Find the index of a given permutation in the sorted list of the permutations of a given string

Question

We\'re given a string and a permutation of the string.

For example, an input string sandeep and a permutation psdenae.

Find the pos

Answer 1

I tried to implement this in js. It works for string that have no repeated letters but I get a wrong count otherwise. Here is my code:

function x(str) {
var sOrdinata = str.split('').sort()
console.log('sOrdinata = '+ sOrdinata)
var str = str.split('')
console.log('str = '+str)
console.log('\n')
var pos = 1;

for(var j in str){
//console.log(j)

for(var i in sOrdinata){
if(sOrdinata[i]==str[j]){
  console.log('found, position: '+ i)
  sOrdinata.splice(i,1)
  console.log('Nuovo sOrdinata = '+sOrdinata)
  console.log('\n')
  break;
}
else{
  //calculate number of permutations
  console.log('valore di j: '+j)

  //console.log('lunghezza stringa da permutare: '+str.slice(~~j+1).length);
  if(str.slice(j).length >1 ){sub = str.slice(~~j+1)}else {sub = str.slice(j)}
  console.log('substring to be used for permutation: '+ sub)

  prep = nrepC(sub.join(''))
  console.log('prep = '+prep)

  num = factorial(sub.length)
  console.log('num = '+num)

  den = denom(prep)
  console.log('den = '+ den)

  pos += num/den
  console.log(num/den)
  console.log('\n')
 }
 }
}
console.log(pos)
return pos
}



/* ------------ functions used by main --------------- */ 

function nrepC(str){
var obj={}
var repeats=[]
var res= [];

for(x = 0, length = str.length; x < length; x++) {
var l = str.charAt(x)
obj[l] = (isNaN(obj[l]) ? 1 : obj[l] + 1);
}
//console.log(obj)

for (var i in obj){
if(obj[i]>1) res.push(obj[i])
}
if(res.length==0){res.push(1); return res}
else return res
}

function num(vect){
var res =  1

}


function denom(vect){
var res = 1
for(var i in vect){
res*= factorial(vect[i])
}
return res
}


function factorial (n){
if (n==0 || n==1){
return 1;
}
return factorial(n-1)*n;
}