97. Interleaving String

Problem

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

Example 1:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false

Solution

/**
 * @param {string} s1
 * @param {string} s2
 * @param {string} s3
 * @return {boolean}
 */
var isInterleave = function(s1, s2, s3) {
  var dp = {};
  if (s3.length !== s1.length + s2.length) return false;
  return helper(s1, s2, s3, 0, 0, 0, dp);
};

var helper = function (s1, s2, s3, i, j, k, dp) {
  var res = false;

  if (k >= s3.length) return true;
  if (dp['' + i + j + k] !== undefined) return dp['' + i + j + k];

  if (s3[k] === s1[i] && s3[k] === s2[j]) {
    res = helper(s1, s2, s3, i + 1, j, k + 1, dp) || helper(s1, s2, s3, i, j + 1, k + 1, dp);
  } else if (s3[k] === s1[i]) {
    res = helper(s1, s2, s3, i + 1, j, k + 1, dp);
  } else if (s3[k] === s2[j]) {
    res = helper(s1, s2, s3, i, j + 1, k + 1, dp);
  }

  dp['' + i + j + k] = res;

  return res;
};

Explain:

nope.

Complexity:

• Time complexity : O(n^2). n 为 s3.length。
• Space complexity : O(n^2).