Simple C++ solution without DP


#1
int Solution::anytwo(string A) {
    string B = A;
    int i = 0;
    int j = 0;
    int count = 0;
    int x = 0;
    for(int j = 0;j<B.size();j++) {
        int i = x;
        while(i < A.size()) {
            if(A[i] == B[j] && i != j) {
                x = ++i;
                count++;
                break;
            } else {
                i++;
            }
        }
    }
    return count>=2;
}

#2

for string “abcbca” : your code will set x=A.size() when j=0. due to this, count remains 1 during whole traversal and thus the code is unable to recognise the repeating “bc” substring. hence there needs to be some correction here.


#3

It is wrong although it is accepted.
try “abbbba” it gives output:0
but it should be 1