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 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. There can be multiple ways to restore the array.
Return the number of possible array that can be printed as a string s using the mentioned program.
The number of ways could be very large so return it modulo 10^9 + 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]
Example 4:
Input: s = "2020", k = 30 Output: 1 Explanation: The only possible array is [20,20]. [2020] is invalid because 2020 > 30. [2,020] is ivalid because 020 contains leading zeros.
Example 5:
Input: s = "1234567890", k = 90 Output: 34
Constraints:
1 <= s.length <= 10^5.sconsists of only digits and doesn't contain leading zeros.1 <= k <= 10^9.
Related Topics:
Dynamic Programming
// OJ: https://leetcode.com/problems/restore-the-array/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
class Solution {
typedef long long LL;
public:
int numberOfArrays(string s, int k) {
if (s[0] - '0' > k) return 0;
int cnt = 0, tmp = k;
while (tmp) {
tmp /= 10;
++cnt;
}
int N = s.size(), mod = 1e9+7;
vector<int> dp(N + 1);
dp[0] = dp[1] = 1;
for (int i = 2; i <= N; ++i) {
LL p = 1, n = 0;
for (int j = i - 1; j >= 0; --j) {
n += (s[j] - '0') * p;
p *= 10;
if (n > k || i - j > cnt) break;
if (n == 0 || s[j] == '0') continue;
dp[i] = (dp[i] + dp[j]) % mod;
}
}
return dp[N];
}
};