Home Coding Leetcode Hard

⭐️⭐️⭐️

# 題目敘述

A program was supposed to print an array of integers. The program forgot to print whitespaces and the array is printed as a string of digits s and all we know is that all integers in the array were in the range [1, k] and there are no leading zeros in the array.

Given the string s and the integer k , return the number of the possible arrays that can be printed as s using the mentioned program. Since the answer may be very large, return it modulo 1e9 + 7 .

# Example 1

Input: s = “1000”, k = 10000
Output: 1
Explanation: The only possible array is [1000]

# Example 2

Input: s = “1000”, k = 10
Output: 0
Explanation: There cannot be an array that was printed this way and has all integer >= 1 and <= 10.

# Example 3

Input: s = “1317”, k = 2000
Output: 8
Explanation: Possible arrays are [1317],[131,7],[13,17],[1,317],[13,1,7],[1,31,7],[1,3,17],[1,3,1,7]

# 解題思路

Let dp[i] denote the number of ways to partition the substring s[i:n] into valid numbers. The base case is dp[n] = 1 , since there is only one way to partition an empty string.

To compute dp[i] for a given index i , we consider all possible substrings that can be formed starting from i , such that the length of the substring is less than or equal to the number of digits in k . We convert each substring to an integer and check if it is less than or equal to k . If it is, we add the number of ways to partition the remaining string s[j+1:n] to dp[i] , where j is the index of the last character in the substring. We sum up the values of dp[i] for all valid substrings to obtain the final value of dp[i] .

# Solution

class Solution {
public:
    int numberOfArrays(string s, int k) {
        int n = s.length();
        vector<long long> dp(n + 1, 0);
        dp[n] = 1;
        
        for (int i = n - 1; i >= 0; i--) {
            if (s[i] == '0') {
                continue;
            }
            long long num = 0;
            int j = i;
            while (j < n) {
                try {
                    int x = stoi(s.substr(i, j-i+1));
                    if (x > k) {
                        break;
                    }
                    num += dp[j+1];
                }
                catch (std::out_of_range& e) {
                    break;
                }
                j++;
            }
            dp[i] = num % 1000000007;
        }
        
        return dp[0];
    }
};


class Solution:
    def numberOfArrays(self, s: str, k: int) -> int:
        n = len(s)
        dp = [0] * (n + 1)
        dp[-1] = 1
        
        for i in range(n - 1, -1, -1):
            if s[i] == '0':
                continue
            num = 0
            j = i
            while j < n and int(s[i:j+1]) <= k:
                num += dp[j+1]
                j += 1
            dp[i] = num % (10 ** 9 + 7)
        
        return dp[0]