🌱 Keegan's Digital Garden

97. Interleaving String

4 min read

📝 Problem Details

input strings s1 s2 s3 find whether s3 is formed by an interleaving of s1 and s2

What is an interleaving string?

  • A string s3 is an interleaving of s1 and s2 if it maintains the relative order of characters from both s1 and s2, but they can be interleaved together.
  • It does not mean the characters are shuffled randomly.
  • Example: s1 = "abc" s2 = "def" s3 = "adbcef" ✅ (valid interleaving) s3 = "abdecf" ❌ (invalid, order changed)

💡 Explanation of Solution

Intuitive Subproblem:
- We need to check if `s3` can be formed by **interleaving** `s1` and `s2`.
- Interleaving means characters from `s1` and `s2` appear in the **same relative order**, but they can be interspersed.
- Example:
    s1 = "abc", s2 = "def"
    s3 = "adbcef" ✅ (valid interleaving)
    s3 = "abdecf" ❌ (invalid because order changed)
- At every step, we either:
  1. Take the next character from `s1`.
  2. Take the next character from `s2`.

Top-Down or Bottom Up?: 
- **Bottom-Up Approach is Preferred** because:
  - It avoids deep recursion (stack overflow risk in large inputs).
  - We can efficiently fill a DP table iteratively.
  - It allows **space optimization** to `O(m)`, reducing memory usage.
- **Top-Down Approach (Recursion + Memoization) is also possible**, but has higher overhead due to recursion.

DP State:
- use a 2d dp table where dp[i][j] is:
	- i represents how many characters we have taken from s1
	- j represents how many characters we have taken from s2
	- dp[i][j] is true if the first i+j characters of s3 can be formed by interleaving the first i characters of s1 and first j characters of s2

State Transition:
- If `dp[i-1][j]` is `true` and `s1[i-1] == s3[i+j-1]`, then `dp[i][j] = true`.
- If `dp[i][j-1]` is `true` and `s2[j-1] == s3[i+j-1]`, then `dp[i][j] = true`.
- the answer should be at dp[s1.size()][s2.size()]

⌛ Complexity Analysis

Time Complexity: O(n * m)
Space Complexity: O(n * m)

💻 Implementation of Solution

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        int n = s1.size(), m = s2.size(), l = s3.size();
        
        // If the total length of s1 and s2 doesn't match s3, it's impossible to interleave them
        if (n + m != l) return false; 
        
        // Create a 2D DP table, initialized to false
        // dp[i][j] = true means the first i characters of s1 and the first j characters of s2
        // can successfully interleave to form the first i + j characters of s3.
        vector<vector<bool>> dp(n + 1, vector<bool>(m + 1, false));
        
        // Base Case: If all strings are empty, they trivially interleave
        dp[0][0] = true; 
        
        // Fill the DP table
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= m; j++) {
                // Check if we can take a character from s1
                if (i > 0 && s1[i - 1] == s3[i + j - 1]) {
                    // If we were previously able to interleave up to dp[i-1][j], then we can extend it
                    dp[i][j] = dp[i][j] || dp[i - 1][j]; 
                }
                
                // Check if we can take a character from s2
                if (j > 0 && s2[j - 1] == s3[i + j - 1]) {
                    // If we were previously able to interleave up to dp[i][j-1], then we can extend it
                    dp[i][j] = dp[i][j] || dp[i][j - 1]; 
                }
            }
        }
 
        // Our answer is in the bottom-right corner of the table
        // If dp[n][m] is true, s3 is a valid interleaving of s1 and s2
        return dp[n][m];
    }
};

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