• Time:O(mn)
• Space:O(mn)

## C++

``````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];
}
};
``````

## 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];
}
}
``````

## 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]
``````

• Time:O(mn)
• Space:O(n)

## C++

``````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;

vector<bool> dp(n + 1);

for (int i = 0; i <= m; ++i)
for (int j = 0; j <= n; ++j)
if (i == 0 && j == 0)
dp[j] = true;
else if (i == 0)
dp[j] = dp[j - 1] && s2[j - 1] == s3[j - 1];
else if (j == 0)
dp[j] = dp[j] && s1[i - 1] == s3[i - 1];
else
dp[j] = dp[j] && s1[i - 1] == s3[i + j - 1] ||
dp[j - 1] && s2[j - 1] == s3[i + j - 1];

return dp[n];
}
};
``````

## 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;

boolean[] dp = new boolean[n + 1];

for (int i = 0; i <= m; ++i)
for (int j = 0; j <= n; ++j)
if (i == 0 && j == 0)
dp[j] = true;
else if (i == 0)
dp[j] = dp[j - 1] && s2.charAt(j - 1) == s3.charAt(j - 1);
else if (j == 0)
dp[j] = dp[j] && s1.charAt(i - 1) == s3.charAt(i - 1);
else
dp[j] = dp[j] && s1.charAt(i - 1) == s3.charAt(i + j - 1) ||
dp[j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);

return dp[n];
}
}
``````

## 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 = [False] * (n + 1)

for i in range(m + 1):
for j in range(n + 1):
if i == 0 and j == 0:
dp[j] = True
elif i == 0:
dp[j] = dp[j - 1] and s2[j - 1] == s3[j - 1]
elif j == 0:
dp[j] = dp[j] and s1[i - 1] == s3[i - 1]
else:
dp[j] = (dp[j] and s1[i - 1] == s3[i + j - 1]) or \
(dp[j - 1] and s2[j - 1] == s3[i + j - 1])

return dp[n]
``````