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1977. Number of Ways to Separate Numbers
Description
You wrote down many positive integers in a string called num
. However, you realized that you forgot to add commas to seperate the different numbers. You remember that the list of integers was non-decreasing and that no integer had leading zeros.
Return the number of possible lists of integers that you could have written down to get the string num
. Since the answer may be large, return it modulo 109 + 7
.
Example 1:
Input: num = "327" Output: 2 Explanation: You could have written down the numbers: 3, 27 327
Example 2:
Input: num = "094" Output: 0 Explanation: No numbers can have leading zeros and all numbers must be positive.
Example 3:
Input: num = "0" Output: 0 Explanation: No numbers can have leading zeros and all numbers must be positive.
Constraints:
1 <= num.length <= 3500
num
consists of digits'0'
through'9'
.
Solutions
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class Solution { private static final int MOD = (int) 1e9 + 7; public int numberOfCombinations(String num) { int n = num.length(); int[][] lcp = new int[n + 1][n + 1]; for (int i = n - 1; i >= 0; --i) { for (int j = n - 1; j >= 0; --j) { if (num.charAt(i) == num.charAt(j)) { lcp[i][j] = 1 + lcp[i + 1][j + 1]; } } } int[][] dp = new int[n + 1][n + 1]; dp[0][0] = 1; for (int i = 1; i <= n; ++i) { for (int j = 1; j <= i; ++j) { int v = 0; if (num.charAt(i - j) != '0') { if (i - j - j >= 0) { int x = lcp[i - j][i - j - j]; if (x >= j || num.charAt(i - j + x) >= num.charAt(i - j - j + x)) { v = dp[i - j][j]; } } if (v == 0) { v = dp[i - j][Math.min(j - 1, i - j)]; } } dp[i][j] = (dp[i][j - 1] + v) % MOD; } } return dp[n][n]; } }
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class Solution { public: const int mod = 1e9 + 7; int numberOfCombinations(string num) { int n = num.size(); vector<vector<int>> lcp(n + 1, vector<int>(n + 1)); for (int i = n - 1; i >= 0; --i) { for (int j = n - 1; j >= 0; --j) { if (num[i] == num[j]) { lcp[i][j] = 1 + lcp[i + 1][j + 1]; } } } auto cmp = [&](int i, int j, int k) { int x = lcp[i][j]; return x >= k || num[i + x] >= num[j + x]; }; vector<vector<int>> dp(n + 1, vector<int>(n + 1)); dp[0][0] = 1; for (int i = 1; i <= n; ++i) { for (int j = 1; j <= i; ++j) { int v = 0; if (num[i - j] != '0') { if (i - j - j >= 0 && cmp(i - j, i - j - j, j)) { v = dp[i - j][j]; } else { v = dp[i - j][min(j - 1, i - j)]; } } dp[i][j] = (dp[i][j - 1] + v) % mod; } } return dp[n][n]; } };
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class Solution: def numberOfCombinations(self, num: str) -> int: def cmp(i, j, k): x = lcp[i][j] return x >= k or num[i + x] >= num[j + x] mod = 10**9 + 7 n = len(num) lcp = [[0] * (n + 1) for _ in range(n + 1)] for i in range(n - 1, -1, -1): for j in range(n - 1, -1, -1): if num[i] == num[j]: lcp[i][j] = 1 + lcp[i + 1][j + 1] dp = [[0] * (n + 1) for _ in range(n + 1)] dp[0][0] = 1 for i in range(1, n + 1): for j in range(1, i + 1): v = 0 if num[i - j] != '0': if i - j - j >= 0 and cmp(i - j, i - j - j, j): v = dp[i - j][j] else: v = dp[i - j][min(j - 1, i - j)] dp[i][j] = (dp[i][j - 1] + v) % mod return dp[n][n]
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func numberOfCombinations(num string) int { n := len(num) lcp := make([][]int, n+1) dp := make([][]int, n+1) for i := range lcp { lcp[i] = make([]int, n+1) dp[i] = make([]int, n+1) } for i := n - 1; i >= 0; i-- { for j := n - 1; j >= 0; j-- { if num[i] == num[j] { lcp[i][j] = 1 + lcp[i+1][j+1] } } } cmp := func(i, j, k int) bool { x := lcp[i][j] return x >= k || num[i+x] >= num[j+x] } dp[0][0] = 1 var mod int = 1e9 + 7 for i := 1; i <= n; i++ { for j := 1; j <= i; j++ { v := 0 if num[i-j] != '0' { if i-j-j >= 0 && cmp(i-j, i-j-j, j) { v = dp[i-j][j] } else { v = dp[i-j][min(j-1, i-j)] } } dp[i][j] = (dp[i][j-1] + v) % mod } } return dp[n][n] }