CareerCup

/* This solution is definitely not optimal. */
static void printNumbersWithoutRepeatingDigits(int start, int end){
		
	String strInt;
	List<String> arrlist = new ArrayList<String>();
		
	while(start <= end){
		
		strInt = String.valueOf(start);
			
		if(doesNotContainRepeatingDigits(strInt))
			arrlist.add(strInt);
			
		start++;
	}
		
	for(int i = 0; i < arrlist.size(); i++){

		System.out.print(arrlist.get(i));
	
		if((i+1) < arrlist.size())
			System.out.print(", ");
	}
}

static boolean doesNotContainRepeatingDigits(String s){
	int firstOccurrence, secondOccurrence;
	
	for(int i = 0; i < 10; i++){
			
		firstOccurrence = s.indexOf((char)(i+48));
		secondOccurrence = s.indexOf((char)(i+48), firstOccurrence+1);
			
		if(secondOccurrence != -1)
			return false;
	}
	
	return true;
}

duplicate

question?id=5982716570894336

What if possible range is from 0 to 987654321 it makes this problem more complex

void rem_dup(int min, int max)
{
int i,val,dig;
int Isloop ,num[10] = {0};
for(i = min; i <= max; i++)
{
val = i;
Isloop = 0;
memset(num, 0, sizeof(num));
while(val != 0)
{
dig = val % 10;
val = val / 10;
num[dig]++;
if(num[dig] > 1)
{
Isloop = 1;
break;
}
}
if(!Isloop)
printf("%d\t", i);
}

}

start = 98
end = 103
results = []

for n in range(start + 1, end):
    t = list(str(n))

    if len(t) == len(set(t)):
        results.append(n)

print 'Results:', results

Results: [102]

for(int num=98;num<=104;num++)

{
int n = num;
List<Integer>list=new ArrayList<Integer>();
int rev=0,rmd;
while(n > 0)
{
rmd = n % 10;
list.add(rmd);
rev = rev * 10 + rmd;
n = n / 10;
}
int count=0;
for(int j=0;j<list.size();j++)
{

for(int k=j+1;k<list.size();k++)
{
if(list.get(j)==list.get(k))
{
count++;
break;
}

}

}

if(count==0)
System.out.println(num);
}
}