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interleaving string dynamic programming leetcode 97

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certainly! the problem "interleaving string" from leetcode (problem 97) asks whether a string `s3` can be formed by interleaving two strings `s1` and `s2` while maintaining the order of characters from `s1` and `s2`.
problem statement
given three strings `s1`, `s2`, and `s3`, return `true` if `s3` is formed by an interleaving of `s1` and `s2`.
example
- input: `s1 = "aab", s2 = "axy", s3 = "aaxaby"`
- output: `true`
- input: `s1 = "aab", s2 = "axy", s3 = "aaabxy"`
- output: `false`
constraints
- the lengths of `s1`, `s2`, and `s3` are all less than or equal to 100.
dynamic programming approach
we can solve this problem using dynamic programming. the idea is to create a 2d table `dp` where `dp[i][j]` represents whether the first `i` characters of `s1` and the first `j` characters of `s2` can form the first `i+j` characters of `s3`.
steps to solve the problem
1. **initialize the dp table**: create a 2d boolean array `dp` of size `(len(s1)+1) x (len(s2)+1)`. the element `dp[i][j]` will be `true` if the first `i` characters of `s1` and the first `j` characters of `s2` can form the first `i+j` characters of `s3`.
2. **base case**: set `dp[0][0]` to `true` since two empty strings can form an empty string.
3. **fill the dp table**:
- iterate through the lengths of `s1` and `s2`.
- for each `dp[i][j]`, check:
- if `s1[i-1]` matches `s3[i+j-1]`, then set `dp[i][j]` to `dp[i-1][j]`.
- if `s2[j-1]` matches `s3[i+j-1]`, then set `dp[i][j]` to `dp[i][j-1]`.
4. **result**: the value at `dp[len(s1)][len(s2)]` will indicate whether `s3` can be formed by interleaving `s1` and `s2`.
code implementation
here is the python code implementing the above logic:
explanation of the code
- the code first checks if the combined length of `s1` and `s2` matches `s3`. if not, it returns `false`.
- it initializes a 2d list `dp` and sets the base case.
- it fills the first row and column based ...
#InterleavingString #DynamicProgramming #LeetCode97
interleaving string
dynamic programming
LeetCode 97
string manipulation
recursive solution
substring check
DP table
backtracking
character matching
problem-solving
algorithm optimization
time complexity
space complexity
input validation
coding interview
problem statement
given three strings `s1`, `s2`, and `s3`, return `true` if `s3` is formed by an interleaving of `s1` and `s2`.
example
- input: `s1 = "aab", s2 = "axy", s3 = "aaxaby"`
- output: `true`
- input: `s1 = "aab", s2 = "axy", s3 = "aaabxy"`
- output: `false`
constraints
- the lengths of `s1`, `s2`, and `s3` are all less than or equal to 100.
dynamic programming approach
we can solve this problem using dynamic programming. the idea is to create a 2d table `dp` where `dp[i][j]` represents whether the first `i` characters of `s1` and the first `j` characters of `s2` can form the first `i+j` characters of `s3`.
steps to solve the problem
1. **initialize the dp table**: create a 2d boolean array `dp` of size `(len(s1)+1) x (len(s2)+1)`. the element `dp[i][j]` will be `true` if the first `i` characters of `s1` and the first `j` characters of `s2` can form the first `i+j` characters of `s3`.
2. **base case**: set `dp[0][0]` to `true` since two empty strings can form an empty string.
3. **fill the dp table**:
- iterate through the lengths of `s1` and `s2`.
- for each `dp[i][j]`, check:
- if `s1[i-1]` matches `s3[i+j-1]`, then set `dp[i][j]` to `dp[i-1][j]`.
- if `s2[j-1]` matches `s3[i+j-1]`, then set `dp[i][j]` to `dp[i][j-1]`.
4. **result**: the value at `dp[len(s1)][len(s2)]` will indicate whether `s3` can be formed by interleaving `s1` and `s2`.
code implementation
here is the python code implementing the above logic:
explanation of the code
- the code first checks if the combined length of `s1` and `s2` matches `s3`. if not, it returns `false`.
- it initializes a 2d list `dp` and sets the base case.
- it fills the first row and column based ...
#InterleavingString #DynamicProgramming #LeetCode97
interleaving string
dynamic programming
LeetCode 97
string manipulation
recursive solution
substring check
DP table
backtracking
character matching
problem-solving
algorithm optimization
time complexity
space complexity
input validation
coding interview