395. Longest Substring with At Least K Repeating Characters 1

题意

给定一个字符串，求得子字符串的最长长度，使得每一个字符最小重复K次

题解

可以先统计出每个字符的次数，然后遍历字符串，对于字符次数小于K的就跳过，重新设置起始点，对于字符次数大于K的end+1,end初始值为start+1;

class Solution {
public:int res=0;int longestSubstring(string s, int k) {if(k>s.length()){//even if all character are same they can't have freq k//as k>length of arrayreturn 0;}return helper(s,0,s.length(),k);}int helper(string s,int start,int end,int k){if(end-start<k)return  0;//we get the freq of the every charcter in s between start and endvector<int> freq(26,0);for(int i=start;i<end;i++){freq[s[i]-'a']++;   }//get the substring again where freq of every character is greater than kfor(int i=start;i<end;i++){if(freq[s[i]-'a']<k){//this cannot form result//now find the last index upto which we cannot make resultint j=i+1;while(j<end && freq[s[j]-'a']<k)j++;//cal the max substr from left and rightreturn max(helper(s,start,i,k),helper(s,j,end,k));}}//if all character freq >=k return end-start;}
};

class Solution {
public:int longestSubstring(string s, int k) {int n = s.length();// if length of string is 0 or it is less than k then there will no longest substring so we will return 0.if(n == 0 or n < k) return 0;//k ==1 means that all the characters will be unique so we will return entire length.if(k <= 1) return n;//count map to store count of charactersunordered_map<char,int> countMap;for(char c : s) countMap[c]++;int left=0;while(left<n && countMap[s[left]] >=k) left++;if(left > n-1) return left;int l1 = longestSubstring(s.substr(0, left), k);while(left < n && countMap[s[left]]<k) left++;// to check for longest Substring in part after leftint l2 = left < n ? longestSubstring(s.substr(left),k) : 0;//return max of l1 and l2return max(l1,l2);}
};

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395. Longest Substring with At Least K Repeating Characters 1

题意

题解

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