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.
Problem
Given strings s1, s2, 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 + ... + snt = t1 + t2 + ... + tm|n - m| <= 1- The interleaving is
s1 + t1 + s2 + t2 + s3 + t3 + ...ort1 + s1 + t2 + s2 + t3 + s3 + ...
Note: a + b is the concatenation of strings a and b.
Example 1:
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 <= 1000 <= s3.length <= 200s1,s2, ands3consist of lowercase English letters.
Follow up: Could you solve it using only O(s2.length) additional memory space?
Now, let’s 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.
