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recency

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1308 Discussions

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  • vizjereix
     + 0 comments

    sorted() with reverse=True is reversing list completely! So you letters with same freq are on inverted order - changing freq to negative do the trick

  • devesh_sharma
     + 0 comments
    def char_count_dict(s):
    d = {}
    for c in s:
        d[c] = 1 if c not in d else d[c] + 1
    return d

if __name__ == '__main__':
    s = input()
    char_counts = char_count_dict(s)
    sorted_dict = dict(sorted(char_counts.items()))
    reverse_dict = {}
    for k,v in sorted_dict.items():
        if v not in reverse_dict: reverse_dict[v] = []
        reverse_dict[v].append(k)
    reverse_sorted_dict = dict(sorted(reverse_dict.items(), reverse=True))
    line_count = 0
    break_outer = False
    for v,k in reverse_sorted_dict.items():
        for i, ki in enumerate(reverse_sorted_dict[v]):
            print(str(reverse_sorted_dict[v][i]) + " " + str(v))
            line_count += 1
            if line_count == 3: 
                break_outer = True
                break
        if break_outer:
            break

  • kashyapnitin9141
     + 0 comments

    from collections import Counter s=input() freq_let=Counter(s) for entry in freq_let.most_common(3): print(*entry)

    In this question by using most_common makes the code very understandable and is less time taking....

  • cdhanush057
     + 0 comments

    u just calculate the words frequency convert them to list sort them alphabetically later sort them value based return the top three items

    s = input()
    d={}
    for i in s:
        if i not in d:
            d[i]=1
        else:
            d[i]+=1
    items = list(d.items())

    items.sort()
    for i in range(len(items)):
        for j in range(len(items)-1-i):
            if items[j][1] < items[j+1][1]:
                items[j], items[j+1] = items[j+1], items[j]
    for ch, freq in items[:3]:
        print(ch, freq)

  • jpconcerman
     + 1 comment
    from collections import Counter


if __name__ == '__main__':
s = input().strip()
ctr = Counter(s)
for key, item in sorted(ctr.items(), key=lambda x: (-x[1], x[0]))[:3]:
    print(f"{key} {item}")