1342. Number of Steps to Reduce a Number to Zero




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.

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"


  • 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).


First Thought

I can come up with a simple solution immediately, that is to create a new string with same length as original one, and then map all characters from original string based on the indices.

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

The execution time is 16ms (73.18%), and memory usage is 15.5MB (30.16%). To be honest, it is not a bad solution in terms of the execution time. However, the memory usage is a little poor. We’re using non-in-place algorithm here, and fairly speaking, that is the least memory usage for a non-in-place solution. If we want to improve the memory usage, we must use an inplace solution, epecially when the first argument of the function is a copy of original string s.

In-Place Solution

An in-place solution must guarantee that when we map an element to a new place, the element at the new place must be mapped right next; otherwise, we will lose the element.

Let’s take a look at the first example. When we process the first element c, it should be mapped to the 4th slot. Therefore, we replace the 4th element l with c. Now we need to process the l immediately, so it should go to the first slot, and we replace the first element with l. But what now? We’re back to the start point. Are we going to do it again? Of course not, because we would be stuck in an infinite loop. We need somehow to tell we’ve already processed a specific element.

Here is the trick. When mapping an element to its new place, instead of just replace the element at new place, we swap them. Then the element at new place goes to the old place. Recall that elements in indices represent that the ith element should be mapped to indices[i]th slot. After the swap, the indices[i]th element is swapped to ith slot. If we also swap ith and indices[i]th element in indices, we can keep this relation unchanged. Let’s still use the first example. After the swap, s becomes lodeceet, and indicesbecomes [0,5,6,7,4,2,1,3]. Now if we look at the indices, basically it says the first element goes to the first slot, which means we have already reach an end here. Everytime we do a swap, the starting point reaches to its final position, which means i == indices[i]. Therefore, we can take i == indices[i] as a condition to stop. That’s the essence of cyclic sort, that is everytime we find a right position for a specific element.


class Solution {
    for (int i = 0; i < indices.size(); i++) {
      while (indices[i] != i) {
        swap(s[i], s[indices[i]]);
        swap(indices[i], indices[indices[i]]);

    return s;

Runtime: 12 ms, faster than 91.66% of C++ online submissions for Shuffle String. Memory Usage: 15.2 MB, less than 93.44% of C++ online submissions for Shuffle String.

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