CareerCup

int repeatedDigitCount(int num,int k){

int i,count = 0;
for(i=0;i<=num;i++){

if(i==0 && k==0)
{
count++;
continue;
}

if(i==1 && k==1)
{
count++;
continue;
}

int val = i;
while(val>1){

if(val%10 == k){
count++;
}
val = val/10;
if(val==k && val!=0){
count++;
}
}


}
return count;
}

public int getTotalOccurrences(int[] list, int k) {
		int occurrences = 0;
		for(int i : list) {
			if(i == k) {
				occurrences++;
			}
			if(i > 9) {
				while(i > 9) {
					int reminder = i % 10;
					if(reminder == k) {
						occurrences++;
					}
					i = i / 10;
					if(i == k) {
						occurrences++;
					}
				}
			}
		}
		return occurrences;
	}

private static int countRep(int[] numbers, int toCheck) {
		int count = 0;
		for (int element : numbers) {
			if (element < 9) {
				if (element == toCheck) {
					count++;
				}
			} else {
				int rem = 0;
				int org = element;

				while (org > 0) {
					rem = org % 10;
					if (rem == toCheck) {
						count++;
					}

					org /= 10;
				}
			}
		}

		return count;
	}

N = 50
alist = [x for x in range(N)]
print(alist)
#print(alist)
#print('HI')
k = 3
count=0
for i in range(len(alist)):
    if alist[i]==k: count+=1
    elif alist[i] > 9:
        temp = alist[i]
        while (temp>9):
            if temp%10==k: count+=1
            temp = temp//10
            if temp==k: count+=1
print(count)

ZoomBA. Spec is not very clear - but this is what is approximating the problem :

def find_occurences( numbers , k ){
  sk = str(k)
  sum ( numbers ) -> { sk @ str($.o) ? 1 : 0 }
}
println( find_occurences ( [0,10] , 0 ) )