CTRL+A, CTRL+C, CTRL+V

Imagine you have a special keyboard with the following keys:

1. A
2. Ctrl+A
3. Ctrl+C
4. Ctrl+V

where CTRL+A, CTRL+C, CTRL+V each acts as one function key for “Select All”, “Copy”, and “Paste” operations respectively.
If you can only press the keyboard for N times (with the above four keys), please write a program to produce maximum numbers of A. If possible, please also print out the sequence of keys.
That is to say, the input parameter is N (No. of keys that you can press), the output is M (No. of As that you can produce).

I do believe there should be an O(n) solution, but sort of tricky. The O(n^2) solution is quite straightforward in DP problems.

public static int maxChar(int n) {
 int[] dp = new int[n];
 for (int i = 0; i < n; ++i) {
  dp[i] = i + 1;    // type A
 }
 for (int i = 0; i < n; ++i) {
  for (int j = i + 4; j < n; ++j) {
   dp[j] = Math.max(dp[j], dp[i] * (j - i - 3));    // using Ctrl + V
  }
 }
 return dp[n - 1];
}

Interleaving String

Interleaving StringAug 31 '12

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

For example,
Given:
s1 = "aabcc",
s2 = "dbbca",

When s3 = "aadbbcbcac", return true.
When s3 = "aadbbbaccc", return false.

dp[i][j] == true means the substring of s3 from start to 'i + j'th char is the interleaved string of s2 till 'i'th char and s1 till 'j'th char

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function    
        if (s1.empty()) return s2 == s3;
        else if (s2.empty()) return s1 == s3;
        else if (s3.empty()) return s1.empty() && s2.empty();
        else if (s1.size() + s2.size() != s3.size()) return false;
        
        int lengthS1 = s1.size() + 1, lengthS2 = s2.size() + 1;
        vector<vector<bool>> dp(lengthS2, vector<bool>(lengthS1, false));
        dp[0][0] = true;
        
        for (int i = 1; i < lengthS2; ++i) {
            dp[i][0] = s2[i - 1] == s3[i - 1] ? dp[i - 1][0] : false;
        }
        
        for (int j = 1; j < lengthS1; ++j) {
            dp[0][j] = s1[j - 1] == s3[j - 1] ? dp[0][j - 1] : false;
        }
        
        for (int i = 1; i < lengthS2; ++i) {
            for (int j = 1; j < lengthS1; ++j) {
                if (s1[j - 1] == s3[i + j - 1]) {
                    dp[i][j] = dp[i][j] || dp[i][j - 1];
                }
                if (s2[i - 1] == s3[i + j - 1]) {
                    dp[i][j] = dp[i][j] || dp[i - 1][j];
                }
            }
        }
        
        return dp[lengthS2 - 1][lengthS1 - 1];
    }
};