Interleaving String


Problem Statement :


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 
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:

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.


Solution :



title-img 


                            Solution in C :

bool backtrack(char *s1, int i, char *s2, int j, char *s3, int k);
int memo[101][101];
bool isInterleave(char * s1, char * s2, char * s3){
    
    int i,j;
    for (i=0; i < 101; i++)
    {
        for (j=0; j < 101; j++)
        {
            memo[i][j]=-1;
        }
    }
        
    return backtrack(s1,0,s2,0,s3,0);

}

bool backtrack(char *s1, int i, char *s2, int j, char *s3, int k)
{
    if (s1[i]==NULL && s2[j]==NULL && s3[k]==NULL)
    {
        return memo[i][j]=1;
    }
    if (s3[k] == NULL)
        return memo[i][j]=0;
    
    if (memo[i][j]!=-1)
        return memo[i][j];
    
    if (s1[i]==s3[k] && s2[j]==s3[k])
    {
        return memo[i][j] = backtrack(s1,i+1,s2,j,s3,k+1) || backtrack(s1,i,s2,j+1,s3,k+1);
    }
    if (s1[i]==s3[k])
        return memo[i][j]=backtrack(s1,i+1,s2,j,s3,k+1);
    if (s2[j]==s3[k])
        return memo[i][j]=backtrack(s1,i,s2,j+1,s3,k+1);
    
    return memo[i][j]=0;
}
                        


                        Solution in C++ :

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

        vector<bool> dp(n + 1, false);
        dp[0] = true;

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

        for (int i = 1; i <= m; ++i) {
            dp[0] = dp[0] && s1[i - 1] == s3[i - 1];
            for (int j = 1; j <= n; ++j) {
                dp[j] = (dp[j] && s1[i - 1] == s3[i + j - 1]) || (dp[j - 1] && s2[j - 1] == s3[i + j - 1]);
            }
        }
        
        return dp[n];
    }
};
                    


                        Solution in Java :

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

        boolean[] dp = new boolean[n + 1];
        dp[0] = true;

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

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


                        Solution in Python : 
                            
class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        m, n, l = len(s1), len(s2), len(s3)
        if m + n != l:
            return False
        
        if m < n:
            return self.isInterleave(s2, s1, s3)
        
        dp = [False] * (n + 1)
        dp[0] = True
        
        for j in range(1, n + 1):
            dp[j] = dp[j-1] and s2[j-1] == s3[j-1]
        
        for i in range(1, m + 1):
            dp[0] = dp[0] and s1[i-1] == s3[i-1]
            for j in range(1, n + 1):
                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]


View More Similar Problems

Array and simple queries

Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty

View Solution →

Median Updates

The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o

View Solution →

Maximum Element

You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each

View Solution →

Balanced Brackets

A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and (). A matching pair of brackets is not balanced if the set of bra

View Solution →

Equal Stacks

ou have three stacks of cylinders where each cylinder has the same diameter, but they may vary in height. You can change the height of a stack by removing and discarding its topmost cylinder any number of times. Find the maximum possible height of the stacks such that all of the stacks are exactly the same height. This means you must remove zero or more cylinders from the top of zero or more of

View Solution →

Game of Two Stacks

Alexa has two stacks of non-negative integers, stack A = [a0, a1, . . . , an-1 ] and stack B = [b0, b1, . . . , b m-1] where index 0 denotes the top of the stack. Alexa challenges Nick to play the following game: In each move, Nick can remove one integer from the top of either stack A or stack B. Nick keeps a running sum of the integers he removes from the two stacks. Nick is disqualified f

View Solution →