Frequency Count - Programmers Heaven

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# Frequency Count

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]

• 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]
:
• 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]
: :
: