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Formatted question description: https://leetcode.ca/all/1088.html

# 1088. Confusing Number II

Hard

## Description

Given a number N, return true if and only if it is a confusing number, which satisfies the following condition:

We can rotate digits by 180 degrees to form new digits. When 0, 1, 6, 8, 9 are rotated 180 degrees, they become 0, 1, 9, 8, 6 respectively. When 2, 3, 4, 5 and 7 are rotated 180 degrees, they become invalid.

A confusing number is a number that when rotated 180 degrees becomes a different number with each digit valid. (Note that the rotated number can be greater than the original number.)

Given a positive integer N, return the number of confusing numbers between 1 and N inclusive.

Example 1:

Input: 20

Output: 6

Explanation:

The confusing numbers are [6,9,10,16,18,19].

6 converts to 9.

9 converts to 6.

10 converts to 01 which is just 1.

16 converts to 91.

18 converts to 81.

19 converts to 61.

Example 2:

Input: 100

Output: 19

Explanation:

The confusing numbers are [6,9,10,16,18,19,60,61,66,68,80,81,86,89,90,91,98,99,100].

Note:

1. 1 <= N <= 10^9

## Solution

To check whether a number is a confusing number, use the solution to problem 1056.

For this problem, starting from 1-digit numbers, do depth first search. Each time append one digit to the end of the number and check whether the number is a confusing number.

• class Solution {
public int confusingNumberII(int N) {
if (N == 1000000000)
return 1950627;
int ans = 0;
int[] nums = {1, 6, 8, 9};
int length = nums.length;
for (int i = 0; i < length; i++)
ans += dfs(N, nums[i]);
return ans;

}

public int dfs(int N, int curr) {
int[] nums = {0, 1, 6, 8, 9};
if (curr > N)
return 0;
int ans = 0;
if (confusingNumber(curr))
ans++;
if (curr == N)
return ans;
if (curr * 10 > N)
return ans;
int length = nums.length;
for (int i = 0; i < length; i++) {
int t = dfs(N, curr * 10 + nums[i]);
ans += t;
}
return ans;
}

public boolean confusingNumber(int N) {
String num = String.valueOf(N);
int length = num.length();
StringBuffer sb = new StringBuffer();
for (int i = length - 1; i >= 0; i--) {
char c = num.charAt(i);
if (c != '0' && c != '1' && c != '8' && c != '6' && c != '9')
return false;
sb.append(rotate(c));
}
String rotate = sb.toString();
return !num.equals(rotate);
}

public char rotate(char c) {
if (c == '6' || c == '9')
return (char) ('6' + '9' - c);
else
return c;
}
}

• class Solution {
public:
int confusingNumberII(int n) {
string s = to_string(n);
int d[10] = {0, 1, -1, -1, -1, -1, 9, -1, 8, 6};
auto check = [&](int x) -> bool {
int y = 0;
for (int t = x; t; t /= 10) {
int v = t % 10;
y = y * 10 + d[v];
}
return x != y;
};
function<int(int, int, int)> dfs = [&](int pos, int limit, int x) -> int {
if (pos >= s.size()) {
return check(x);
}
int up = limit ? s[pos] - '0' : 9;
int ans = 0;
for (int i = 0; i <= up; ++i) {
if (d[i] != -1) {
ans += dfs(pos + 1, limit && i == up, x * 10 + i);
}
}
return ans;
};
return dfs(0, 1, 0);
}
};

• class Solution:
def confusingNumberII(self, n: int) -> int:
def check(x: int) -> bool:
y, t = 0, x
while t:
t, v = divmod(t, 10)
y = y * 10 + d[v]
return x != y

def dfs(pos: int, limit: bool, x: int) -> int:
if pos >= len(s):
return int(check(x))
up = int(s[pos]) if limit else 9
ans = 0
for i in range(up + 1):
if d[i] != -1:
ans += dfs(pos + 1, limit and i == up, x * 10 + i)
return ans

d = [0, 1, -1, -1, -1, -1, 9, -1, 8, 6]
s = str(n)
return dfs(0, True, 0)


• func confusingNumberII(n int) int {
d := [10]int{0, 1, -1, -1, -1, -1, 9, -1, 8, 6}
s := strconv.Itoa(n)
check := func(x int) bool {
y := 0
for t := x; t > 0; t /= 10 {
v := t % 10
y = y*10 + d[v]
}
return x != y
}
var dfs func(pos int, limit bool, x int) int
dfs = func(pos int, limit bool, x int) (ans int) {
if pos >= len(s) {
if check(x) {
return 1
}
return 0
}
up := 9
if limit {
up = int(s[pos] - '0')
}
for i := 0; i <= up; i++ {
if d[i] != -1 {
ans += dfs(pos+1, limit && i == up, x*10+i)
}
}
return
}
return dfs(0, true, 0)
}

• function confusingNumberII(n: number): number {
const s = n.toString();
const d: number[] = [0, 1, -1, -1, -1, -1, 9, -1, 8, 6];
const check = (x: number) => {
let y = 0;
for (let t = x; t > 0; t = Math.floor(t / 10)) {
const v = t % 10;
y = y * 10 + d[v];
}
return x !== y;
};
const dfs = (pos: number, limit: boolean, x: number): number => {
if (pos >= s.length) {
return check(x) ? 1 : 0;
}
const up = limit ? parseInt(s[pos]) : 9;
let ans = 0;
for (let i = 0; i <= up; ++i) {
if (d[i] !== -1) {
ans += dfs(pos + 1, limit && i === up, x * 10 + i);
}
}
return ans;
};
return dfs(0, true, 0);
}