Determine If String Halves Are Alike

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Complexity

Runtime

$O(N)$

Space

$O(1)$

Code

C++

class Solution {
public:
    bool halvesAreAlike(string s) {
        int n = s.length();
        int c1 = 0;
        int c2 = 0;
        std::set<char> vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};
 
        for (size_t i = 0; i < n; ++i) {
            if (vowels.find(s[i]) != vowels.end()) {
                if (i < n / 2) {
                    c1 += 1;
                } else {
                    c2 += 1;
                }
            }
        }
 
        return c1 == c2;
    }
};

C

bool in(char c) {
    char vowels[] = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};
    for(int i = 0; i < 10; i++) {
        if (c == vowels[i]) {
            return true;
        } // end if
    } // end for
 
    return false;
} // end func
 
bool halvesAreAlike(char* s) {
    int c1 = 0;
    int c2 = 0;
 
    int n = strlen(s);
    int mid = n / 2;
    for (int i = 0; i < n; i++) {
        if (in(s[i])) {
            if (i < mid) {
                c1 += 1;
            } else {
                c2 += 1;
            }
        }
    } // end for
 
    return c1 == c2;
} // end function

Python

class Solution:
    def halvesAreAlike(self, s: str) -> bool:
        vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
        c1, c2 = 0, 0
        mid = len(s) // 2
        for i, c in enumerate(s):
            if c in vowels:
                if i < mid:
                    c1 += 1
                else:
                    c2 += 1
        
        return c1 == c2

Notes

• s → a 1st half, b 2nd half
• alike: vowel count is same

Cards

START Basic Front: Determine If String Halves Are Alike Back: Hard coded vowels to check in $O(1)$ time first and 2nd halves, respectively.

END

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References

• Determine If String Halves Are Alike