Skip to content

3144. Minimum Substring Partition of Equal Character Frequency 👍

  • Time: $O(n^2)$
  • Space: $O(n)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
class Solution {
 public:
  int minimumSubstringsInPartition(string s) {
    const int n = s.length();
    // dp[i] := the minimum number of substrings in s[0..i]
    vector<int> dp(n, n);

    for (int i = 0; i < n; ++i) {
      vector<int> count(26);
      for (int j = i; j >= 0; --j) {
        ++count[s[j] - 'a'];
        if (isBalanced(count))  // word[j..i] is balanced.
          dp[i] = j > 0 ? min(dp[i], 1 + dp[j - 1]) : 1;
      }
    }

    return dp.back();
  }

 private:
  static constexpr int kMax = 1001;

  // Returns true if all non-zero frequencies are the same.
  bool isBalanced(const vector<int>& count) {
    int minfreq = kMax;
    int maxfreq = 0;
    for (const int freq : count)
      if (freq > 0) {
        minfreq = min(minfreq, freq);
        maxfreq = max(maxfreq, freq);
      }
    return minfreq == maxfreq;
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
class Solution {
  public int minimumSubstringsInPartition(String s) {
    final int n = s.length();
    // dp[i] := the minimum number of substrings in s[0..i]
    int[] dp = new int[n];
    Arrays.fill(dp, n);

    for (int i = 0; i < n; ++i) {
      int[] count = new int[26];
      for (int j = i; j >= 0; --j) {
        ++count[s.charAt(j) - 'a'];
        if (isBalanced(count)) // word[j..i] is balanced.
          dp[i] = j > 0 ? Math.min(dp[i], 1 + dp[j - 1]) : 1;
      }
    }

    return dp[n - 1];
  }

  private static final int kMax = 1001;

  // Returns true if all non-zero frequencies are the same.
  private boolean isBalanced(int[] count) {
    int minfreq = kMax;
    int maxfreq = 0;
    for (final int freq : count)
      if (freq > 0) {
        minfreq = Math.min(minfreq, freq);
        maxfreq = Math.max(maxfreq, freq);
      }
    return minfreq == maxfreq;
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
class Solution:
  def minimumSubstringsInPartition(self, s: str) -> int:
    n = len(s)
    # dp[i] := the minimum number of substrings in s[0..i]
    dp = [n] * n

    for i in range(n):
      count = collections.Counter()
      for j in range(i, -1, -1):
        count[s[j]] += 1
        # word[j..i] is balanced.
        if min(count.values()) == max(count.values()):
          dp[i] = min(dp[i], 1 + dp[j - 1] if j > 0 else 1)

    return dp[-1]