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451. Sort Characters By Frequency

Description

Given a string s, sort it in decreasing order based on the frequency of the characters. The frequency of a character is the number of times it appears in the string.

Return the sorted string. If there are multiple answers, return any of them.

 

Example 1:

Input: s = "tree"
Output: "eert"
Explanation: 'e' appears twice while 'r' and 't' both appear once.
So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.

Example 2:

Input: s = "cccaaa"
Output: "aaaccc"
Explanation: Both 'c' and 'a' appear three times, so both "cccaaa" and "aaaccc" are valid answers.
Note that "cacaca" is incorrect, as the same characters must be together.

Example 3:

Input: s = "Aabb"
Output: "bbAa"
Explanation: "bbaA" is also a valid answer, but "Aabb" is incorrect.
Note that 'A' and 'a' are treated as two different characters.

 

Constraints:

  • 1 <= s.length <= 5 * 105
  • s consists of uppercase and lowercase English letters and digits.

Solutions

  • class Solution {
    public String frequencySort(String s) {
        Map<Character, Integer> cnt = new HashMap<>(52);
        for (int i = 0; i < s.length(); ++i) {
            cnt.merge(s.charAt(i), 1, Integer::sum);
        }
        List<Character> cs = new ArrayList<>(cnt.keySet());
        cs.sort((a, b) -> cnt.get(b) - cnt.get(a));
        StringBuilder ans = new StringBuilder();
        for (char c : cs) {
            for (int v = cnt.get(c); v > 0; --v) {
                ans.append(c);
            }
        }
        return ans.toString();
    }
}

  • class Solution {
public:
    string frequencySort(string s) {
        unordered_map<char, int> cnt;
        for (char& c : s) {
            ++cnt[c];
        }
        vector<char> cs;
        for (auto& [c, _] : cnt) {
            cs.push_back(c);
        }
        sort(cs.begin(), cs.end(), [&](char& a, char& b) {
            return cnt[a] > cnt[b];
        });
        string ans;
        for (char& c : cs) {
            ans += string(cnt[c], c);
        }
        return ans;
    }
};

  • class Solution:
    def frequencySort(self, s: str) -> str:
        cnt = Counter(s)
        return ''.join(c * v for c, v in sorted(cnt.items(), key=lambda x: -x[1]))


  • func frequencySort(s string) string {
	cnt := map[byte]int{}
	for i := range s {
		cnt[s[i]]++
	}
	cs := make([]byte, 0, len(s))
	for c := range cnt {
		cs = append(cs, c)
	}
	sort.Slice(cs, func(i, j int) bool { return cnt[cs[i]] > cnt[cs[j]] })
	ans := make([]byte, 0, len(s))
	for _, c := range cs {
		ans = append(ans, bytes.Repeat([]byte{c}, cnt[c])...)
	}
	return string(ans)
}

  • function frequencySort(s: string): string {
    const cnt: Map<string, number> = new Map();
    for (const c of s) {
        cnt.set(c, (cnt.get(c) || 0) + 1);
    }
    const cs = Array.from(cnt.keys()).sort((a, b) => cnt.get(b)! - cnt.get(a)!);
    const ans: string[] = [];
    for (const c of cs) {
        ans.push(c.repeat(cnt.get(c)!));
    }
    return ans.join('');
}


  • class Solution {
    /**
     * @param String $s
     * @return String
     */
    function frequencySort($s) {
        for ($i = 0; $i < strlen($s); $i++) {
            $hashtable[$s[$i]] += 1;
        }
        arsort($hashtable);
        $keys = array_keys($hashtable);
        for ($j = 0; $j < count($keys); $j++) {
            $rs = $rs . str_repeat($keys[$j], $hashtable[$keys[$j]]);
        }
        return $rs;
    }
}

  • use std::collections::HashMap;
impl Solution {
    pub fn frequency_sort(s: String) -> String {
        let mut cnt = HashMap::new();
        for c in s.chars() {
            cnt.insert(c, cnt.get(&c).unwrap_or(&0) + 1);
        }
        let mut cs = cnt.into_iter().collect::<Vec<(char, i32)>>();
        cs.sort_unstable_by(|(_, a), (_, b)| b.cmp(&a));
        cs.into_iter()
            .map(|(c, v)| vec![c; v as usize].into_iter().collect::<String>())
            .collect()
    }
}

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