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791. Custom Sort String

Description

You are given two strings order and s. All the characters of order are unique and were sorted in some custom order previously.

Permute the characters of s so that they match the order that order was sorted. More specifically, if a character x occurs before a character y in order, then x should occur before y in the permuted string.

Return any permutation of s that satisfies this property.

 

Example 1:

Input: order = "cba", s = "abcd"
Output: "cbad"
Explanation: 
"a", "b", "c" appear in order, so the order of "a", "b", "c" should be "c", "b", and "a". 
Since "d" does not appear in order, it can be at any position in the returned string. "dcba", "cdba", "cbda" are also valid outputs.

Example 2:

Input: order = "cbafg", s = "abcd"
Output: "cbad"

 

Constraints:

  • 1 <= order.length <= 26
  • 1 <= s.length <= 200
  • order and s consist of lowercase English letters.
  • All the characters of order are unique.

Solutions

  • class Solution {
    public String customSortString(String order, String s) {
        int[] cnt = new int[26];
        for (int i = 0; i < s.length(); ++i) {
            ++cnt[s.charAt(i) - 'a'];
        }
        StringBuilder ans = new StringBuilder();
        for (int i = 0; i < order.length(); ++i) {
            char c = order.charAt(i);
            while (cnt[c - 'a']-- > 0) {
                ans.append(c);
            }
        }
        for (int i = 0; i < 26; ++i) {
            while (cnt[i]-- > 0) {
                ans.append((char) ('a' + i));
            }
        }
        return ans.toString();
    }
}

  • class Solution {
public:
    string customSortString(string order, string s) {
        int cnt[26] = {0};
        for (char& c : s) ++cnt[c - 'a'];
        string ans;
        for (char& c : order)
            while (cnt[c - 'a']-- > 0) ans += c;
        for (int i = 0; i < 26; ++i)
            if (cnt[i] > 0) ans += string(cnt[i], i + 'a');
        return ans;
    }
};

  • class Solution:
    def customSortString(self, order: str, s: str) -> str:
        cnt = Counter(s)
        ans = []
        for c in order:
            ans.append(c * cnt[c])
            cnt[c] = 0
        for c, v in cnt.items():
            ans.append(c * v)
        return ''.join(ans)


  • func customSortString(order string, s string) string {
	cnt := [26]int{}
	for _, c := range s {
		cnt[c-'a']++
	}
	ans := []rune{}
	for _, c := range order {
		for cnt[c-'a'] > 0 {
			ans = append(ans, c)
			cnt[c-'a']--
		}
	}
	for i, v := range cnt {
		for j := 0; j < v; j++ {
			ans = append(ans, rune('a'+i))
		}
	}
	return string(ans)
}

  • function customSortString(order: string, s: string): string {
    const toIndex = (c: string) => c.charCodeAt(0) - 'a'.charCodeAt(0);
    const count = new Array(26).fill(0);
    for (const c of s) {
        count[toIndex(c)]++;
    }
    const ans: string[] = [];
    for (const c of order) {
        const i = toIndex(c);
        ans.push(c.repeat(count[i]));
        count[i] = 0;
    }
    for (let i = 0; i < 26; i++) {
        if (!count[i]) continue;
        ans.push(String.fromCharCode('a'.charCodeAt(0) + i).repeat(count[i]));
    }
    return ans.join('');
}


  • impl Solution {
    pub fn custom_sort_string(order: String, s: String) -> String {
        let mut count = [0; 26];
        for c in s.as_bytes() {
            count[(c - b'a') as usize] += 1;
        }
        let mut ans = String::new();
        for c in order.as_bytes() {
            for _ in 0..count[(c - b'a') as usize] {
                ans.push(char::from(*c));
            }
            count[(c - b'a') as usize] = 0;
        }
        for i in 0..count.len() {
            for _ in 0..count[i] {
                ans.push(char::from(b'a' + (i as u8)));
            }
        }
        ans
    }
}

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