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  1. Prepare
  2. Algorithms
  3. Strings
  4. Making Anagrams
  5. Discussions

Making Anagrams

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772 Discussions

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  • kavyahegde242
     + 0 comments

    public static int makingAnagrams(String s1, String s2) { // Write your code here int freq[] = new int[26]; for(char s : s1.toCharArray()){ freq[s-'a']++; } for(char z : s2.toCharArray()){ freq[z-'a']--; } int count=0; for(int f : freq){ if(f!=0){ count= count+Math.abs(f); } } return count; }

    }

  • savra_sv
     + 0 comments

    //TC: O(N + M), SC: O(1)

    public static int makingAnagrams(String s1, String s2) {

        int[] ar1 = new int[26];
    int[] ar2 = new int[26];

    for (int i = 0; i < s1.length(); i++) {
        ar1[s1.charAt(i) - 'a']++;
    }

    for (int i = 0; i < s2.length(); i++) {
        ar2[s2.charAt(i) - 'a']++;
    }

    int result = 0;

    for (int i = 0; i < 26; i++) {
        result += Math.abs(ar1[i] - ar2[i]);
    }

    return result;
}

  • afzal_saiyed931
     + 0 comments
    def makingAnagrams(s1, s2):
    # Write your code here
    temp = {}
    for i in s1:
        if i in temp:
            temp[i] += 1
        else:
            temp[i] = 1
            
    c = 0      
    for i in s2:
        if temp.get(i):
            temp[i] -= 1
        else:
            c += 1
         
    for i in temp.values():
        c += i
    return c

  • spriyamalar
     + 0 comments
    public static int makingAnagrams(String s1, String s2) {
    Map<String, Long> s1map = Arrays.stream(s1.split("")).collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
    Map<String, Long> s2map = Arrays.stream(s2.split("")).collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

    long delcount = 0;

    Set<Entry<String, Long>> s1entryset = s1map.entrySet();
    Iterator<Entry<String, Long>> s1iterator = s1entryset.iterator();
    while(s1iterator.hasNext()) {
        Entry<String, Long> entry = s1iterator.next();

        String s1key = entry.getKey();
        Long s1val = entry.getValue();
        if(s2map.get(s1key) != null) {
            if(s1val > s2map.get(s1key)) delcount += s1val - s2map.get(s1key);
            else if(s2map.get(s1key) > s1val) delcount += s2map.get(s1key) - s1val;
        }
        else if(s2map.get(s1key) == null) {
            delcount += s1val;
        }
    }

    Set<Entry<String, Long>> s2entryset = s2map.entrySet();
    Iterator<Entry<String, Long>> s2iterator = s2entryset.iterator();
    while(s2iterator.hasNext()) {
        Entry<String, Long> entry = s2iterator.next();
        String s2key = entry.getKey();
        Long s2val = entry.getValue();
        if(s1map.get(s2key) == null) {
            delcount += s2val;
        }
    }
    return (int)delcount;
}

  • geovanni_luna
     + 0 comments

    This python solution is O(n + m) time and O(k) space.

    O(k) because in the worse case all keys will will be unique and it will result in a space of O(n + m)

    def makingAnagrams(s1, s2):
    w1 = {}
    w2 = {}
    n = max(len(s1), len(s2))
    
    for i in range(n):
        if i < len(s1):
            if s1[i] not in w1:
                w1[s1[i]] = 0
            w1[s1[i]] += 1

        if i < len(s2):
            if s2[i] not in w2:
                w2[s2[i]] = 0
            w2[s2[i]] += 1
            
    all_keys = set(w1.keys()) | set(w2.keys())
    res = {}
    for key in all_keys:
        if key in w1 and key in w2:
            res[key] = abs(w1[key] - w2[key])
        elif key in w1:
            res[key] = w1[key]
        else:
            res[key] = w2[key]
            
    return sum(res.values())