Question

17.06. Number Of 2s In Range

中文文档

Description

Write a method to count the number of 2s that appear in all the numbers between 0 and n (inclusive).

Example:

Input: 25

Output: 9

Explanation: (2, 12, 20, 21, 22, 23, 24, 25)(Note that 22 counts for two 2s.)

Note:

• n <= 10^9

Solutions

Solution 1

class Solution:
  def numberOf2sInRange(self, n: int) -> int:
    @cache
    def dfs(pos, cnt, limit):
      if pos <= 0:
        return cnt
      up = a[pos] if limit else 9
      ans = 0
      for i in range(up + 1):
        ans += dfs(pos - 1, cnt + (i == 2), limit and i == up)
      return ans

    a = [0] * 12
    l = 0
    while n:
      l += 1
      a[l] = n % 10
      n //= 10
    return dfs(l, 0, True)

class Solution {
  private int[] a = new int[12];
  private int[][] dp = new int[12][12];

  public int numberOf2sInRange(int n) {
    int len = 0;
    while (n > 0) {
      a[++len] = n % 10;
      n /= 10;
    }
    for (var e : dp) {
      Arrays.fill(e, -1);
    }
    return dfs(len, 0, true);
  }

  private int dfs(int pos, int cnt, boolean limit) {
    if (pos <= 0) {
      return cnt;
    }
    if (!limit && dp[pos][cnt] != -1) {
      return dp[pos][cnt];
    }
    int up = limit ? a[pos] : 9;
    int ans = 0;
    for (int i = 0; i <= up; ++i) {
      ans += dfs(pos - 1, cnt + (i == 2 ? 1 : 0), limit && i == up);
    }
    if (!limit) {
      dp[pos][cnt] = ans;
    }
    return ans;
  }
}

class Solution {
public:
  int a[12];
  int dp[12][12];

  int numberOf2sInRange(int n) {
    int len = 0;
    while (n) {
      a[++len] = n % 10;
      n /= 10;
    }
    memset(dp, -1, sizeof dp);
    return dfs(len, 0, true);
  }

  int dfs(int pos, int cnt, bool limit) {
    if (pos <= 0) {
      return cnt;
    }
    if (!limit && dp[pos][cnt] != -1) {
      return dp[pos][cnt];
    }
    int ans = 0;
    int up = limit ? a[pos] : 9;
    for (int i = 0; i <= up; ++i) {
      ans += dfs(pos - 1, cnt + (i == 2), limit && i == up);
    }
    if (!limit) {
      dp[pos][cnt] = ans;
    }
    return ans;
  }
};

func numberOf2sInRange(n int) int {
  a := make([]int, 12)
  dp := make([][]int, 12)
  for i := range dp {
    dp[i] = make([]int, 12)
    for j := range dp[i] {
      dp[i][j] = -1
    }
  }
  l := 0
  for n > 0 {
    l++
    a[l] = n % 10
    n /= 10
  }
  var dfs func(int, int, bool) int
  dfs = func(pos, cnt int, limit bool) int {
    if pos <= 0 {
      return cnt
    }
    if !limit && dp[pos][cnt] != -1 {
      return dp[pos][cnt]
    }
    up := 9
    if limit {
      up = a[pos]
    }
    ans := 0
    for i := 0; i <= up; i++ {
      t := cnt
      if i == 2 {
        t++
      }
      ans += dfs(pos-1, t, limit && i == up)
    }
    if !limit {
      dp[pos][cnt] = ans
    }
    return ans
  }
  return dfs(l, 0, true)
}