Day 7 - Edit Distance.md

Difficulty	Hard
Source	160 Days of Problem Solving
Tags	Dynamic Programming Strings

🚀 Day 7. Edit Distance 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given two strings s1 and s2, you need to find the minimum number of operations required to convert s1 into s2. The valid operations are:

1. Insert a character at any position.
2. Delete any character from the string.
3. Replace any character with another character.

🔍 Example Walkthrough:

Example 1:

Input:

s1 = "geek"
s2 = "gesek"

Output:

1

Explanation:

We can insert 's' between the two 'e' in "geek" to form "gesek".

Example 2:

Input:

s1 = "gfg"
s2 = "gfg"

Output:

0

Explanation:

The strings are already the same, so 0 operations are required.

Example 3:

Input:

s1 = "abcd"
s2 = "bcfe"

Output:

3

Explanation:

• Remove 'a' from "abcd" → "bcd"
• Replace 'd' with 'f' → "bcf"
• Insert 'e' at the end → "bcfe"

Constraints:

• $(1 \leq s1.length(), s2.length() \leq 10^3)$
• Both strings are in lowercase.

🎯 My Approach:

Space-Optimized Dynamic Programming

1. We use two 1D arrays (prev and curr) of size ( n+1 ) (where ( n ) is the length of ( s2 )).
2. Initialize the prev array such that prev[j] = j, meaning if ( s1 ) is empty, it takes j operations (all inserts) to match ( s2[0..j-1] ).
3. Iterate over each character in ( s1 ) (index i) and fill curr:
   • curr[0] = i (if ( s2 ) is empty, we need i deletions to match).
   • For each j in ( s2 ), compare s1[i-1] with s2[j-1]:
     • If they match, carry over prev[j-1].
     • Otherwise, curr[j] = 1 + min({ replace, delete, insert }).
4. Swap prev and curr after each iteration to save space.
5. Finally, prev[n] holds the edit distance.

Algorithm Steps:

1. Let $( m = \text{length}(s1), n = \text{length}(s2) )$.
2. Initialize prev array of size ( n+1 ) with prev[j] = j.
3. Loop i from 1 to m:
   • Set curr[0] = i.
   • For each j from 1 to n:
     • If s1[i-1] == s2[j-1], then curr[j] = prev[j-1].
     • Else curr[j] = 1 + min( prev[j-1], prev[j], curr[j-1] ).
   • Swap prev and curr.
4. Answer is prev[n].

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: $( O(M \times N) )$, as we iterate through each character of $( s1 )$ and $( s2 )$.
• Expected Auxiliary Space Complexity: $( O(N) )$, because we only keep two 1D arrays of size $( n+1 )$.

📝 Solution Code

Code (C++)

class Solution {
public:
    int editDistance(string& s1, string& s2) {
        int m = s1.size(), n = s2.size();
        vector<int> prev(n + 1), curr(n + 1);
        iota(prev.begin(), prev.end(), 0);
        for (int i = 1; i <= m; i++) {
            curr[0] = i;
            for (int j = 1; j <= n; j++)
                curr[j] = s1[i-1] == s2[j-1] ? prev[j-1] : 1 + min({prev[j-1], prev[j], curr[j-1]});
            swap(prev, curr);
        }
        return prev[n];
    }
};

⚡ Alternative Approaches

2️⃣ Dynamic Programming (O(M * N) Time, O(M * N) Space)

Idea:

• Create a 2D DP array where dp[i][j] represents the minimum operations to convert s1[0...i-1] to s2[0...j-1].
• If characters match, carry forward the diagonal value.
• Otherwise, consider the minimum of insert, delete, and replace.

class Solution {
public:
    int editDistance(string& s1, string& s2) {
        int m = s1.size(), n = s2.size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1));
        for (int i = 0; i <= m; i++) dp[i][0] = i;
        for (int j = 0; j <= n; j++) dp[0][j] = j;
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (s1[i-1] == s2[j-1]) dp[i][j] = dp[i-1][j-1];
                else dp[i][j] = 1 + min({dp[i-1][j-1], dp[i-1][j], dp[i][j-1]});
            }
        }
        return dp[m][n];
    }
};

🔹 More intuitive for understanding
🔹 Easier to debug

📊 Comparison of Approaches

Approach	⏱️ Time Complexity	🗂️ Space Complexity	✅ Pros	⚠️ Cons
Space Optimized DP (1D)	🟡 O(M * N)	🟢 O(N)	Efficient for large data	Slightly harder to debug
DP (2D Table)	🟡 O(M * N)	🟡 O(M * N)	Easier to understand	More memory-intensive

💡 Best Choice?

• ✅ For learning concepts: Use the 2D DP approach.
• ✅ For optimal performance: Use the Space Optimized DP approach.

Code (Java)

class Solution {
    public int editDistance(String s1, String s2) {
        int m = s1.length(), n = s2.length();
        int[] prev = new int[n + 1], curr = new int[n + 1];
        for (int i = 0; i <= n; i++) prev[i] = i;
        for (int i = 1; i <= m; i++) {
            curr[0] = i;
            for (int j = 1; j <= n; j++)
                curr[j] = s1.charAt(i-1) == s2.charAt(j-1) ? prev[j-1] : 1 + Math.min(prev[j-1], Math.min(prev[j], curr[j-1]));
            int[] temp = prev;
            prev = curr;
            curr = temp;
        }
        return prev[n];
    }
}

Code (Python)

class Solution:
    def editDistance(self, s1, s2):
        m, n = len(s1), len(s2)
        prev, curr = list(range(n + 1)), [0] * (n + 1)
        for i in range(1, m + 1):
            curr[0] = i
            for j in range(1, n + 1):
                curr[j] = prev[j-1] if s1[i-1] == s2[j-1] else 1 + min(prev[j-1], prev[j], curr[j-1])
            prev, curr = curr, prev
        return prev[n]

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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