Given a range of numbers [l, r] and an integer q. The task is to count all such number in the given range such that any digit of the number does not match with any digit in its product with the given number q.
Example 1:
Input: l = 10, r = 12, q = 2
Output: 1
Explanation:
10*2 = 20 which has 0 as same digit
12*2 = 24 which as 2 as same digit
11*2 = 22 no same digitExample 2:
Input: l = 5, r = 15, q = 2
Output: 9Solution
static boolean areInDigitInX(int n, int from) {
int product = n * from;
Set<Integer> numberDigits = digits(n);
Set<Integer> productDigits = digits(product);
for (Integer i : productDigits) {
if (numberDigits.contains(i))
return false;
}
return true;
}
static Set<Integer> digits(int item) {
Set<Integer> set = new HashSet<>();
int n = item;
if (n == 0) {
set.add(0);
return set;
}
while (n > 0) {
int digit = n % 10;
set.add(digit);
n = n / 10;
}
return set;
}
// Complete the nonRepeatingDigitProductCount function below.
static int nonRepeatingDigitProductCount(int x, int y, int z) {
int count = 0;
for (int i = y; i <= z; i++) {
if (areInDigitInX(i, x))
count++;
}
return count;
}Comments (16)