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1528. Shuffle String

Description

You are given a string s and an integer array indices of the same length. The string s will be shuffled such that the character at the ith position moves to indices[i] in the shuffled string.

Return the shuffled string.

 

Example 1:

Input: s = "codeleet", indices = [4,5,6,7,0,2,1,3]
Output: "leetcode"
Explanation: As shown, "codeleet" becomes "leetcode" after shuffling.

Example 2:

Input: s = "abc", indices = [0,1,2]
Output: "abc"
Explanation: After shuffling, each character remains in its position.

 

Constraints:

  • s.length == indices.length == n
  • 1 <= n <= 100
  • s consists of only lowercase English letters.
  • 0 <= indices[i] < n
  • All values of indices are unique.

Solutions

Solution 1: Simulation

We create a character array or string \(\textit{ans}\) of the same length as the input string, then iterate through the string \(\textit{s}\) and place each character \(\textit{s}[i]\) at position \(\textit{indices}[i]\) in \(\textit{ans}\). Finally, we join the character array or string \(\textit{ans}\) to form the final result and return it.

The time complexity is \(O(n)\) and the space complexity is \(O(n)\), where \(n\) is the length of the string.

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class Solution:
    def restoreString(self, s: str, indices: List[int]) -> str:
        ans = [None] * len(s)
        for c, j in zip(s, indices):
            ans[j] = c
        return "".join(ans)

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class Solution {
    public String restoreString(String s, int[] indices) {
        int n = s.length();
        char[] ans = new char[n];
        for (int i = 0; i < n; ++i) {
            ans[indices[i]] = s.charAt(i);
        }
        return new String(ans);
    }
}

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class Solution {
public:
    string restoreString(string s, vector<int>& indices) {
        int n = s.size();
        string ans(n, 0);
        for (int i = 0; i < n; ++i) {
            ans[indices[i]] = s[i];
        }
        return ans;
    }
};

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func restoreString(s string, indices []int) string {
    ans := make([]rune, len(s))
    for i, c := range s {
        ans[indices[i]] = c
    }
    return string(ans)
}

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function restoreString(s: string, indices: number[]): string {
    const ans: string[] = [];
    for (let i = 0; i < s.length; i++) {
        ans[indices[i]] = s[i];
    }
    return ans.join('');
}

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impl Solution {
    pub fn restore_string(s: String, indices: Vec<i32>) -> String {
        let n = s.len();
        let mut ans = vec![' '; n];
        let chars: Vec<char> = s.chars().collect();
        for i in 0..n {
            ans[indices[i] as usize] = chars[i];
        }
        ans.iter().collect()
    }
}

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/**
 * @param {string} s
 * @param {number[]} indices
 * @return {string}
 */
var restoreString = function (s, indices) {
    const ans = [];
    for (let i = 0; i < s.length; i++) {
        ans[indices[i]] = s[i];
    }
    return ans.join('');
};

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