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Frequency Count

texmommyof4texmommyof4 Posts: 13Member
Trying to work with 3 arrays. Taking an original finding the duplicates and counting the dupes and the new elements in the dupe array. Here is the code I have so far. It doesn't work. Asking for suggestions to where I am going wrong.
[code]

void frequencyCount(short original[],short nonduplicate[], short count[],short& countNew,short size)
{
short value=0;
short k=0;
short m=0;

nonduplicate[0] = original[0];//since I know first element is not yet duplicated I assign it to element 0
count[0]=1; //also nothing in Count yet so 0 element of count will count be 1 until element 0 is duped the should inc.
countNew=1;//new element in dupe array so element count is 1;

for(k=1;k<size;k++)//assign 1 to k because want to start now with that element in original array.
{
value=original[k];//do I even need the variable value?
if(value==nonduplicate[m])
count[m++];//if the value is a duplicate inc the value in the element of count array
else
m++; //otherwise put the nondupe in nondupe array
nonduplicate[m]=value;
count[m]=1;//make the count 1 for new num
countNew++;//new element count

}

}
[code]

Comments

  • stoberstober Posts: 9,765Member ✭✭✭
    I think it is missiong some parentheses

    : [code]
    :
    : void frequencyCount(short original[],short nonduplicate[], short count[],short& countNew,short size)
    : {
    : short value=0;
    : short k=0;
    : short m=0;
    :
    : nonduplicate[0] = original[0];//since I know first element is not yet duplicated I assign it to element 0
    : count[0]=1; //also nothing in Count yet so 0 element of count will count be 1 until element 0 is duped the should inc.
    : countNew=1;//new element in dupe array so element count is 1;
    :
    : for(k=1;k<size;k++)//assign 1 to k because want to start now with that element in original array.
    : {
    : value=original[k];//do I even need the variable value?
    : if(value==nonduplicate[m])
    : count[m++];//if the value is a duplicate inc the value in the element of count array
    : else
    [red] {[/red]
    : m++; //otherwise put the nondupe in nondupe array
    : nonduplicate[m]=value;
    : count[m]=1;//make the count 1 for new num
    : countNew++;//new element count
    [red] }[/red]
    :
    : }
    :
    : }
    : [/code]
    :
  • texmommyof4texmommyof4 Posts: 13Member
    :[blue]Thank you! :)[/blue]
    I think it is missiong some parentheses
    :
    : : [code]
    : :
    : : void frequencyCount(short original[],short nonduplicate[], short count[],short& countNew,short size)
    : : {
    : : short value=0;
    : : short k=0;
    : : short m=0;
    : :
    : : nonduplicate[0] = original[0];//since I know first element is not yet duplicated I assign it to element 0
    : : count[0]=1; //also nothing in Count yet so 0 element of count will count be 1 until element 0 is duped the should inc.
    : : countNew=1;//new element in dupe array so element count is 1;
    : :
    : : for(k=1;k<size;k++)//assign 1 to k because want to start now with that element in original array.
    : : {
    : : value=original[k];//do I even need the variable value?
    : : if(value==nonduplicate[m])
    : : count[m++];//if the value is a duplicate inc the value in the element of count array
    : : else
    : [red] {[/red]
    : : m++; //otherwise put the nondupe in nondupe array
    : : nonduplicate[m]=value;
    : : count[m]=1;//make the count 1 for new num
    : : countNew++;//new element count
    : [red] }[/red]
    : :
    : : }
    : :
    : : }
    : : [/code]
    : :
    :

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