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1849. Splitting a String Into Descending Consecutive Values

Description

You are given a string s that consists of only digits.

Check if we can split s into two or more non-empty substrings such that the numerical values of the substrings are in descending order and the difference between numerical values of every two adjacent substrings is equal to 1.

  • For example, the string s = "0090089" can be split into ["0090", "089"] with numerical values [90,89]. The values are in descending order and adjacent values differ by 1, so this way is valid.
  • Another example, the string s = "001" can be split into ["0", "01"], ["00", "1"], or ["0", "0", "1"]. However all the ways are invalid because they have numerical values [0,1], [0,1], and [0,0,1] respectively, all of which are not in descending order.

Return true if it is possible to split s​​​​​​ as described above, or false otherwise.

A substring is a contiguous sequence of characters in a string.

 

Example 1:

Input: s = "1234"
Output: false
Explanation: There is no valid way to split s.

Example 2:

Input: s = "050043"
Output: true
Explanation: s can be split into ["05", "004", "3"] with numerical values [5,4,3].
The values are in descending order with adjacent values differing by 1.

Example 3:

Input: s = "9080701"
Output: false
Explanation: There is no valid way to split s.

 

Constraints:

  • 1 <= s.length <= 20
  • s only consists of digits.

Solutions

Solution 1: DFS

Thinking

The string must split into at least two parts whose integer values decrease by exactly \(1\). Leading zeros are allowed but do not change the value. The number of cuts is exponential, yet \(n\le 20\) makes search feasible.

Extend the current part from the left, accumulating \(y\). The first part is free; later parts must be exactly one less than the previous value. The first part must not consume the whole string. DFS succeeds if we reach the end.

We can start from the first character of the string and try to split it into one or more substrings, then recursively process the remaining part.

Specifically, we design a function \(\textit{dfs}(i, x)\), where \(i\) represents the current position being processed, and \(x\) represents the last split value. Initially, \(x = -1\), indicating that we have not split out any value yet.

In \(\textit{dfs}(i, x)\), we first calculate the current split value \(y\). If \(x = -1\), or \(x - y = 1\), then we can try to use \(y\) as the next value and continue to recursively process the remaining part. If the result of the recursion is \(\textit{true}\), we have found a valid split method and return \(\textit{true}\).

After traversing all possible split methods, if no valid split method is found, we return \(\textit{false}\).

The time complexity is \(O(n^2)\), and the space complexity is \(O(n)\), where \(n\) is the length of the string.

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class Solution:
    def splitString(self, s: str) -> bool:
        def dfs(i: int, x: int) -> bool:
            if i >= len(s):
                return True
            y = 0
            r = len(s) - 1 if x < 0 else len(s)
            for j in range(i, r):
                y = y * 10 + int(s[j])
                if (x < 0 or x - y == 1) and dfs(j + 1, y):
                    return True
            return False

        return dfs(0, -1)

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class Solution {
    private char[] s;

    public boolean splitString(String s) {
        this.s = s.toCharArray();
        return dfs(0, -1);
    }

    private boolean dfs(int i, long x) {
        if (i >= s.length) {
            return true;
        }
        long y = 0;
        int r = x < 0 ? s.length - 1 : s.length;
        for (int j = i; j < r; ++j) {
            y = y * 10 + s[j] - '0';
            if ((x < 0 || x - y == 1) && dfs(j + 1, y)) {
                return true;
            }
        }
        return false;
    }
}

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class Solution {
public:
    bool splitString(string s) {
        auto dfs = [&](this auto&& dfs, int i, long long x) -> bool {
            if (i >= s.size()) {
                return true;
            }
            long long y = 0;
            int r = x < 0 ? s.size() - 1 : s.size();
            for (int j = i; j < r; ++j) {
                y = y * 10 + s[j] - '0';
                if (y > 1e10) {
                    break;
                }
                if ((x < 0 || x - y == 1) && dfs(j + 1, y)) {
                    return true;
                }
            }
            return false;
        };
        return dfs(0, -1);
    }
};

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func splitString(s string) bool {
    var dfs func(i, x int) bool
    dfs = func(i, x int) bool {
        if i >= len(s) {
            return true
        }
        y := 0
        r := len(s)
        if x < 0 {
            r--
        }
        for j := i; j < r; j++ {
            y = y*10 + int(s[j]-'0')
            if (x < 0 || x-y == 1) && dfs(j+1, y) {
                return true
            }
        }
        return false
    }
    return dfs(0, -1)
}

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function splitString(s: string): boolean {
    const dfs = (i: number, x: number): boolean => {
        if (i >= s.length) {
            return true;
        }
        let y = 0;
        const r = x < 0 ? s.length - 1 : s.length;
        for (let j = i; j < r; ++j) {
            y = y * 10 + +s[j];
            if ((x < 0 || x - y === 1) && dfs(j + 1, y)) {
                return true;
            }
        }
        return false;
    };
    return dfs(0, -1);
}

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