Question

600. Non-negative Integers without Consecutive Ones

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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

Solution 1

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)

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;
  }
};

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)
}