Formatted question description: https://leetcode.ca/all/1525.html

1525. Number of Good Ways to Split a String (Medium)

You are given a string s, a split is called good if you can split s into 2 non-empty strings p and q where its concatenation is equal to s and the number of distinct letters in p and q are the same.

Return the number of good splits you can make in s.

 

Example 1:

Input: s = "aacaba"
Output: 2
Explanation: There are 5 ways to split "aacaba" and 2 of them are good. 
("a", "acaba") Left string and right string contains 1 and 3 different letters respectively.
("aa", "caba") Left string and right string contains 1 and 3 different letters respectively.
("aac", "aba") Left string and right string contains 2 and 2 different letters respectively (good split).
("aaca", "ba") Left string and right string contains 2 and 2 different letters respectively (good split).
("aacab", "a") Left string and right string contains 3 and 1 different letters respectively.

Example 2:

Input: s = "abcd"
Output: 1
Explanation: Split the string as follows ("ab", "cd").

Example 3:

Input: s = "aaaaa"
Output: 4
Explanation: All possible splits are good.

Example 4:

Input: s = "acbadbaada"
Output: 2

 

Constraints:

  • s contains only lowercase English letters.
  • 1 <= s.length <= 10^5

Related Topics:
String, Bit Manipulation

Solution 1.

Simply use a linear scan and increment the answer if the left unique count equals the right unique count. If the left unique count is greater than the left one, break the loop.

// OJ: https://leetcode.com/problems/number-of-good-ways-to-split-a-string/

// Time: O(N)
// Space: O(1)
class Solution {
public:
    int numSplits(string s) {
        int total[26] = {}, left[26] = {}, ans = 0;
        for (char c : s) total[c - 'a']++;
        for (char c : s) {
            ++left[c - 'a'];
            int L = 0, R = 0;
            for (int i = 0; i < 26; ++i) {
                if (left[i]) ++L;
                if (total[i] - left[i]) ++R;
            }
            if (L == R) ++ans;
            if (L > R) break;
        }
        return ans;
    }
};

Another implementation that doesn’t require the O(26) check

// OJ: https://leetcode.com/problems/number-of-good-ways-to-split-a-string/

// Time: O(N)
// Space: O(1)
class Solution {
public:
    int numSplits(string s) {
        int right[26] = {}, left[26] = {}, ans = 0, L = 0, R = 0;
        for (char c : s ) R += right[c - 'a']++ == 0;
        for (char c : s) {
            L += left[c - 'a']++ == 0;
            R -= right[c - 'a']-- == 1;
            ans += L == R;
            if (L > R) break;
        }
        return ans;
    }
};

Java

class Solution {
    public int numSplits(String s) {
        Map<Character, Integer> map1 = new HashMap<Character, Integer>();
        Map<Character, Integer> map2 = new HashMap<Character, Integer>();
        int length = s.length();
        for (int i = 0; i < length; i++) {
            char c = s.charAt(i);
            int count = map2.getOrDefault(c, 0) + 1;
            map2.put(c, count);
        }
        int goodSplits = 0;
        for (int i = 0; i < length - 1; i++) {
            char c = s.charAt(i);
            int count1 = map1.getOrDefault(c, 0) + 1;
            map1.put(c, count1);
            int count2 = map2.get(c) - 1;
            if (count2 > 0)
                map2.put(c, count2);
            else
                map2.remove(c);
            if (map1.size() == map2.size())
                goodSplits++;
            else if (map1.size() > map2.size())
                break;
        }
        return goodSplits;
    }
}

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