least frequent common number from a int array - c++

Question

I have to find least common number from an int array , I have written code but it is not working properly ,
Here is my logic,
1. sort the array
2. get min common counter updated
3. get if all are unique
and the code below,
static int min_loc ; //minimum value location
static int min_cnt ;
int all_uniqFlag = true;
void leastCommon(int data[],int n)
{
int rcount = 0; //Repeated number counter
int mcount = n; // minimum repetetion counter;
// The array is already sorted we need to only find the least common value.
for(int i = 0 ; i < n-1 ; i++)
{
//Case A : 1 1 2 2 2 3 3 3 3 4 5 5 5 5 : result should be 4
//Case B : 1 2 3 4 5 6 7 (All unique number and common values so all values should be printed
// and )
//Case C : 1 1 2 2 3 3 4 4 (all numbers have same frequency so need to display all )
cout << "data[i] : " << data[i] << " data[i+1] : " << data[i+1] << "i = " << i << endl;
if(data[i] != data[i+1])
{
//mcount = 0;
//min_loc = i;
//return;
}
if(data[i] == data[i+1])
{
all_uniqFlag = false;
rcount++;
}
else if(rcount < mcount)
{
mcount = rcount;
min_loc = i ;//data[i];
}
}
min_cnt = mcount;
}
As mentioned in the comment only Case B works and Case A and C is not working could you help me fix the issue ?

Answer 1

scan through the list
compare each element in the list with the last element in the out array
If the element matches, then increment its count by 1
If the element doesn't match then add the new element into out
array and increment index by 1
Once the scan is done, the out array will have all the distinct elementsout[][0] and their frequencies out[][1]
Scan through the frequency list (out[][1]) to find the lowest frequency
Finally do another scan through the element list out[][0] and print elements whose frequency matches with the lowest frequency
.
#include<stdio.h>
#include<stdlib.h>
#define N 8
int main()
{
//int data[N]={1,2,3,4,5,6,7};
int data[N]={1,1,2,2,3,3,4,4};
//int data[N]={1,1,2,2,2,3,3,3,3,4,5,5,5,5};
int out[N][2];
int i=0,index=0;
for(i=0;i<N;i++)
{
out[i][0]=0;
out[i][1]=0;
}
out[0][0] = data[0];
out[0][1]=1;
for(i=1;i<N;i++)
{
if(data[i] != out[index][0])
{
index++;
out[index][0] = data[i];
out[index][1] = 1;
}
else
{
out[index][1]++;
}
}
int min=65536;
for(i=0;i<N;i++)
{
if(out[i][1] == 0)
{
break;
}
if(out[i][1] < min)
{
min = out[i][1];
}
}
for(i=0;i<N;i++)
{
if(out[i][1] == min)
{
printf("%d\t",out[i][0]);
}
}
printf("\n");
}

Answer 2

You can use a map for this:
#include <string>
#include <map>
#include <iostream>
typedef std::map<int, int> Counter;
void leastCommon(int data[],int n) {
Counter counter;
int min = n;
for (int i = 0; i < n; i++)
counter[data[i]]++;
for (Counter::iterator it = counter.begin(); it != counter.end(); it++) {
if (min > it->second) min = it->second;
}
for (int i = 0; i < n; i++) {
if (counter[data[i]] == min) {
std::cout << data[i] << std::endl;
counter[data[i]]++;
}
}
}
int main() {
int data[] = {1, 1,3,4,4,2,4,3,2};
leastCommon(data, 9);
return 0;
}

Answer 3

Approach is-
select 1st element from the sorted array, and while consecutive elements to it are same, store them in output[] until the loop breaks
store the frequency of element in leastFrequency
select next element, check with its consecutive ones and store them in same output[] until the loop breaks
check frequency of this with the leastFrequency
if same, do nothing (let these be added in the output[])
if less, clear output[] and store the element same no. of times
if more, change the effective output[] length to previous length before iterating for this element
similarly iterate for all distinct elements and finally get the result from output[] from 0 to effective length
void leastCommon(int data[], int len) {
if ( len > 0) {
int output[] = new int[len];
int outlen = 0; // stores the size of useful-output array
int leastFrequency = len; // stores the lowest frequency of elements
int i=0;
int now = data[i];
while (i < len) {
int num = now;
int count = 0;
do {
output[outlen] = now;
outlen++;
count++;
if((++i == len)){
break;
}
now = data[i];
} while (num == now); // while now and next are same it adds them to output[]
if (i - count == 0) { // avoids copy of same values to output[] for 1st iteration
leastFrequency = count;
} else if (count < leastFrequency) { // if count for the element is less than the current minimum then re-creates the output[]
leastFrequency = count;
output = new int[len];
outlen = 0;
for (; outlen < leastFrequency; outlen++) {
output[outlen] = num; // populates the output[] with lower frequent element, to its count
}
} else if (count > leastFrequency) {
outlen -= count; // marks outlen to its same frequent numbers, i.e., discarding higher frequency values from output[]
}
}
//for(int j = 0; j < outlen; j++) {
// print output[] to console
//}
}
}
Plz suggest for improvements.