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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:

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.

• Java
• Python
• C++
• Go
• Javascript

• 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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