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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;
}
};

• 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;
}
}

############

class Solution {
public int numSplits(String s) {
Map<Character, Integer> cnt = new HashMap<>();
for (char c : s.toCharArray()) {
cnt.merge(c, 1, Integer::sum);
}
Set<Character> vis = new HashSet<>();
int ans = 0;
for (char c : s.toCharArray()) {
if (cnt.merge(c, -1, Integer::sum) == 0) {
cnt.remove(c);
}
if (vis.size() == cnt.size()) {
++ans;
}
}
return ans;
}
}

• // 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;
}
};

• class Solution:
def numSplits(self, s: str) -> int:
cnt = Counter(s)
vis = set()
ans = 0
for c in s:
cnt[c] -= 1
if cnt[c] == 0:
cnt.pop(c)
ans += len(vis) == len(cnt)
return ans


• func numSplits(s string) (ans int) {
cnt := map[rune]int{}
for _, c := range s {
cnt[c]++
}
vis := map[rune]bool{}
for _, c := range s {
vis[c] = true
cnt[c]--
if cnt[c] == 0 {
delete(cnt, c)
}
if len(vis) == len(cnt) {
ans++
}
}
return
}