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
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/
// Author: github.com/lzl124631x
// 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/
// Author: github.com/lzl124631x
// 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;
}
};