if si then
begin
si:=false;
compara(1,1,-1);
end;
if si then
writeln('YES ',res)
else
writeln('NO');
end;
end.
```

Disclaimer: This problem (Separate the Numbers) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.