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rrohitramsenrrohitramsen
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added dp problems
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‎ConvertIntoPalindrome.java‎

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package com.program.dynamic.programming;
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/**
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* @author rrohit
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*/
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public class ConvertIntoPalindrome {
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/**
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* This method will add characters to make the given word a palindrome.
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* Time : O(n+n) = O(n)
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* Space: O(n)
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* @param word
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* @return
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*/
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public static String convertIntoPalindrome(String word) {
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if (word == null || word.isEmpty()) {
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return "a";
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}
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int wordlen = word.length();
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if (wordlen == 1){
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return word;
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}
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boolean match[] = new boolean[wordlen];
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int left = 0, right = wordlen-1, matchCount = 0;
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while (left < right) {
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if (word.charAt(left) == word.charAt(right)) {
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match[left] = match[right] = true;
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matchCount += 2;
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left++;
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right--;
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} else {
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match[right] = false;
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right--;
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}
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}
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/**
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* In case word it self is a palindrome.
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*/
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if (matchCount == wordlen || matchCount == wordlen-1 && match[wordlen-1]) {
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return word;
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}
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int j = wordlen-1;
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StringBuilder sb = new StringBuilder();
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while (j > right) {
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if (!match[j]) {
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sb.append(word.charAt(j));
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}
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j--;
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}
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String palindrome = sb.append(word).toString();
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return palindrome;
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}
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/**
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* Condition 1 : If First and Last chars should match.
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* Condition 2 : Substring (excluding first and last chars) must be palindrome.
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* @param word
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* @return
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*/
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public static boolean palindrome(String word) {
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if (word == null) {
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return false;
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}
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if (word.isEmpty() || word.length() == 1) {
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return true;
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}
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if (word.charAt(0) != word.charAt(word.length()-1)){
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return false;
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}
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return palindrome(word.substring(1, word.length()-1));
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}
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public static void main(String[] args) {
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//System.out.println(convertIntoPalindrome("poaop"));
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System.out.println(convertIntoPalindrome("madam"));
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}
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}

‎Knapsack.java‎

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package com.program.dynamic.programming;
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public class Knapsack {
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private static int table[][];
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private static int n;
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/**
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* Time : O(N * C )
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* Space : O (N * C)
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* @param capacity
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* @param w
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* @param v
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* @return
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*/
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public static int knapsackSol(int capacity, int [] w, int v []) {
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table = new int[v.length+1][capacity+1];
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n = w.length;
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return knapsackDP(0, capacity, w, v);
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}
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static int max(int a, int b) {
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return a > b ? a : b;
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}
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/**
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* Recursive Solution.
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*
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* Time Complexity : O(2^n).
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* @param w
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* @param wt
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* @param val
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* @param n
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* @return
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*/
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public static int knapsack(int w, int wt[], int val[], int n) {
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if (w == 0 || n == 0) {
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return 0;
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}
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if (wt[n-1] > w) {
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return knapSack(w, wt, val, n-1);
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} else {
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return max(val[n-1] + knapSack(w - wt[n-1], wt, val, n-1),
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knapsack(w, wt, val, n-1));
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}
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}
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static int knapsackDP(int i, int w, int [] wt, int []v){
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if (i >= n) {
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table[i][w] = 0;
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return table[i][w];
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}
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if (wt[i] > w)
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return table[i+1][w] = knapsackDP(i + 1, w, wt, v);
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else
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return table[i+1][w] = max(knapsackDP(i + 1, w, wt, v), knapsackDP(i + 1, w - wt[i], wt, v) + v[i]);
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}
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// Returns the maximum value that can be put in a knapsack of capacity W
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static int knapSack(int W, int wt[], int val[], int n) {
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int i, w;
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int table[][] = new int[n+1][W+1];
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// Build table K[][] in bottom up manner
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for (i = 0; i <= n; i++) {
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for (w = 0; w <= W; w++)
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{
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if (i==0 || w==0)
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table[i][w] = 0;
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else if (wt[i-1] <= w)
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table[i][w] = max(val[i-1] + table[i-1][w-wt[i-1]], table[i-1][w]);
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else
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table[i][w] = table[i-1][w];
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}
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}
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return table[n][W];
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}
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public static void main(String[] args) {
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int weight[] = new int[7];
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weight[0] = 9;
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weight[1] = 6;
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weight[2] = 1;
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weight[3] = 2;
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weight[4] = 5;
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weight[5] = 4;
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int value[] = new int[7];
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value[0] = 11;
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value[1] = 10;
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value[2] = 5;
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value[3] = 7;
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value[4] = 12;
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value[5] = 8;
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/*
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* Set all values of 2D matrix CostTable to Minus 1
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*/
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int capacity = 10;
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System.out.println(knapsackSol(capacity, weight, value));
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System.out.println(knapSack(capacity, weight, value, weight.length));
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System.out.println( knapsack(capacity, weight, value, weight.length-1));
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}
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}

‎LongestCommonSubsequence.java‎

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package com.program.dynamic.programming;
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/*
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* We have two strings X = {"ABCBDAB"} and Y = {"BDCABA"}
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* Output must be : LCS(X, Y) = {"BCBA", "BDAB", "BCAB"}
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*
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* Solution :
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* i = X.length() and j = Y.lenght()
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* LCS(i,j),
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* if X[i-1] == Y[j-1], then LCS(X[i-1], Y[j-1])
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* if X[i-1] != Y[j-1], then Max(LCS(X[i-1], Y[j]), LCS(X[i], Y[j-1]))
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*
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* @author rrohit
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* */
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public class LongestCommonSubsequence {
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public static int LCSLength(String X, String Y){
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return LCSLength(X.toCharArray(), X.length(), Y.toCharArray(), Y.length());
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}
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/*
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* This is only Recursive Solution, very time consuming O(2^n). Now we need to add memorization
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* technique convert this into Dynamic Programming. And reduce the time complexity to polynomial instead of
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* exponential.
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*/
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private static int LCSLength(char[] X, int i, char[] Y, int j) {
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if (i==0 || j==0){
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return 0;
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} else if (X[i-1] == Y[j-1]) {
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return 1 + LCSLength(X, i-1, Y, j-1);
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} else {
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return Math.max(LCSLength(X, i-1, Y, j), LCSLength(X, i, Y, j-1));
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}
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}
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/* Dynamic Programming Solution
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* https://www.youtube.com/watch?v=wJ-rP9hJXO0
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*
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* Using memorization, LCS[][] = new int[m][n]; where m = X.lenth() and n = Y.length()
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* LCS[0][j] = 0, for all j, coz we are not considering X only taking Y
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* LCS[i][0] = 0, for all i, coz we are not considering Y only taking X
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* LCS[i][j] = 1 + LCS(i-1, j-1) if X[i-1] == Y[j-1]
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* LCS[i][j] = max(LCS(i-1,j), LCS(i, j-1)) if X[i-1] != Y[j-1]
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*
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* Bottom Up Construction
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* Time : O(m*n)
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* Space : O(m*n)
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*/
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public static int LCSLengthDP(char[] X, char[] Y) {
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int m = X.length;
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int n = Y.length;
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int [][] LCS = new int[m+1][n+1];
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int i,j;
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for (i=0; i<=m; i++){
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LCS[i][0] = 0;
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}
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for (j=0; j<=n; j++){
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LCS[0][j] = 0;
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}
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for (i=1; i<=m; i++){
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for (j=1; j<=n; j++){
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if (X[i-1]==Y[j-1]) {
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LCS[i][j] = 1 + LCS[i-1][j-1];
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}else{
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LCS[i][j] = Math.max(LCS[i-1][j], LCS[i][j-1]);
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}
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}
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}
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return LCS[m][n];
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}
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public static void main(String[] args) {
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String X = "ABCBDAB";
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String Y = "BDCABA";
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System.out.println(LCSLength(X, Y));
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System.out.println(LCSLengthDP(X.toCharArray(), Y.toCharArray()));
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}
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}

‎Pair.java‎

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package com.program.dynamic.programming;
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/**
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* Created by rrohit on 13/10/16.
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*/
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public class Pair {
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public static int pairCount(int arr[], int target) {
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int pairCount = 0;
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if (arr.length == 0 || arr.length == 1) {
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return pairCount;
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}
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int start = 0, end = arr.length-1;
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int currentSum = 0;
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while (start < end) {
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currentSum = arr[start] + arr[end];
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if (currentSum >= target) {
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pairCount += end - start;
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end--;
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} else {
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start++;
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}
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}
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return pairCount;
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}
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public static void main(String args[]) {
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int arr[] = {2, 3, 4, 10, 12, 30, 45};
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System.out.println(pairCount(arr, 40));
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}
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}

‎PascalTriangle.java‎

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package com.program.dynamic.programming;
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/**
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* Created by rrohit on 11/10/16.
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*/
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public class PascalTriangle {
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public static void main(String[] args) {
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PascalTriangle main = new PascalTriangle();
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main.printPascalTriangle(7);
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}
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public int pascal(int i, int j) {
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if (j == 0 || i == j) {
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return 1;
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}
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return pascal(i - 1, j - 1) + pascal(i - 1, j);
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}
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/**
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*
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* @param level
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*/
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public void printPascalTriangle(int level) {
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for (int i = 0; i < level; i++) {
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for (int j = 0; j <= i; j++) {
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System.out.print("\t"+pascal(i, j)+" ");
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}
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System.out.println();
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}
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}
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}

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