L788.java

import java.util.HashMap;
import java.util.Map;
import java.util.function.IntPredicate;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

class Solution788 {
    class Solution {
        private final Map<Character, Character> ROTATION_MAP = new HashMap<Character, Character>() {{
            put('0', '0');
            put('1', '1');
            put('2', '5');
            put('5', '2');
            put('6', '9');
            put('8', '8');
            put('9', '6');
        }};

        private final IntPredicate IS_GOOD_NUMBER = (N) -> {
            String inputChars = String.valueOf(N);
            String mappedChars = inputChars.chars().
                    filter(c -> ROTATION_MAP.containsKey((char) c)).
                    mapToObj(c -> String.valueOf(ROTATION_MAP.get((char) c))).
                    collect(Collectors.joining());
            if (mappedChars.length() != inputChars.length()) {
                return false;
            } else if (mappedChars.equals(inputChars)) {
                return false;
            }
            return true;
        };

        /**
         * 788. Rotated Digits https://leetcode.com/problems/rotated-digits/description/
         *
         * @param N The number N
         * @return Number of good numbers in the range [1, N]
         * @timeComplexity O(N * log N) in terms of N. O(2^N * N) in terms of input size.
         * For each number upto N, we perform a linear time operation on each digit to check if it is good number
         * @spaceComplexity O(1) - The hashmap has constant number of records.
         */
        public int rotatedDigits(int N) {
            return (int) IntStream.range(1, N + 1).filter(IS_GOOD_NUMBER).count();
        }
    }
}