Day-55 Dynamic programming problems 2

Author: mandliya

README.md

@@ -4,9 +4,9 @@
 
 | Current Status|     Stats     |
 | :------------: | :----------: |
-| Total Problems | 75 |
-| Current Streak | 54 |
-| Longest Streak | 54 ( August 17, 2015 - October 9, 2015 ) |
+| Total Problems | 77 |
+| Current Streak | 55 |
+| Longest Streak | 55 ( August 17, 2015 - October 10, 2015 ) |
 
 </center>
 
@@ -79,6 +79,11 @@ Include contains single header implementation of data structures and some algori
 | Problem 6: Implement a method to perform basic string compression. Example string **aabcccccaaa** should be compressed to **a2b1c5a3**, however if compressed string is bigger than original string, return original string| [1-6-string-compression.cpp](cracking_the_coding_interview_problems/1-6-string-compression.cpp)|
 | Problem 7: Rotate the matrix clockwise( & anticlockwise) by 90 degrees| [1-7-matrix-rotation.cpp](cracking_the_coding_interview_problems/1-7-matrix-rotation.cpp)|
 
+#Dynamic Programming Problems
+| Problem | Solution |
+| :------------ | :----------: |
+| N<sup>th</th> Fibonacci term using different memoization techniques | [fibonacci.cpp](dynamic_programming_problems/fibonacci.cpp)|
+| Longest Common Subsequence Problem | [lcs.cpp](dynamic_programming_problems/lcs.cpp) | 
 
 ### Tree Problems
 | Problem | Solution |

dynamic_programming_problems/fibonacci.cpp

@@ -0,0 +1,58 @@
+/**
+ * Calculating Fibonacci series term of index N using DP memoization.
+ */
+
+#include <iostream>
+#include <vector>
+
+//bottom up
+
+int fib1(int n) {
+	std::vector<int> fib(n, 0);
+	if ( n == 0 || n == 1 ) {
+		return 1;
+	}
+	fib[0] = 1;
+	fib[1] = 1;
+	for ( int i = 2; i < n; ++i ) {
+		fib[i] = fib[i-1] + fib[i-2];
+	}
+	return fib[n-1];
+}
+
+//top down
+
+std::vector<int> fib(1000, 0);
+int fib2( int n ) {
+	if ( n == 0 ) {
+		return 0;
+	}
+	if ( n == 1 || n == 2 ) {
+		return 1;
+	}
+	if ( fib[n] != 0 ) {
+		return fib[n];
+	}
+	fib[n] = fib2(n-1) + fib2(n-2);
+	return fib[n];
+}
+
+//leverage the fact we are just using last 2 term
+int fib3( int n ) {
+	int a = 0;
+	int b = 1;
+	for ( int i = 2; i <=  n; ++i ) {
+		int c = a + b;
+		a = b;
+		b = c;
+	}
+	return b;
+}
+
+
+int main()
+{
+	std::cout << "Demonstrating fibonacci term calculation:\n";
+	std::cout << fib1(9) << " " << fib2(9) << " " << fib3(9) << std::endl;
+	return 0;
+}

dynamic_programming_problems/lcs.cpp

@@ -0,0 +1,85 @@
+/**
+ * Given two strings X(1..m) and Y(1..n). Find the longest common substring
+ * that appears left to right (but not necessarily in a contiguous manner)
+ * in both strings. For example X = "ABCBDAB" and Y = "BDCABA"
+ * LCS(X,Y) = {"BCBA", "BDAB", "BCAB"}.
+ * LCS-LENGTH(X,Y) = 4
+ */
+
+
+#include <iostream>
+#include <string>
+#include <vector>
+
+size_t max( size_t a, size_t b ) {
+	return a > b ? a : b;
+}
+
+size_t max(size_t a, size_t b, size_t c) {
+	return (max(a,b) > c ) ? max(a,b) : c ;
+}
+
+
+
+
+void longest_common_subsequence(std::vector<std::vector<size_t>> & lcs,
+		std::string s1, std::string s2)
+{
+	size_t m = s1.length();
+	size_t n = s2.length();
+
+	for ( size_t j = 0; j <= n; ++j ) {
+		lcs[0][j] = 0;
+	}
+
+	for ( size_t i = 0; i <= m; ++i ) {
+		lcs[i][0] = 0;
+	}
+
+	for( size_t i = 0; i < m; ++i ) {
+		for( size_t j = 0; j < n; ++j ) {
+			lcs[i+1][j+1] = lcs[i][j];  //getting previous max val
+			if (s1[i] == s2[j]) {
+				lcs[i+1][j+1]++;
+			} else {
+				lcs[i+1][j+1] = max(lcs[i][j+1], lcs[i+1][j]);
+			}
+		}
+	}
+	std::cout << "Longest common subsequence length - " << lcs[m][n] << std::endl;
+
+	size_t i = m;
+	size_t j = n;
+	size_t index = lcs[m][n];
+	std::string seq(lcs[m][n], ' ');
+	while( i > 0 && j > 0 ) {
+		if ( s1[i-1] == s2[j-1] ) {
+			seq[index -1 ] = s1[i-1];
+			--index;
+			--i;
+			--j;
+		} else if ( lcs[i-1][j] > lcs[i][j-1] ) {
+			--i;
+		} else {
+			--j;
+		}
+
+	}
+	std::cout << "One of possible many Longest Common Subsequence is : " << seq << std::endl;
+
+}
+
+int main()
+{
+	std::string str1, str2;
+	std::cout << "Longest common subsequence Problem:\n";
+	std::cout << "Enter string 1:";
+	std::cin >> str1;
+	std::cout << "Enter string 2:";
+	std::cin >> str2;
+
+	std::vector<std::vector<size_t>> lcs( str1.length() + 1, std::vector<size_t>(str2.length() + 1));
+
+	longest_common_subsequence(lcs, str1, str2);
+	return 0;
+}