CareerCup

Void printFrequency(int arr[], int length)
{
  int current = a[0];
  int count = 1;
  
  for(int i = 1; i < length; i++)
  {
    if(a[i] != current || i == (length - 1))
    {
      cout<<a[i-1]<<":"<<count"\n";
      current = a[i];
      count = 1;
    }
    else
    {
      count++;
    }
  }
}

function stringrep ( str ) {
  var collection = str.split(' ');
  var seen = {};
  collection.forEach( function ( item ) {
    if ( item in seen ) {
      seen[item]++
    } else {
      seen[item] = 1;
    }
  });
  var resultingCollection = []
  for (var key in seen) {
    resultingCollection.push(key + ":" + seen[key])
  }
  return resultingCollection.join(", ")
}

Language - Ruby

def find_freq(str)
  result = {}
  str.split(' ').map do |val|
    (result[val]) ? (result[val] = result[val] + 1) : result[val] = 1
  end
  result
end

In Scala:

val m = str.split('  ').toList.groupBy(x=>x)
val k = m.keySet
k.foreach(x=>print(" " + x+":"+k(x).size))

Python:

#!/usr/bin/python
def recurse(list,dic,forbidden):
  if (len(list) == 0):
    print dic
    return
  else:
      try:
        item = int(list.pop())
      except :
        ''' try with hex '''
        item = int(list.pop(),16)
      if item in forbidden:
          recurse(list,dic,forbidden)
          return
      try:
        dic[item] += 1
      except KeyError:
        dic[item] = 1
      recurse(list,dic,forbidden)

if __name__  == "__main__":
  forbidden = [9,11]
  number = "1 0 1 11 19 19 19 12 9 9 11 0x4 0x4"
  result={}
  recurse(number.split(),result,forbidden)

public void q1(String str) {
        if (str==null || str.length()==0) return;
        String[] s = str.split(" ");
        int[] l = new int[s.length];
        for (int i=0; i<s.length; i++) {
            int h = l[Integer.valueOf(s[i])-1];
            l[Integer.valueOf(s[i])-1] = (h==0?1:h+1);
        }
        for (int i=0; i<l.length; i++) {
            if (l[i]>0)
                System.out.print((i+1)+":"+l[i]+" ");
        }
    }