Interleaving String Leetcode Solution

In this post, we are going to solve the Interleaving String Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Interleaving String Leetcode Solution
Interleaving String Leetcode Solution

Problem

Given strings s1s2, and s3, find whether s3 is formed by an interleaving of s1 and s2.

An interleaving of two strings s and t is a configuration where s and t are divided into n and m non-empty substrings respectively, such that:

  • s = s1 + s2 + ... + sn
  • t = t1 + t2 + ... + tm
  • |n - m| <= 1
  • The interleaving is s1 + t1 + s2 + t2 + s3 + t3 + ... or t1 + s1 + t2 + s2 + t3 + s3 + ...

Note: a + b is the concatenation of strings a and b.

Example 1:

Interleaving String Leetcode Solution
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
Explanation: One way to obtain s3 is:
Split s1 into s1 = "aa" + "bc" + "c", and s2 into s2 = "dbbc" + "a".
Interleaving the two splits, we get "aa" + "dbbc" + "bc" + "a" + "c" = "aadbbcbcac".
Since s3 can be obtained by interleaving s1 and s2, we return true.

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
Explanation: Notice how it is impossible to interleave s2 with any other string to obtain s3.

Example 3:

Input: s1 = "", s2 = "", s3 = ""
Output: true

Constraints:

  • 0 <= s1.length, s2.length <= 100
  • 0 <= s3.length <= 200
  • s1s2, and s3 consist of lowercase English letters.

Follow up: Could you solve it using only O(s2.length) additional memory space?

Now, lets see the leetcode solution of Interleaving String Traversal Leetcode Solution.

Interleaving String Leetcode Solution in Python

class Solution:
  def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
    m = len(s1)
    n = len(s2)
    if m + n != len(s3):
      return False

    # dp[i][j] := true if s3[0..i + j) is formed by the interleaving of
    #             s1[0..i) and s2[0..j)
    dp = [[False] * (n + 1) for _ in range(m + 1)]
    dp[0][0] = True

    for i in range(1, m + 1):
      dp[i][0] = dp[i - 1][0] and s1[i - 1] == s3[i - 1]

    for j in range(1, n + 1):
      dp[0][j] = dp[0][j - 1] and s2[j - 1] == s3[j - 1]

    for i in range(1, m + 1):
      for j in range(1, n + 1):
        dp[i][j] = (dp[i - 1][j] and s1[i - 1] == s3[i + j - 1]) or \
            (dp[i][j - 1] and s2[j - 1] == s3[i + j - 1])

    return dp[m][n]

Interleaving String Leetcode Solution in CPP

class Solution {
 public:
  bool isInterleave(string s1, string s2, string s3) {
    const int m = s1.length();
    const int n = s2.length();
    if (m + n != s3.length())
      return false;

    // dp[i][j] := true if s3[0..i + j) is formed by the interleaving of
    //             s1[0..i) and s2[0..j)
    vector<vector<bool>> dp(m + 1, vector<bool>(n + 1));
    dp[0][0] = true;

    for (int i = 1; i <= m; ++i)
      dp[i][0] = dp[i - 1][0] && s1[i - 1] == s3[i - 1];

    for (int j = 1; j <= n; ++j)
      dp[0][j] = dp[0][j - 1] && s2[j - 1] == s3[j - 1];

    for (int i = 1; i <= m; ++i)
      for (int j = 1; j <= n; ++j)
        dp[i][j] = dp[i - 1][j] && s1[i - 1] == s3[i + j - 1] ||
                   dp[i][j - 1] && s2[j - 1] == s3[i + j - 1];

    return dp[m][n];
  }
};

Interleaving String Leetcode Solution in Java

class Solution {
  public boolean isInterleave(String s1, String s2, String s3) {
    final int m = s1.length();
    final int n = s2.length();
    if (m + n != s3.length())
      return false;

    // dp[i][j] := true if s3[0..i + j) is formed by the interleaving of
    //             s1[0..i) and s2[0..j)
    boolean[][] dp = new boolean[m + 1][n + 1];
    dp[0][0] = true;

    for (int i = 1; i <= m; ++i)
      dp[i][0] = dp[i - 1][0] && s1.charAt(i - 1) == s3.charAt(i - 1);

    for (int j = 1; j <= n; ++j)
      dp[0][j] = dp[0][j - 1] && s2.charAt(j - 1) == s3.charAt(j - 1);

    for (int i = 1; i <= m; ++i)
      for (int j = 1; j <= n; ++j)
        dp[i][j] = dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1) ||
                   dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);

    return dp[m][n];
  }
}

Note: This problem Interleaving String is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

NEXT: Validate Binary Search Tree Leetcode Solution

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