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Formatted question description: https://leetcode.ca/all/1528.html

1528. Shuffle String

Level

Easy

Description

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 i-th position moves to indices[i] in the shuffled string.

Return the shuffled string.

Example 1:

Image text

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.

Example 3:

Input: s = “aiohn”, indices = [3,1,4,2,0]

Output: “nihao”

Example 4:

Input: s = “aaiougrt”, indices = [4,0,2,6,7,3,1,5]

Output: “arigatou”

Example 5:

Input: s = “art”, indices = [1,0,2]

Output: “rat”

Constraints:

  • s.length == indices.length == n
  • 1 <= n <= 100
  • s contains only lower-case English letters.
  • 0 <= indices[i] < n
  • All values of indices are unique (i.e. indices is a permutation of the integers from 0 to n - 1).

Solution

Create an array array of type char with length n. For each i from 0 to n - 1, set array[indices[i]] = s.charAt(i). Finally, convert array to a string and return.

  • class Solution {
        public String restoreString(String s, int[] indices) {
            int length = indices.length;
            char[] array = new char[length];
            for (int i = 0; i < length; i++) {
                char c = s.charAt(i);
                int index = indices[i];
                array[index] = c;
            }
            return new String(array);
        }
    }
    
    ############
    
    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 String.valueOf(ans);
        }
    }
    
  • class Solution:
        def restoreString(self, s: str, indices: List[int]) -> str:
            ans = [0] * len(s)
            for i, c in enumerate(s):
                ans[indices[i]] = c
            return ''.join(ans)
    
    
    
  • 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;
        }
    };
    
  • func restoreString(s string, indices []int) string {
    	ans := make([]rune, len(s))
    	for i, c := range s {
    		ans[indices[i]] = c
    	}
    	return string(ans)
    }
    
  • /**
     * @param {string} s
     * @param {number[]} indices
     * @return {string}
     */
    var restoreString = function (s, indices) {
        let rs = [];
        for (let i = 0; i < s.length; i++) {
            rs[indices[i]] = s[i];
        }
        return rs.join("");
    };
    
    

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