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Formatted question description: https://leetcode.ca/all/1067.html

1067. Digit Count in Range

Level

Hard

Description

Given an integer d between 0 and 9, and two positive integers low and high as lower and upper bounds, respectively. Return the number of times that d occurs as a digit in all integers between low and high, including the bounds low and 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.

Note:

  1. 0 <= d <= 9
  2. 1 <= low <= high <= 2×10^8

Solution

Calculate the number of times that d occurs as a digit in all integers between 1 and high, and calculate the number of times that d occurs as a digit in all integers between 1 and low - 1. Then calculate the difference between the two numbers of times to obtain the result.

If d is not 0, the number of times is calculated as follows.

  1. Loop over i from 1 to n. Each time set i *= 10.
  2. The value n / (i * 10) * i represents the number of digit d at digit i * 10.
  3. The value Math.min(Math.max(n % (i * 10) - d * i + 1, 0), i) represents the extra number of digit d at digit i * 10.
  4. Calculate the number of times of d by calculating the sum of the values in the previous two steps.

If d is 0, the number of times is calculated as follows.

  1. Loop over i from 1 to n / 10. Each time set i *= 10.
  2. The value (n / (i * 10) - 1) * i represents the number of digit d at digit i * 10.
  3. The value Math.min(Math.max(n % (i * 10) - d * i + 1, 0), i) represents the extra number of digit d at digit i * 10.
  4. Calculate the number of times of d by calculating the sum of the values in the previous two steps.

The difference exists because digit 0 cannot appear at the highest digit, except the number 0. However, in this problem, both low and high are positive integers, so 0 is never in the range.

  • class Solution {
    public int digitsCount(int d, int low, int high) {
        long highCount = countDigits(high, d);
        long lowCount = countDigits(low - 1, d);
        int count = (int) (highCount - lowCount);
        return count;
    }

    public long countDigits(int n, int digit) {
        if (n <= 0)
            return 0;
        long count = 0;
        if (digit == 0) {
            for (long i = 1; i <= n / 10; i *= 10) {
                long unit = i * 10;
                count += (n / unit - 1) * i + Math.min(i, Math.max(0, n % unit - digit * i + 1));
            }
        } else {
            for (long i = 1; i <= n; i *= 10) {
                long unit = i * 10;
                count += n / unit * i + Math.min(i, Math.max(0, n % unit - digit * i + 1));
            }
        }
        return count;
    }
}

  • 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 {
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)
}

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