Problem
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* 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]
Constraints:
1 <= s.length <= 10^5sconsists of only digits and does not contain leading zeros.1 <= k <= 10^9
Solution
class Solution {
public int numberOfArrays(String s, int k) {
int kMod = 1000000007;
int n = s.length();
int[] dp = new int[n];
if (s.charAt(n - 1) != '0') {
dp[n - 1] = 1;
}
for (int i = n - 2; i >= 0; i--) {
if (s.charAt(i) == '0') {
continue;
}
long temp = 0;
int j = i;
while (j < n && temp * 10 + s.charAt(j) - '0' <= k) {
temp = temp * 10 + s.charAt(j) - '0';
if (j == n - 1) {
dp[i] += 1;
} else {
dp[i] += dp[j + 1];
}
dp[i] %= kMod;
j++;
}
}
return dp[0];
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).