87. Scramble String

Problem

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = "great":

great
   /    \
  gr    eat
 / \    /  \
g   r  e   at
           / \
          a   t

To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".

rgeat
   /    \
  rg    eat
 / \    /  \
r   g  e   at
           / \
          a   t

We say that "rgeat" is a scrambled string of "great".

Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".

rgtae
   /    \
  rg    tae
 / \    /  \
r   g  ta  e
       / \
      t   a

We say that "rgtae" is a scrambled string of "great".

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

Example 1:

Input: s1 = "great", s2 = "rgeat"
Output: true

Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false

Solution

/**
 * @param {string} s1
 * @param {string} s2
 * @return {boolean}
 */
var isScramble = function(s1, s2) {
  return helper({}, s1, s2);
};

var helper = function (dp, s1, s2) {
  var map = {};

  if (dp[s1 + s2] !== undefined) return dp[s1 + s2];
  if (s1 === s2) return true;

  for (var j = 0; j < s1.length; j++) {
    if (map[s1[j]] === undefined) map[s1[j]] = 0;
    if (map[s2[j]] === undefined) map[s2[j]] = 0;
    map[s1[j]]++;
    map[s2[j]]--;
  }

  for (var key in map) {
    if (map[key] !== 0) {
      dp[s1 + s2] = false;
      return false;
    }
  }

  for (var i = 1; i < s1.length; i++) {
    if ((helper(dp, s1.substr(0, i), s2.substr(0, i))
         && helper(dp, s1.substr(i), s2.substr(i))) ||
        (helper(dp, s1.substr(0, i), s2.substr(s2.length - i))
         && helper(dp, s1.substr(i), s2.substr(0, s2.length - i)))) {
      dp[s1 + s2] = true;
      return true;
    }
  }

  dp[s1 + s2] = false;
  return false;
};

Explain:

nope.

Complexity:

• Time complexity : O(n^4).
• Space complexity : O(n^4).