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- Strings: Making Anagrams
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Strings: Making Anagrams
Strings: Making Anagrams
+ 0 comments My solution Java 8
public static int makeAnagram(String a, String b) { Map<Character, Integer> charFreq = new HashMap<>(); int count = 0; for (char c : a.toCharArray()) { charFreq.put(c, charFreq.get(c) != null ? charFreq.get(c) + 1 : 1); } for (char c : b.toCharArray()) { charFreq.put(c, charFreq.get(c) != null ? charFreq.get(c) - 1 : -1); } for (char key : charFreq.keySet()) { if (charFreq.get(key) != 0) { count += charFreq.get(key) < 0 ? charFreq.get(key) * -1 : charFreq.get(key); } } return count; }
+ 0 comments def makeAnagram(a, b): count_a, count_b = Counter(a), Counter(b) diff_a = count_a - count_b diff_b = count_b - count_a return sum(diff_a.values()) + sum(diff_b.values())
+ 0 comments int result = 0; HashMap map1 = new HashMap<>(); HashMap map2 = new HashMap<>();
for(char c:a.toCharArray()){ map1.put(c, map1.getOrDefault(c, 0)+1); } for(char c:b.toCharArray()){ map2.put(c, map2.getOrDefault(c, 0)+1); } for(Map.Entry<Character,Integer> set : map1.entrySet()){ if(map2.containsKey(set.getKey())){ if(set.getValue()>=map2.get(set.getKey())){ result += map2.get(set.getKey())*2; }else if(set.getValue()<=map2.get(set.getKey())){ result += set.getValue()*2; } } } return (a.length()+b.length())- result;
+ 0 comments C++ solution Using HashMap
int makeAnagram(string a, string b) {
map<char,int> m; int sum=0; for(int i=0;i<a.size();i++){ m[a[i]]++; } for(int i=0;i<b.size();i++){ m[b[i]]--; } for(auto i:m){ sum+=abs(i.second); } return sum;
}
+ 0 comments Javascript solution
function makeAnagram(a, b) { let lettersToRemove = 0; for (let i = 0; i < a.length; i++) { if (b.includes(a[i])) { b = b.replace(a[i], '') } else { lettersToRemove += 1; } } return lettersToRemove + b.length; }
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