69 lines
2.3 KiB
C++
69 lines
2.3 KiB
C++
// Source : https://leetcode.com/problems/minimum-length-of-string-after-deleting-similar-ends/
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// Author : Hao Chen
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// Date : 2021-02-12
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/*****************************************************************************************************
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*
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* Given a string s consisting only of characters 'a', 'b', and 'c'. You are asked to apply the
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* following algorithm on the string any number of times:
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*
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* Pick a non-empty prefix from the string s where all the characters in the prefix are equal.
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* Pick a non-empty suffix from the string s where all the characters in this suffix are equal.
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* The prefix and the suffix should not intersect at any index.
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* The characters from the prefix and suffix must be the same.
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* Delete both the prefix and the suffix.
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*
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* Return the minimum length of s after performing the above operation any number of times (possibly
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* zero times).
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*
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* Example 1:
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*
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* Input: s = "ca"
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* Output: 2
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* Explanation: You can't remove any characters, so the string stays as is.
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*
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* Example 2:
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*
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* Input: s = "cabaabac"
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* Output: 0
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* Explanation: An optimal sequence of operations is:
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* - Take prefix = "c" and suffix = "c" and remove them, s = "abaaba".
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* - Take prefix = "a" and suffix = "a" and remove them, s = "baab".
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* - Take prefix = "b" and suffix = "b" and remove them, s = "aa".
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* - Take prefix = "a" and suffix = "a" and remove them, s = "".
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*
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* Example 3:
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*
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* Input: s = "aabccabba"
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* Output: 3
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* Explanation: An optimal sequence of operations is:
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* - Take prefix = "aa" and suffix = "a" and remove them, s = "bccabb".
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* - Take prefix = "b" and suffix = "bb" and remove them, s = "cca".
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*
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* Constraints:
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*
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* 1 <= s.length <= 105
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* s only consists of characters 'a', 'b', and 'c'.
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******************************************************************************************************/
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class Solution {
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public:
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int minimumLength(string s) {
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char ch;
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int left=0, right=s.size()-1;
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while(left < right) {
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ch = s[left];
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if (s[right] != ch) break;
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while (s[left] == ch) left++;
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if (left >= right ) return 0;
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while (s[right] == ch) right--;
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}
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return right - left + 1;
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}
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};
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