Question

1067. Digit Count in Range

中文文档

Description

Given a single-digit integer d and two integers low and high, return _the number of times that _d_ occurs as a digit in all integers in the inclusive range _[low, high].

 

Example 1:

Input: d = 1, low = 1, high = 13
Output: 6
Explanation: The digit d = 1 occurs 6 times in 1, 10, 11, 12, 13.
Note that the digit d = 1 occurs twice in the number 11.

Example 2:

Input: d = 3, low = 100, high = 250
Output: 35
Explanation: The digit d = 3 occurs 35 times in 103,113,123,130,131,...,238,239,243.

 

Constraints:

• 0 <= d <= 9
• 1 <= low <= high <= 2 * 108

Solutions

Solution 1

class Solution:
  def digitsCount(self, d: int, low: int, high: int) -> int:
    return self.f(high, d) - self.f(low - 1, d)

  def f(self, n, d):
    @cache
    def dfs(pos, cnt, lead, limit):
      if pos <= 0:
        return cnt
      up = a[pos] if limit else 9
      ans = 0
      for i in range(up + 1):
        if i == 0 and lead:
          ans += dfs(pos - 1, cnt, lead, limit and i == up)
        else:
          ans += dfs(pos - 1, cnt + (i == d), False, limit and i == up)
      return ans

    a = [0] * 11
    l = 0
    while n:
      l += 1
      a[l] = n % 10
      n //= 10
    return dfs(l, 0, True, True)

class Solution {
  private int d;
  private int[] a = new int[11];
  private int[][] dp = new int[11][11];

  public int digitsCount(int d, int low, int high) {
    this.d = d;
    return f(high) - f(low - 1);
  }

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

  private int dfs(int pos, int cnt, boolean lead, boolean limit) {
    if (pos <= 0) {
      return cnt;
    }
    if (!lead && !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) {
      if (i == 0 && lead) {
        ans += dfs(pos - 1, cnt, lead, limit && i == up);
      } else {
        ans += dfs(pos - 1, cnt + (i == d ? 1 : 0), false, limit && i == up);
      }
    }
    if (!lead && !limit) {
      dp[pos][cnt] = ans;
    }
    return ans;
  }
}

class Solution {
public:
  int d;
  int a[11];
  int dp[11][11];

  int digitsCount(int d, int low, int high) {
    this->d = d;
    return f(high) - f(low - 1);
  }

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

  int dfs(int pos, int cnt, bool lead, bool limit) {
    if (pos <= 0) {
      return cnt;
    }
    if (!lead && !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) {
      if (i == 0 && lead) {
        ans += dfs(pos - 1, cnt, lead, limit && i == up);
      } else {
        ans += dfs(pos - 1, cnt + (i == d), false, limit && i == up);
      }
    }
    if (!lead && !limit) {
      dp[pos][cnt] = ans;
    }
    return ans;
  }
};

func digitsCount(d int, low int, high int) int {
  f := func(n int) int {
    a := make([]int, 11)
    dp := make([][]int, 11)
    for i := range dp {
      dp[i] = make([]int, 11)
      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, bool) int
    dfs = func(pos, cnt int, lead, limit bool) int {
      if pos <= 0 {
        return cnt
      }
      if !lead && !limit && dp[pos][cnt] != -1 {
        return dp[pos][cnt]
      }
      up := 9
      if limit {
        up = a[pos]
      }
      ans := 0
      for i := 0; i <= up; i++ {
        if i == 0 && lead {
          ans += dfs(pos-1, cnt, lead, limit && i == up)
        } else {
          t := cnt
          if d == i {
            t++
          }
          ans += dfs(pos-1, t, false, limit && i == up)
        }
      }
      if !lead && !limit {
        dp[pos][cnt] = ans
      }
      return ans
    }

    return dfs(l, 0, true, true)
  }
  return f(high) - f(low-1)
}