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1 month ago+ 0 comments

JS version all pass

function substrCount(n, s) {
  let count = 0, strArr = s.split("")
  for (let i = 0; i < n; i++) {
    count++
    let goLeft = i - 1, goRight = i + 1, value = strArr[i + 1]
    while (goLeft >= 0 && goRight < n) {
      if (strArr[goLeft--] == value && strArr[goRight++] == value)
        count++
      else break
    }
    goLeft = i, goRight = i + 1
    while (goLeft >= 0 && goRight < n) {
      if (strArr[goLeft--] == value && strArr[goRight++] == value)
        count++
      else break
    }
  }
  return count
}

5 months ago+ 1 comment

easy understand java code with O(n^2) time complexity please refer to leetcode647:palindromic substrings for this templates.

public class Solution {
    static int count;
    // Complete the substrCount function below.
    static long substrCount(int n, String s) {
         count = 0;
         //O(n^2)
        for(int i = 0 ; i < n;i++){//O(n)
            helperOdd(i,i,s);//0(n)
            helperEven(i,i+1,s);//O(n)
        }
        return count;
    }
   
    static void helperEven(int i, int j, String s){
        if(i >=0 && j < s.length()){
        char c = s.charAt(i);
        while(i >=0 && j < s.length() && s.charAt(i) == s.charAt(j)&& s.charAt(i) == c){
            count++;
            i--;
            j++;
        } 
        }
        
    }
    static void helperOdd(int i, int j, String s){
        count++;
				i--;
        j++;
       helperEven(i,j,s);
    }

9 months ago+ 2 comments

Java

static long substrCount(int n, String s) {
        long count = n;
        char[] arr = s.toCharArray();
        for (int i = 0; i < n - 1; i++) {
           for (int k = i + 1; k < n; k++) {
               if (arr[i] == arr[k]) count++;
               if (arr[i] != arr[k]) {
                   int lastIndex = k * 2 - i;
                   if (lastIndex < n && s.substring(i, k).equals(s.substring(k + 1, lastIndex + 1))) {
                       count++;
                       
                   }
                   break;
               }
           } 
        }
        
        return count;
    }

10 months ago+ 0 comments

C++ Simple solution

long substrCount(int n, string s) {
    long ans = 0;
    char c = s[0];
    int j=0;
    while(j<s.length()) {
        char c= s[j];
        long cnt = 1;
        j++;
        while(j<s.length() && s[j]==c)j++, cnt++;
        ans = ans + (cnt+1)*cnt/2;
    }
    for(int i=1;i<n-1;++i){
        if(s[i]==s[i-1]||s[i]==s[i+1])continue;
        int x = i-1;
        int y = i+1;
        char z = s[i-1];
        while(x>=0&&y<n&&s[x]==s[y]&&s[x]==z)x--,y++;
        ans = ans + (i-x-1);
    }
    return ans;
}

10 months ago+ 1 comment

hopefully a readable one:

long substrCount(int n, string s) {

// minimum count is n 
int count = n;

for (int i=0;i<n;i++) {
    char current = s[i];

    // just continuous - equal substrings like "aaa"
    int j = i+1;
    while (j<n && s[j] == current) {
        count++;
        j++;
    }

    // continuous on left and right side from current char
    j = 1;
    while (i-j>=0 && i+j<n && // array bounds
        current != s[i+j] && // next is different than current
        s[i-j] == s[i+j] && // char's on left and right sides are the same
        s[i+j] == s[i+1] // next char must be the same as previous 
    ) {
        count++;
        j++;
    }
}

return count;

}

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