finding sequential in string using javascript

user_11935558 0
Viewed 503

I am going to find sequential string of numbers in string which starts from 1. For example I have this string.

"456000123456009123456780001234000"

Here sequential strings would be

"123456", "12345678", "1234"

How can I get above result using Javascript efficiently? The code looks like this.

findSequential("456000123456009123456780001234000");
//expected output
"123456", "12345678", "1234"

***note: "1" itself is not sequential, for example: 

"3938139" - has no sequence
"39381249" - has "12"

With any language's solution would be appreciated but prefer Javascript or C. Thanks for your help!

algorithm javascript
5 Answers

A simple for loop should be able to achieve this. In JavaScript: 

function findSequential(s) {
    const res = []
    let current = []
    let num = 1
    for(let char of s) {
      if (char == num) {
        current.push(char)
        num ++
      } else if (current.length > 1) {
        res.push(current.reduce((acc, cur) => acc += cur, ''))
        if (char == 1) {
          current = ['1']
          num = 2
        } else {
          current = []
          num = 1
        }
      } else if (current.length === 1) {
        current = []
        num = 1
      }
    }
    if (current.length > 1) {
      res.push(current.reduce((acc, cur) => acc += cur, ''))
    }
    return res
}

console.log(findSequential('31234121'))

user_57401810

You could do it this way:

function findSequential(str) {
  const res = [];
  let currentHigh = 0;
  for (let char of str) {
    if (+char === currentHigh + 1) {
      currentHigh++;
    } else if (currentHigh > 0) {
      res.push(buildSequence(currentHigh))
      currentHigh = +char === 1 ? 1 : 0;
    }
  }
  if (currentHigh > 0) {
    res.push(buildSequence(currentHigh));
  }
  return res;
}

function buildSequence(max) {
  if (max <= 1) { return "1"; }
  else { return buildSequence(max - 1) + max; }
}

console.log(findSequential("456000123456009123456780001234000"));
// ["123456", "12345678", "1234"]

console.log(findSequential("12"));
// ["12"]

console.log(findSequential("1212"));
// ["12", "12"]

console.log(findSequential("23"));
// []

user_19137290

Finally, i have this code for you

var str = "456000123456009123456780001234000";
var length = str.length;
var out= "";

for(i = 0; i < length; i++){
 if(str[i] == 1){
  out += str[i];
  var j = i;
  do{  
   j++;
   out += str[j];
  } while(str[j + 1] > out[out.length - 1])
  out += " ";
 }
}

console.log(out); // Outputs: 123456 12345678 1234

user_136623390

very simple and short code 

s= "456000123456009123456780001234000"
    var p
    var r=''
    for(let n of s ){
      if(n==1||n-p==1){
        r+=n, p=n
        r.length>1&&r.charAt(r.length-1)==1?r=r.substring(0,r.length-1)+'-1':r
      }
    }
    console.log(r.split('-'))

user_99220110

Here's a recursive version that keeps track of the index and length of the current sequence until it actually needs to output it, then calls substr. (Also useful to easily convert to returning just the sequences' locations in the string.)

function f(s, i=1, l=0){
  if (i == s.length)
    return l ? [s.substr(i-l, l)] : [];
    
  if (s[i] != Number(s[i-1]) + 1 && l)
    return [s.substr(i-l, l)].concat(f(s, i+1, 0));
    
  if (s[i] == 2 && s[i-1] == 1)
    return f(s, i+1, 2);
    
  return f(s, i+1, l ? l+1: 0);
}

var strs = [
  "456000123456009123456780001234000",
  "1212",
  "12",
  "23"
];

for (let s of strs){
  console.log(s);
  console.log(JSON.stringify(f(s)));
  console.log('');
}

user_20347870
Related