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  1. Prepare
  2. Algorithms
  3. Strings
  4. Anagram
  5. Discussions

Anagram

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976 Discussions

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  • ilovemylittlecr1
     + 0 comments

    include

    include

    using namespace std;

    /*Think of it this way: e.g.xaxb|bbxx left : 2x 1a 1b right: 2x 0a 2b , and they cancle out. What is left in the right is the answer.

    O(n) per query*/

    int arr[26];

    int main() { ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); 

    int q;
cin >> q; 

string str; 
while(q--) {
    cin >> str; 
    if(str.size() % 2) { cout << -1 << '\n'; continue; }

    int mid = str.size() >> 1; 

    for(int i = 0; i < mid; ++i) ++arr[str[i] - 'a']; 

    int ret = 0; 
    for(int i = mid; i < str.size(); ++i) {
        if(arr[str[i] - 'a']) --arr[str[i] - 'a']; 
        else ++ret; 
    }

    cout << ret << '\n'; 
    memset(arr, 0, sizeof(arr)); 
}

return 0;

    }

  • whiland
     + 0 comments

    Minimalist O(n) solution:

        public static int anagram(String s) {
        if (s.length() % 2 != 0) return -1;
        
        int size = s.length() / 2;
        int[] s1 = new int[26];
        int base = 'a';
        int changes = 0;        

        //build first half
        for (int i = 0; i < size; i++) {
            s1[s.charAt(i) - base] ++;
        }

        //check second half
        for (int i = size; i < size*2; i++) {
            if (s1[s.charAt(i) - base] > 0) s1[s.charAt(i) - base]--;
            else changes++;
        }

        return changes;
    
    }

  • greivin_lopez
     + 0 comments
    # Anagram
from collections import Counter
def anagram(s):
    len_s = len(s)
    result = -1
    if len_s % 2 == 0:
        l, r = Counter(s[:len_s // 2]), Counter(s[len_s // 2:])
        result = sum((r - l).values())
    return result

  • eliaskabbani1
     + 0 comments
    fun anagram(s: String): Int {
    // Write your code here
    if(s.length % 2 !=0)
        return -1
    var res = 0
    val middle = s.length/2
    var sub1 = s.subSequence(0, middle)
    var sub2 = s.subSequence(middle,s.length)
    var s1 = StringBuilder(sub1)
    var s2 = StringBuilder(sub2)
    for(c in s1){
        if(s2.contains(c)){
            val index = s2.indexOf(c)
            s2.deleteCharAt(index)
        }else
            res++
    }
    return res
}

  • enix12enix
     + 0 comments

    rust answer

    fn anagram(s: &str) -> i32 {
    let mut counter = vec![0;26];
    let chs: Vec<char> = s.chars().collect();
    let mut res = 0;
    let mut idx:usize = 0;
    
    if chs.len() % 2 == 1 {
        return -1;
    }
    
    for i in 0..chs.len()/2 {
        idx = (chs[i] as u32 - 'a' as u32) as usize;
        counter[idx] = counter[idx] + 1;
    }
    for i in chs.len()/2..chs.len() {
        idx = (chs[i] as u32 - 'a' as u32) as usize;
        if counter[idx] == 0 {
            res += 1;
        } else {
            counter[idx] = counter[idx] - 1;
        }
    }
    
    res

}