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1067. Digit Count in Range

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

• Java
• C++
• Python
• Go

• 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;
      }
  };
  

• 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)
  
  

• 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)
  }
  

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