Sort 1840 Discussions, By:

16 hours ago+ 0 comments

def isValid(s):

c = set(s)
l = []
result = "YES"
for i in c:
    k = s.count(i)
    l.append(k)
if(max(l) == min(l)):
    return result
if(max(l)-min(l)== 1):
    if(l.count(min(l))<2 or l.count(max(l))<2):
        return result
    else:
        result = "NO"
        return result
if(min(l) == 1 and l.count(min(l)==1) and l.count(max(l))==len(l)-1):
    return result
else:
    result = "NO"  
    return result

6 days ago+ 0 comments

JAVA 8 simple solution:

public static String isValid(String s) 
    {
        Map<Character,Integer> map = new HashMap<>();
        for(Character c : s.toCharArray())
        {           
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
       List<Integer> list  = new ArrayList<>(map.values());
       
       Collections.sort(list);
       int min =list.get(0);
       int max= list.get(list.size() - 1);
       if(min == max)
        {
           return "YES";
        }
        int minCount = 0;
        int maxCount = 0;

        for(int val : list)
        {
            if(min == val)
            {
                minCount++;
            }
            if(max == val)
            {
                maxCount++;
            }
        }
        System.out.println("max:"+max+"\n min:"+min);
        System.out.println("maxCount:"+maxCount+"\n minCount:"+minCount);
        if(max - min > 1 )
        {
            if(minCount != 1)
            {
                return "NO";
            }
        }
        return ((maxCount + minCount == list.size() ) && (maxCount < 2 || minCount <2 )  ? "YES": "NO"  );
    }

2 weeks ago+ 0 comments

Counter is nice.

def isValid(s):
    counts = dict(Counter(s))
    count_freqs = dict(Counter(counts.values()))
    if len(count_freqs) == 1:
        return 'YES'
    elif len(count_freqs) > 2:
        return 'NO'
    elif count_freqs.get(1, 0) == 1:
        return 'YES'
    elif abs(list(count_freqs.keys())[-1] - list(count_freqs.keys())[0]) == 1 and count_freqs[max(list(count_freqs.keys()))] == 1:
        return 'YES'
    
    return 'NO'

2 weeks ago+ 0 comments

Python:

def isValid(s):
    Dict = createFirstDict(s)
    Dict2 = createSecondDict(Dict)
    res = determineIfIsValid(Dict2)
    return res
    
def createFirstDict(s):
    res = {}
    for ch in s:
        if ch not in res:
            res[ch] = 1
        else:
            res[ch] +=1
            
    return res

def createSecondDict(Dict):
    res = {}
    for key1 in Dict:
         if Dict[key1] not in res:
            res[Dict[key1]] = 1
         else:
            res[Dict[key1]] += 1
            
    return res
    
def determineIfIsValid(Dict2):
    if len(Dict2) == 1:
        return "YES"
    elif len(Dict2) > 2:
        return "NO"
    else:
        keyList = list(Dict2.keys())
        key1 = keyList[0]
        key2 = keyList[1]
        value1 = Dict2[key1]
        value2 = Dict2[key2]
        maxKey = max(key1, key2)
        minKey = min(key1, key2)
        if maxKey - minKey == 1 and Dict2[maxKey] == 1:
            return "YES"
        elif key1 == value1 == 1 or key2 == value2 == 1:
				
            return "YES"
        
    return "NO"

2 weeks ago+ 1 comment

I tested the following scenario with one of the editorial solutions:

assert isValid("aaabbbcc") == "NO"

This scenario fails one solution from the editorial. I don't get it why this is not being considered on the full test cases, is there something I am missing from the description?

Load more conversations