Day 58 Prim's algo and CTCI question

Author: mandliya

.gitignore

@@ -31,3 +31,6 @@
 
 #build director
 build/
+
+#vim backup
+*.cpp~

.gitignore~

@@ -0,0 +1,33 @@
+# Created by https://www.gitignore.io/api/c++
+### C++ ###
+# Compiled Object files
+*.slo
+*.lo
+*.o
+*.obj
+
+# Precompiled Headers
+*.gch
+*.pch
+
+# Compiled Dynamic libraries
+*.so
+*.dylib
+*.dll
+
+# Fortran module files
+*.mod
+
+# Compiled Static libraries
+*.lai
+*.la
+*.a
+*.lib
+
+# Executables
+*.exe
+*.out
+*.app
+
+#build director
+build/

README.md

@@ -4,9 +4,9 @@
 
 | Current Status|     Stats     |
 | :------------: | :----------: |
-| Total Problems | 80 |
-| Current Streak | 57 |
-| Longest Streak | 57 ( August 17, 2015 - October 12, 2015 ) |
+| Total Problems | 82 |
+| Current Streak | 58 |
+| Longest Streak | 58 ( August 17, 2015 - October 13, 2015 ) |
 
 </center>
 
@@ -69,17 +69,18 @@ Include contains single header implementation of data structures and some algori
 ### Cracking the coding interview problems
 | Problem | Solution |
 | :------------ | :----------: |
-| Problem 1 : Edition 6: Write an algorithm to determine whether a string has unique characters or not. Can we do it without using addtional data structures? | [1-1-hasUniqueChars.cpp](cracking_the_coding_interview_problems/1-1-hasUniqueChars.cpp)|
-| Problem 2 : Edition 5: Reverse a string when you are a pass a null terminated C string.|[1-2-edi5-reverseString.cpp ](cracking_the_coding_interview_problems/1-2-edi5-reverseString.cpp)|
-| Problem 2 : Edition 6: Given two strings, determine if one is permutation of other.|[1-2-perm-strings.cpp](cracking_the_coding_interview_problems/1-2-perm-strings.cpp)|
-| Problem 3 : Edition 5: Write an algorithm to remove duplicate chars from a string.|[1-3-edi5-removeDuplicates.cpp](cracking_the_coding_interview_problems/1-3-edi5-removeDuplicates.cpp)|
-| Problem 3 : Edition 6: URLify: Replace all the spaces in a string with '%20'. Preferebly Inplace |[1-3-URLify.cpp](cracking_the_coding_interview_problems/1-3-URLify.cpp)|
-| Problem 4 : Edition 6: Given a string, write a function to check if it is a permutation of a pallindrome.|[1-4-pallindrome-permutations.cpp ](cracking_the_coding_interview_problems/1-4-pallindrome-permutations.cpp)|
-| Problem 5 : Edition 6: There are three possible edits that can be performed on a string - Insert a char, Delete a char, Replace a char. Given two strings, determine if they are one or 0 edit away.|[1-5-one-edit-away.cpp ](cracking_the_coding_interview_problems/1-5-one-edit-away.cpp)|
-| 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)|
-| Problem 8: Write an algorithm such that if an element of MxN matrix is 0, its entire row and column is set to 0. | [1-8-zero-matrix.cpp](cracking_the_coding_interview_problems/1-8-zero-matrix.cpp)|
-| Problem 9: Given two strings s1 and s2, determine s2 is rotation of s1 using only one call to a function which checks whether one string is rotation of another.|[1-9-string-rotation.cpp](cracking_the_coding_interview_problems/1-9-string-rotation.cpp)|
+| Problem 1-1 : Edition 6: Write an algorithm to determine whether a string has unique characters or not. Can we do it without using addtional data structures? | [1-1-hasUniqueChars.cpp](cracking_the_coding_interview_problems/1-1-hasUniqueChars.cpp)|
+| Problem 1-2 : Edition 5: Reverse a string when you are a pass a null terminated C string.|[1-2-edi5-reverseString.cpp ](cracking_the_coding_interview_problems/1-2-edi5-reverseString.cpp)|
+| Problem 1-2 : Edition 6: Given two strings, determine if one is permutation of other.|[1-2-perm-strings.cpp](cracking_the_coding_interview_problems/1-2-perm-strings.cpp)|
+| Problem 1-3 : Edition 5: Write an algorithm to remove duplicate chars from a string.|[1-3-edi5-removeDuplicates.cpp](cracking_the_coding_interview_problems/1-3-edi5-removeDuplicates.cpp)|
+| Problem 1-3 : Edition 6: URLify: Replace all the spaces in a string with '%20'. Preferebly Inplace |[1-3-URLify.cpp](cracking_the_coding_interview_problems/1-3-URLify.cpp)|
+| Problem 1-4 : Edition 6: Given a string, write a function to check if it is a permutation of a pallindrome.|[1-4-pallindrome-permutations.cpp ](cracking_the_coding_interview_problems/1-4-pallindrome-permutations.cpp)|
+| Problem 1-5 : Edition 6: There are three possible edits that can be performed on a string - Insert a char, Delete a char, Replace a char. Given two strings, determine if they are one or 0 edit away.|[1-5-one-edit-away.cpp ](cracking_the_coding_interview_problems/1-5-one-edit-away.cpp)|
+| Problem 1-6: 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 1-7: Rotate the matrix clockwise( & anticlockwise) by 90 degrees| [1-7-matrix-rotation.cpp](cracking_the_coding_interview_problems/1-7-matrix-rotation.cpp)|
+| Problem 1-8: Write an algorithm such that if an element of MxN matrix is 0, its entire row and column is set to 0. | [1-8-zero-matrix.cpp](cracking_the_coding_interview_problems/1-8-zero-matrix.cpp)|
+| Problem 1-9: Given two strings s1 and s2, determine s2 is rotation of s1 using only one call to a function which checks whether one string is rotation of another.|[1-9-string-rotation.cpp](cracking_the_coding_interview_problems/1-9-string-rotation.cpp)|
+| Problem 2-1: Remove duplicates from an *unsorted* linked list. What if no temporary buffer is allowed.|[2-1-remove-dups.cpp](cracking_the_coding_interview_problems/2-1-remove-dups.cpp)|
 
 ###Dynamic Programming Problems
 | Problem | Solution |
@@ -141,6 +142,7 @@ Include contains single header implementation of data structures and some algori
 | Depth First Traversal of a Graph | [dfsDemo.cpp](graph_problems/dfsDemo.cpp) |
 | Breadth First Traversal of a Graph | [bfsDemo.cpp](graph_problems/bfsDemo.cpp) |
 | calculate the shortest distance from the start position (Node S) to all of the other nodes in the graph using Dijkstra algorithm. | [dijkstra-shortest-reach.cpp](graph_problems/dijkstra-shortest-reach.cpp)|
+| Calculate total weight of Minimum Spanning Tree of a given graph ( sum of weights of edges which forms MST) using Prim's algorithm | [primsMST.cpp](graph_problems/primsMST.cpp)|
 
 ###Greedy Problems
 | Problem | Solution |

README.md~

@@ -4,9 +4,9 @@
 
 | Current Status|     Stats     |
 | :------------: | :----------: |
-| Total Problems | 78 |
-| Current Streak | 56 |
-| Longest Streak | 56 ( August 17, 2015 - October 11, 2015 ) |
+| Total Problems | 82 |
+| Current Streak | 58 |
+| Longest Streak | 58 ( August 17, 2015 - October 13, 2015 ) |
 
 </center>
 
@@ -69,15 +69,18 @@ Include contains single header implementation of data structures and some algori
 ### Cracking the coding interview problems
 | Problem | Solution |
 | :------------ | :----------: |
-| Problem 1 : Edition 6: Write an algorithm to determine whether a string has unique characters or not. Can we do it without using addtional data structures? | [1-1-hasUniqueChars.cpp](cracking_the_coding_interview_problems/1-1-hasUniqueChars.cpp)|
-| Problem 2 : Edition 5: Reverse a string when you are a pass a null terminated C string.|[1-2-edi5-reverseString.cpp ](cracking_the_coding_interview_problems/1-2-edi5-reverseString.cpp)|
-| Problem 2 : Edition 6: Given two strings, determine if one is permutation of other.|[1-2-perm-strings.cpp](cracking_the_coding_interview_problems/1-2-perm-strings.cpp)|
-| Problem 3 : Edition 5: Write an algorithm to remove duplicate chars from a string.|[1-3-edi5-removeDuplicates.cpp](cracking_the_coding_interview_problems/1-3-edi5-removeDuplicates.cpp)|
-| Problem 3 : Edition 6: URLify: Replace all the spaces in a string with '%20'. Preferebly Inplace |[1-3-URLify.cpp](cracking_the_coding_interview_problems/1-3-URLify.cpp)|
-| Problem 4 : Edition 6: Given a string, write a function to check if it is a permutation of a pallindrome.|[1-4-pallindrome-permutations.cpp ](cracking_the_coding_interview_problems/1-4-pallindrome-permutations.cpp)|
-| Problem 5 : Edition 6: There are three possible edits that can be performed on a string - Insert a char, Delete a char, Replace a char. Given two strings, determine if they are one or 0 edit away.|[1-5-one-edit-away.cpp ](cracking_the_coding_interview_problems/1-5-one-edit-away.cpp)|
-| 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)|
+| Problem 1-1 : Edition 6: Write an algorithm to determine whether a string has unique characters or not. Can we do it without using addtional data structures? | [1-1-hasUniqueChars.cpp](cracking_the_coding_interview_problems/1-1-hasUniqueChars.cpp)|
+| Problem 1-2 : Edition 5: Reverse a string when you are a pass a null terminated C string.|[1-2-edi5-reverseString.cpp ](cracking_the_coding_interview_problems/1-2-edi5-reverseString.cpp)|
+| Problem 1-2 : Edition 6: Given two strings, determine if one is permutation of other.|[1-2-perm-strings.cpp](cracking_the_coding_interview_problems/1-2-perm-strings.cpp)|
+| Problem 1-3 : Edition 5: Write an algorithm to remove duplicate chars from a string.|[1-3-edi5-removeDuplicates.cpp](cracking_the_coding_interview_problems/1-3-edi5-removeDuplicates.cpp)|
+| Problem 1-3 : Edition 6: URLify: Replace all the spaces in a string with '%20'. Preferebly Inplace |[1-3-URLify.cpp](cracking_the_coding_interview_problems/1-3-URLify.cpp)|
+| Problem 1-4 : Edition 6: Given a string, write a function to check if it is a permutation of a pallindrome.|[1-4-pallindrome-permutations.cpp ](cracking_the_coding_interview_problems/1-4-pallindrome-permutations.cpp)|
+| Problem 1-5 : Edition 6: There are three possible edits that can be performed on a string - Insert a char, Delete a char, Replace a char. Given two strings, determine if they are one or 0 edit away.|[1-5-one-edit-away.cpp ](cracking_the_coding_interview_problems/1-5-one-edit-away.cpp)|
+| Problem 1-6: 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 1-7: Rotate the matrix clockwise( & anticlockwise) by 90 degrees| [1-7-matrix-rotation.cpp](cracking_the_coding_interview_problems/1-7-matrix-rotation.cpp)|
+| Problem 1-8: Write an algorithm such that if an element of MxN matrix is 0, its entire row and column is set to 0. | [1-8-zero-matrix.cpp](cracking_the_coding_interview_problems/1-8-zero-matrix.cpp)|
+| Problem 1-9: Given two strings s1 and s2, determine s2 is rotation of s1 using only one call to a function which checks whether one string is rotation of another.|[1-9-string-rotation.cpp](cracking_the_coding_interview_problems/1-9-string-rotation.cpp)|
+| Problem 2-1: Remove duplicates from an *unsorted* linked list. What if no temporary buffer is allowed.|[2-1-remove-dups.cpp](cracking_the_coding_interview_problems/2-1-remove-dups.cpp)|
 
 ###Dynamic Programming Problems
 | Problem | Solution |

cracking_the_coding_interview_problems/2-1-remove-dups.cpp

@@ -0,0 +1,111 @@
+/*
+ * Cracking the coding interview edition 6
+ * Problem 2-1 : Remove duplicates from an unsorted linked list.
+ * Approach 1 : Naive approach of iterating and remove all further duplicates of current node. Space complexity O(1) & time complexity O(n^2)
+ * Approach 2: Use a hash table
+ */
+
+
+#include <iostream>
+#include <unordered_map>
+#include <random>
+
+
+struct Node {
+	int data = 0;
+	Node * next = nullptr;
+};
+
+void insert( Node * & head, int data )
+{
+	Node * newNode = new Node;
+	newNode->data = data;
+	newNode->next = head;
+	head = newNode;
+}
+
+void printList( Node * head ) {
+	while( head ) {
+		std::cout << head->data << "  ";
+		head = head->next;
+	}
+	std::cout << std::endl;
+}
+
+//generate a random int between min and max
+static inline int random_range(const int min, const int max) {
+	std::random_device rd;
+	std::mt19937 mt(rd());
+	std::uniform_int_distribution<int> distribution(min, max);
+	return distribution(mt);
+}
+
+
+// Method 1
+//space complexity O(1)
+// time complexity O(n^2)
+void removeDuplicates( Node * head ) {
+	if ( head == nullptr || ( head && (head->next == nullptr))) {
+		return;
+	}
+	Node * curr = head;
+	while(curr) {
+		Node * runner = curr;
+		while (runner->next != nullptr) {
+			if (runner->next->data == curr->data) {
+				runner->next = runner->next->next;
+			} else {
+				runner = runner->next;
+			}
+		}
+		curr = curr->next;
+	}
+}
+
+// Method 2
+// space complexity - O(n)
+// time complexity - O(n)
+void removeDuplicates1( Node * head ) {
+	if ( head == nullptr || ( head && (head->next == nullptr) )) {
+		return ;
+	}
+	std::unordered_map<int, int> node_map;
+	Node * prev = head;
+	Node * curr = head->next;
+	node_map[head->data] = 1;
+	while( curr != nullptr ) {
+		while (curr && node_map.find(curr->data) != node_map.end()) {
+			curr = curr->next;
+		}
+		prev->next = curr;
+		prev = curr;
+		if (curr) {
+			node_map[curr->data] = 1;
+			curr = curr->next;
+		}
+	}
+}
+
+
+
+int main() {
+	std::cout << "Method 1 : \n";
+	Node * head = nullptr;
+	for ( int i = 0; i < 10; ++i ) {
+		insert(head, random_range(1,7));
+	}
+	printList(head);
+	removeDuplicates(head);
+	printList(head);
+
+	std::cout << "Method 2 : \n";
+	Node * head1 = nullptr;
+	for ( int i = 0; i < 10; ++i ) {
+		insert(head1, random_range(1,7));
+	}
+	printList(head1);
+	removeDuplicates1(head1);
+	printList(head1);
+
+
+}

graph_problems/primsMST.cpp

@@ -0,0 +1,132 @@
+/**
+ * Minimum spanning tree using Prim's algorithm.
+ * Approach greedy
+ * Calculate the weight of minimum spanning tree(sum of weight of edges of MST)
+ * and print the edges which makes MST.
+ * Input format:
+ * ROW 1 :: V - number of vertices, E - number of Edges.
+ * Next E lines give details of each edge in format x y w.
+ * where x is start of edge, y is end of edge and w is weight of xy edge
+ * After that, S - Index of the starting point of finding the MST.
+ * all the input is not 0 indexed.
+ */
+
+
+
+#include <iostream>
+#include <limits>
+
+const int MIN = std::numeric_limits<int>::min();
+const int MAX = std::numeric_limits<int>::max();
+
+
+/**
+ * Print the Minimum spanning tree
+ */
+void printMST( int ** graph, int * MST, int V ) {
+	std::cout << " Edge    Weight \n";
+	for ( int v = 0; v < V; ++v ) {
+		if (MST[v] != -1 && graph[v][MST[v]] != -1 ) {
+			std::cout << v+1 << "<-->" << MST[v]+1 << "   " << graph[v][MST[v]] << std::endl;
+		}
+	}
+}
+
+
+/**
+ * minKey
+ * returns the index of the vertex which is not yet part of MST
+ * and has minimum key.
+ */
+
+int minKey( int * keys, bool * inMST, int V ) {
+	int min = MAX, minIndex;
+	for ( int v = 0; v < V; ++v ) {
+		if ( inMST[v] == false && min > keys[v] ) {
+			minIndex = v;
+			min = keys[v];
+		}
+	}
+	return minIndex;
+}
+
+
+/**
+ * primsMSTWeight
+ * returns the weight of the minimum spanning tree of the graph
+ * Args:
+ * graph: adjacency matrix of the graph
+ * V :    Total number of vertices in the graph
+ * S :	  Starting index of MST
+ */
+int primsMSTWeight( int ** graph, int V, int S, int * MST)
+{
+	int * key = new int[V];
+	bool * inMST = new bool[V];
+
+	for ( int v = 0; v < V; ++v ) {
+		inMST[v] = false;
+		key[v] = MAX;
+	}
+
+	key[S] = 0;
+	for ( int count = 0; count < V-1; ++count ) {
+		int u = minKey(key, inMST, V);
+		inMST[u] = true;
+		for ( int v = 0; v < V; ++v ) {
+			if ( (graph[u][v] != -1) && (inMST[v] == false) && (key[v] > graph[u][v]) ) {
+				key[v] = graph[u][v];
+				MST[v] = u;
+			}
+		}
+	}
+
+	int sum = 0;
+	for ( int v = 0 ;v < V; ++v ) {
+		if ( MST[V] != -1  && graph[v][MST[v]] != -1 ) {
+			sum += graph[v][MST[v]];
+		}
+	}
+	delete[] key;
+	delete[] inMST;
+	return sum;
+}
+
+
+int main()
+{
+	int V, E, x, y, w, S;
+	/**
+	 * V -- number of vertices
+	 * E -- number of edges
+	 * xy-- edge with endpoints x and y
+	 * w -- weight of edge
+	 * S -- Starting index
+	 */
+
+	std::cin >> V >> E;
+	int ** graph = new int*[V];
+	for ( int v = 0; v < V; ++v ) {
+		graph[v] = new int[V];
+		for ( int w = 0; w < V; ++w ) {
+			graph[v][w] = -1;
+		}
+	}
+
+	for ( int i = 0; i < E; ++i ) {
+		std::cin >> x >> y >> w;
+		if ( graph[x-1][y-1] == -1 ||
+			 ( (graph[x-1][y-1] > 0) && (graph[x-1][y-1] > w ))) {
+			graph[x-1][y-1] = w;
+			graph[y-1][x-1] = w;
+		}
+	}
+
+	std::cin >> S;
+	int * MST = new int[V];
+	MST[S-1] = -1;
+	std::cout << "Min Spanning tree weight " << primsMSTWeight(graph, V, S-1, MST) << std::endl;
+	std::cout << "Min Spanning tree:\n";
+	printMST(graph, MST, V);
+	return 0;
+}