2287. Rearrange Characters to Make Target String

Description

You are given two 0-indexed strings s and target. You can take some letters from s and rearrange them to form new strings.

Return the maximum number of copies of target that can be formed by taking letters from s and rearranging them.

Example 1:

Input: s = "ilovecodingonleetcode", target = "code"
Output: 2
Explanation:
For the first copy of "code", take the letters at indices 4, 5, 6, and 7.
For the second copy of "code", take the letters at indices 17, 18, 19, and 20.
The strings that are formed are "ecod" and "code" which can both be rearranged into "code".
We can make at most two copies of "code", so we return 2.

Example 2:

Input: s = "abcba", target = "abc"
Output: 1
Explanation:
We can make one copy of "abc" by taking the letters at indices 0, 1, and 2.
We can make at most one copy of "abc", so we return 1.
Note that while there is an extra 'a' and 'b' at indices 3 and 4, we cannot reuse the letter 'c' at index 2, so we cannot make a second copy of "abc".

Example 3:

Input: s = "abbaccaddaeea", target = "aaaaa"
Output: 1
Explanation:
We can make one copy of "aaaaa" by taking the letters at indices 0, 3, 6, 9, and 12.
We can make at most one copy of "aaaaa", so we return 1.

Constraints:

1 <= s.length <= 100
1 <= target.length <= 10
s and target consist of lowercase English letters.

Note: This question is the same as 1189: Maximum Number of Balloons.

Solutions

Solution 1: Counting

We count the occurrences of each character in the strings \(\textit{s}\) and \(\textit{target}\), denoted as \(\textit{cnt1}\) and \(\textit{cnt2}\). For each character in \(\textit{target}\), we calculate the number of times it appears in \(\textit{cnt1}\) divided by the number of times it appears in \(\textit{cnt2}\), and take the minimum value.

The time complexity is \(O(n + m)\), and the space complexity is \(O(|\Sigma|)\). Where \(n\) and \(m\) are the lengths of the strings \(\textit{s}\) and \(\textit{target}\), respectively. And \(|\Sigma|\) is the size of the character set, which is 26 in this problem.

Python3JavaC++GoTypeScriptRustC

class Solution:
    def rearrangeCharacters(self, s: str, target: str) -> int:
        cnt1 = Counter(s)
        cnt2 = Counter(target)
        return min(cnt1[c] // v for c, v in cnt2.items())

class Solution {
    public int rearrangeCharacters(String s, String target) {
        int[] cnt1 = new int[26];
        int[] cnt2 = new int[26];
        for (int i = 0; i < s.length(); ++i) {
            ++cnt1[s.charAt(i) - 'a'];
        }
        for (int i = 0; i < target.length(); ++i) {
            ++cnt2[target.charAt(i) - 'a'];
        }
        int ans = 100;
        for (int i = 0; i < 26; ++i) {
            if (cnt2[i] > 0) {
                ans = Math.min(ans, cnt1[i] / cnt2[i]);
            }
        }
        return ans;
    }
}

class Solution {
public:
    int rearrangeCharacters(string s, string target) {
        int cnt1[26]{};
        int cnt2[26]{};
        for (char& c : s) {
            ++cnt1[c - 'a'];
        }
        for (char& c : target) {
            ++cnt2[c - 'a'];
        }
        int ans = 100;
        for (int i = 0; i < 26; ++i) {
            if (cnt2[i]) {
                ans = min(ans, cnt1[i] / cnt2[i]);
            }
        }
        return ans;
    }
};

func rearrangeCharacters(s string, target string) int {
    var cnt1, cnt2 [26]int
    for _, c := range s {
        cnt1[c-'a']++
    }
    for _, c := range target {
        cnt2[c-'a']++
    }
    ans := 100
    for i, v := range cnt2 {
        if v > 0 {
            ans = min(ans, cnt1[i]/v)
        }
    }
    return ans
}

function rearrangeCharacters(s: string, target: string): number {
    const idx = (s: string) => s.charCodeAt(0) - 97;
    const cnt1 = new Array(26).fill(0);
    const cnt2 = new Array(26).fill(0);
    for (const c of s) {
        ++cnt1[idx(c)];
    }
    for (const c of target) {
        ++cnt2[idx(c)];
    }
    let ans = 100;
    for (let i = 0; i < 26; ++i) {
        if (cnt2[i]) {
            ans = Math.min(ans, Math.floor(cnt1[i] / cnt2[i]));
        }
    }
    return ans;
}

impl Solution {
    pub fn rearrange_characters(s: String, target: String) -> i32 {
        let mut count1 = [0; 26];
        let mut count2 = [0; 26];
        for c in s.as_bytes() {
            count1[(c - b'a') as usize] += 1;
        }
        for c in target.as_bytes() {
            count2[(c - b'a') as usize] += 1;
        }
        let mut ans = i32::MAX;
        for i in 0..26 {
            if count2[i] != 0 {
                ans = ans.min(count1[i] / count2[i]);
            }
        }
        ans
    }
}

#define min(a, b) (((a) < (b)) ? (a) : (b))

int rearrangeCharacters(char* s, char* target) {
    int count1[26] = {0};
    int count2[26] = {0};
    for (int i = 0; s[i]; i++) {
        count1[s[i] - 'a']++;
    }
    for (int i = 0; target[i]; i++) {
        count2[target[i] - 'a']++;
    }
    int ans = INT_MAX;
    for (int i = 0; i < 26; i++) {
        if (count2[i]) {
            ans = min(ans, count1[i] / count2[i]);
        }
    }
    return ans;
}

2287. Rearrange Characters to Make Target String

Description

Solutions

Solution 1: Counting

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