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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 + ... + sn
t = t1 + t2 + ... + tm
|n - m| <= 1
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:
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
s1
, s2
, and s3
consist of lowercase English letters.Follow up: Could you solve it using only O(s2.length)
additional memory space?
bool isInterleave(string s1, string s2, string s3) {
if(s3.length() != s1.length() + s2.length())
return false;
bool table[s1.length()+1][s2.length()+1];
for(int i=0; i<s1.length()+1; i++)
for(int j=0; j< s2.length()+1; j++){
if(i==0 && j==0)
table[i][j] = true;
else if(i == 0)
table[i][j] = ( table[i][j-1] && s2[j-1] == s3[i+j-1]);
else if(j == 0)
table[i][j] = ( table[i-1][j] && s1[i-1] == s3[i+j-1]);
else
table[i][j] = (table[i-1][j] && s1[i-1] == s3[i+j-1] ) || (table[i][j-1] && s2[j-1] == s3[i+j-1] );
}
return table[s1.length()][s2.length()];
}
def isInterleave1(self, s1, s2, s3):
r, c, l= len(s1), len(s2), len(s3)
if r+c != l:
return False
dp = [[True for _ in xrange(c+1)] for _ in xrange(r+1)]
for i in xrange(1, r+1):
dp[i][0] = dp[i-1][0] and s1[i-1] == s3[i-1]
for j in xrange(1, c+1):
dp[0][j] = dp[0][j-1] and s2[j-1] == s3[j-1]
for i in xrange(1, r+1):
for j in xrange(1, c+1):
dp[i][j] = (dp[i-1][j] and s1[i-1] == s3[i-1+j]) or \
(dp[i][j-1] and s2[j-1] == s3[i-1+j])
return dp[-1][-1]
public boolean isInterleave(String s1, String s2, String s3) {
if ((s1.length()+s2.length())!=s3.length()) return false;
boolean[][] matrix = new boolean[s2.length()+1][s1.length()+1];
matrix[0][0] = true;
for (int i = 1; i < matrix[0].length; i++){
matrix[0][i] = matrix[0][i-1]&&(s1.charAt(i-1)==s3.charAt(i-1));
}
for (int i = 1; i < matrix.length; i++){
matrix[i][0] = matrix[i-1][0]&&(s2.charAt(i-1)==s3.charAt(i-1));
}
for (int i = 1; i < matrix.length; i++){
for (int j = 1; j < matrix[0].length; j++){
matrix[i][j] = (matrix[i-1][j]&&(s2.charAt(i-1)==s3.charAt(i+j-1)))
|| (matrix[i][j-1]&&(s1.charAt(j-1)==s3.charAt(i+j-1)));
}
}
return matrix[s2.length()][s1.length()];
}
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