g0701_0800.s0788_rotated_digits.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0701_0800.s0788_rotated_digits;

// #Medium #Dynamic_Programming #Math #2022_03_26_Time_2_ms_(98.95%)_Space_39.6_MB_(87.35%)

/**
 * 788 - Rotated Digits\.
 *
 * Medium
 *
 * An integer `x` is a **good** if after rotating each digit individually by 180 degrees, we get a valid number that is different from `x`. Each digit must be rotated - we cannot choose to leave it alone.
 *
 * A number is valid if each digit remains a digit after rotation. For example:
 *
 * *   `0`, `1`, and `8` rotate to themselves,
 * *   `2` and `5` rotate to each other (in this case they are rotated in a different direction, in other words, `2` or `5` gets mirrored),
 * *   `6` and `9` rotate to each other, and
 * *   the rest of the numbers do not rotate to any other number and become invalid.
 *
 * Given an integer `n`, return _the number of **good** integers in the range_ `[1, n]`.
 *
 * **Example 1:**
 *
 * **Input:** n = 10
 *
 * **Output:** 4
 *
 * **Explanation:**
 *
 *     There are four good numbers in the range [1, 10] : 2, 5, 6, 9.
 *     Note that 1 and 10 are not good numbers, since they remain unchanged after rotating. 
 *
 * **Example 2:**
 *
 * **Input:** n = 1
 *
 * **Output:** 0 
 *
 * **Example 3:**
 *
 * **Input:** n = 2
 *
 * **Output:** 1 
 *
 * **Constraints:**
 *
 * *   1 <= n <= 10⁴
**/
public class Solution {
    public int rotatedDigits(int n) {
        int[] flag = new int[n + 1];
        flag[0] = 2;
        int[] indexesValueTwo = {1, 8};
        for (int value : indexesValueTwo) {
            if (n >= value) {
                flag[value] = 2;
            }
        }
        int[] indexesValueOne = {2, 5, 6, 9};
        for (int value : indexesValueOne) {
            if (n >= value) {
                flag[value] = 1;
            }
        }
        int rs = 0;
        for (int i = 1; i <= n; i++) {
            int residual = i % 10;
            if (flag[residual] != 0) {
                if ((residual == 1 || residual == 0 || residual == 8) && (flag[i / 10] == 2)) {
                    flag[i] = 2;
                    continue;
                }
                if (flag[i / 10] != 0) {
                    flag[i] = 1;
                    rs++;
                }
            }
        }
        return rs;
    }
}