83 lines
2.3 KiB
C++
83 lines
2.3 KiB
C++
// Source : https://leetcode.com/problems/sum-of-beauty-of-all-substrings/
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// Author : Hao Chen
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// Date : 2021-03-13
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/*****************************************************************************************************
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*
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* The beauty of a string is the difference in frequencies between the most frequent and least
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* frequent characters.
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*
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* For example, the beauty of "abaacc" is 3 - 1 = 2.
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*
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* Given a string s, return the sum of beauty of all of its substrings.
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*
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* Example 1:
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*
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* Input: s = "aabcb"
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* Output: 5
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* Explanation: The substrings with non-zero beauty are ["aab","aabc","aabcb","abcb","bcb"], each with
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* beauty equal to 1.
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*
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* Example 2:
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*
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* Input: s = "aabcbaa"
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* Output: 17
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*
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* Constraints:
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*
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* 1 <= s.length <= 500
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* s consists of only lowercase English letters.
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******************************************************************************************************/
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class Solution {
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private:
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int beauty(string& s, int start, int end) {
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int stat[26] = {0};
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for (int i=start; i<=end; i++){
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stat[s[i]-'a']++;
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}
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int max = INT_MIN, min = INT_MAX;
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for (auto s: stat) {
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if (s == 0 ) continue;
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max = s > max ? s : max;
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min = s < min ? s : min;
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}
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return max - min;
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}
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public:
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int beautySum(string s) {
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return beautySum02(s);
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return beautySum01(s);
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}
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int beautySum01(string& s) {
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int sum = 0;
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for (int i=0; i<s.size()-1; i++) {
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for (int j=i+1; j<s.size(); j++) {
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sum += beauty(s, i, j);
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}
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}
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return sum;
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}
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//same as beautySum01(), but optimazed slightly
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int beautySum02(string& s) {
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int sum = 0;
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for (int i=0; i<s.size()-1; i++) {
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int stat[26] = {0};
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for (int j=i; j<s.size(); j++) {
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stat[s[j]-'a']++;
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int max = INT_MIN, min = INT_MAX;
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for (auto& n: stat) {
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if (n <= 0 ) continue;
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max = n > max ? n : max;
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min = n < min ? n : min;
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}
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//cout << s.substr(i, j-i+1) << " --> "<< max << ":" << min << endl;
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sum += (max - min);
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}
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}
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return sum;
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}
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};
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