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InterLeavingString.cs
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InterLeavingString.cs
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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 they are divided into non-empty substrings 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:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
Example 2:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
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?
*/
public class Solution {
public bool IsInterleave(string s1, string s2, string s3) {
var cache = new Dictionary< Tuple<int,int,int>, bool >();
return IsInterleaveUtil(s1, 0, s2,0, s3, 0, cache);
}
public bool IsInterleaveUtil(string s1, int i, string s2, int j, string s3, int k, Dictionary<Tuple<int,int,int>, bool> cache){
if (k == s3.Length){
return (i == s1.Length && j == s2.Length);
}
var t = new Tuple<int, int, int>(i,j,k);
if (cache.ContainsKey(t)){
return cache[t];
}
bool b1 = false, b2 = false;
if (i < s1.Length && s1[i] == s3[k]){
b1 = IsInterleaveUtil(s1, i+1, s2, j, s3, k+1, cache);
}
if (b1) return true;
if (j < s2.Length && s2[j] == s3[k]){
b2 = IsInterleaveUtil(s1, i, s2, j+1, s3, k+1, cache);
}
if (b2) return true;
cache[t] = false;
return false;
}
}