LeetCode 1977: Number of Ways to Separate Numbers Solution

Master LeetCode problem 1977 (Number of Ways to Separate Numbers), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1977. Number of Ways to Separate Numbers

Hard

String Dynamic Programming Suffix Array

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We will iterate through the input string and keep track of the number of ways to separate numbers up to that point. We will use a dynamic programming array to store this information. At each position, we will consider two cases: either we can append the current digit to the previous number, or we can start a new number with the current digit. We will update the dynamic programming array accordingly.

Initialize a dynamic programming array dp of size n+1 where n is the length of the input string.
Set dp[0] = 1 since there is only one way to separate an empty string.
Iterate through the input string from the first character to the last character.
For each character, consider two cases:
1. If the current character is '0', we cannot form a new number with '0' as a leading digit. So, if the previous character is not '1' or '2', set dp[i+1] = 0.
2. If the current character is not '0', we can either append it to the previous number or start a new number. So, update dp[i+1] accordingly.
Return dp[n] mod 1e9 + 7 as the final answer.

Time Complexity: O(n) Space Complexity: O(n)

Solution Code

class Solution {
    public int numberOfCombinations(String num) {
        int n = num.length();
        int mod = 1000000007;
        long[] dp = new long[n + 1];
        dp[0] = 1;
        
        for (int i = 0; i < n; i++) {
            if (num.charAt(i) == '0') {
                if (i == 0 || num.charAt(i - 1) > '2') {
                    dp[i + 1] = 0;
                } else {
                    dp[i + 1] = dp[i];
                }
            } else {
                dp[i + 1] = dp[i];
                for (int j = i - 1; j >= 0; j--) {
                    if (num.charAt(j) != '0' && Integer.parseInt(num.substring(j, i + 1)) <= i - j + 1) {
                        dp[i + 1] = (dp[i + 1] + dp[j]) % mod;
                    }
                }
            }
        }
        
        return (int) dp[n];
    }
}

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