README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2500-2599/2522.Partition%20String%20Into%20Substrings%20With%20Values%20at%20Most%20K/README_EN.md	1604	Weekly Contest 326 Q3	Greedy String Dynamic Programming

2522. Partition String Into Substrings With Values at Most K

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Description

You are given a string s consisting of digits from 1 to 9 and an integer k.

A partition of a string s is called good if:

• Each digit of s is part of exactly one substring.
• The value of each substring is less than or equal to k.

Return the minimum number of substrings in a good partition of s. If no good partition of s exists, return -1.

Note that:

• The value of a string is its result when interpreted as an integer. For example, the value of "123" is 123 and the value of "1" is 1.
• A substring is a contiguous sequence of characters within a string.

 

Example 1:

Input: s = "165462", k = 60
Output: 4
Explanation: We can partition the string into substrings "16", "54", "6", and "2". Each substring has a value less than or equal to k = 60.
It can be shown that we cannot partition the string into less than 4 substrings.

Example 2:

Input: s = "238182", k = 5
Output: -1
Explanation: There is no good partition for this string.

 

Constraints:

• 1 <= s.length <= 105
• s[i] is a digit from '1' to '9'.
• 1 <= k <= 109

 

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Solutions

Solution 1: Memoization Search

We design a function $dfs(i)$ to represent the minimum number of partitions starting from index $i$ of string $s$. The answer is $dfs(0)$.

The calculation process of the function $dfs(i)$ is as follows:

• If $i \geq n$, it means that it has reached the end of the string, return $0$.
• Otherwise, we enumerate all substrings starting from $i$. If the value of the substring is less than or equal to $k$, then we can take the substring as a partition. Then we can get $dfs(j + 1)$, where $j$ is the end index of the substring. We take the minimum value among all possible partitions, add $1$, and that is the value of $dfs(i)$.

Finally, if $dfs(0) = \infty$, it means there is no good partition, return $-1$. Otherwise, return $dfs(0)$.

To avoid repeated calculations, we can use memoization search.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string $s$.

Python3

class Solution:
    def minimumPartition(self, s: str, k: int) -> int:
        @cache
        def dfs(i):
            if i >= n:
                return 0
            res, v = inf, 0
            for j in range(i, n):
                v = v * 10 + int(s[j])
                if v > k:
                    break
                res = min(res, dfs(j + 1))
            return res + 1

        n = len(s)
        ans = dfs(0)
        return ans if ans < inf else -1

Java

class Solution {
    private Integer[] f;
    private int n;
    private String s;
    private int k;
    private int inf = 1 << 30;

    public int minimumPartition(String s, int k) {
        n = s.length();
        f = new Integer[n];
        this.s = s;
        this.k = k;
        int ans = dfs(0);
        return ans < inf ? ans : -1;
    }

    private int dfs(int i) {
        if (i >= n) {
            return 0;
        }
        if (f[i] != null) {
            return f[i];
        }
        int res = inf;
        long v = 0;
        for (int j = i; j < n; ++j) {
            v = v * 10 + (s.charAt(j) - '0');
            if (v > k) {
                break;
            }
            res = Math.min(res, dfs(j + 1));
        }
        return f[i] = res + 1;
    }
}

C++

class Solution {
public:
    int minimumPartition(string s, int k) {
        int n = s.size();
        int f[n];
        memset(f, 0, sizeof f);
        const int inf = 1 << 30;
        function<int(int)> dfs = [&](int i) -> int {
            if (i >= n) return 0;
            if (f[i]) return f[i];
            int res = inf;
            long v = 0;
            for (int j = i; j < n; ++j) {
                v = v * 10 + (s[j] - '0');
                if (v > k) break;
                res = min(res, dfs(j + 1));
            }
            return f[i] = res + 1;
        };
        int ans = dfs(0);
        return ans < inf ? ans : -1;
    }
};

Go

func minimumPartition(s string, k int) int {
	n := len(s)
	f := make([]int, n)
	const inf int = 1 << 30
	var dfs func(int) int
	dfs = func(i int) int {
		if i >= n {
			return 0
		}
		if f[i] > 0 {
			return f[i]
		}
		res, v := inf, 0
		for j := i; j < n; j++ {
			v = v*10 + int(s[j]-'0')
			if v > k {
				break
			}
			res = min(res, dfs(j+1))
		}
		f[i] = res + 1
		return f[i]
	}
	ans := dfs(0)
	if ans < inf {
		return ans
	}
	return -1
}