600. Non-negative Integers without Consecutive Ones

Description

Given a positive integer n, return the number of the integers in the range [0, n] whose binary representations do not contain consecutive ones.

Example 1:

Input: n = 5
Output: 5
Explanation:
Here are the non-negative integers <= 5 with their corresponding binary representations:
0 : 0
1 : 1
2 : 10
3 : 11
4 : 100
5 : 101
Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule.


Example 2:

Input: n = 1
Output: 2


Example 3:

Input: n = 2
Output: 3


Constraints:

• 1 <= n <= 109

Solutions

• class Solution {
private int[] a = new int[33];
private int[][] dp = new int[33][2];

public int findIntegers(int n) {
int len = 0;
while (n > 0) {
a[++len] = n & 1;
n >>= 1;
}
for (var e : dp) {
Arrays.fill(e, -1);
}
return dfs(len, 0, true);
}

private int dfs(int pos, int pre, boolean limit) {
if (pos <= 0) {
return 1;
}
if (!limit && dp[pos][pre] != -1) {
return dp[pos][pre];
}
int up = limit ? a[pos] : 1;
int ans = 0;
for (int i = 0; i <= up; ++i) {
if (!(pre == 1 && i == 1)) {
ans += dfs(pos - 1, i, limit && i == up);
}
}
if (!limit) {
dp[pos][pre] = ans;
}
return ans;
}
}

• class Solution {
public:
int a[33];
int dp[33][2];

int findIntegers(int n) {
int len = 0;
while (n) {
a[++len] = n & 1;
n >>= 1;
}
memset(dp, -1, sizeof dp);
return dfs(len, 0, true);
}

int dfs(int pos, int pre, bool limit) {
if (pos <= 0) {
return 1;
}
if (!limit && dp[pos][pre] != -1) {
return dp[pos][pre];
}
int ans = 0;
int up = limit ? a[pos] : 1;
for (int i = 0; i <= up; ++i) {
if (!(pre == 1 && i == 1)) {
ans += dfs(pos - 1, i, limit && i == up);
}
}
if (!limit) {
dp[pos][pre] = ans;
}
return ans;
}
};

• class Solution:
def findIntegers(self, n: int) -> int:
@cache
def dfs(pos, pre, limit):
if pos <= 0:
return 1
up = a[pos] if limit else 1
ans = 0
for i in range(up + 1):
if pre == 1 and i == 1:
continue
ans += dfs(pos - 1, i, limit and i == up)
return ans

a = [0] * 33
l = 0
while n:
l += 1
a[l] = n & 1
n >>= 1
return dfs(l, 0, True)


• func findIntegers(n int) int {
a := make([]int, 33)
dp := make([][2]int, 33)
for i := range dp {
dp[i] = [2]int{-1, -1}
}
l := 0
for n > 0 {
l++
a[l] = n & 1
n >>= 1
}
var dfs func(int, int, bool) int
dfs = func(pos, pre int, limit bool) int {
if pos <= 0 {
return 1
}
if !limit && dp[pos][pre] != -1 {
return dp[pos][pre]
}
up := 1
if limit {
up = a[pos]
}
ans := 0
for i := 0; i <= up; i++ {
if !(pre == 1 && i == 1) {
ans += dfs(pos-1, i, limit && i == up)
}
}
if !limit {
dp[pos][pre] = ans
}
return ans
}
return dfs(l, 0, true)
}