Question

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

 

Solutions

Solution 1

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

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;
  }
}

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;
  }
};

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
}