Dynamic programming (#944)

* Dynamic programming

* Changes

* removed executable files

Author: Nabeelcodes110

Java/Algorithm/DynamicProgramming/ClimbStairs/ClimbStairs1.java

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+import java.util.Scanner;
+
+public class ClimbStairs1 {
+
+    public static int solution(int n) {
+        int[] dp = new int[n + 1];
+        // in dp array dp[i] represents no of ways to reach ith stair
+        // we assume to reach 0th steps we need 1 steps
+        dp[0] = 1;
+        for (int i = 1; i <= n; i++) {
+            if (i == 1) {
+                // handling edge case
+                dp[i] = dp[i - 1];
+                continue;
+            }
+            // we can reach the ith step from (i-2)th step by taking 2steps or from (i-1)th
+            // step by taking 1 step
+            dp[i] = dp[i - 1] + dp[i - 2];
+        }
+        return dp[n];
+
+    }
+
+    public static void main(String[] args) {
+        Scanner sc = new Scanner(System.in);
+        int n = sc.nextInt();
+        int ans = solution(n);
+        System.out.println(ans);
+        sc.close();
+
+    }
+
+}

Java/Algorithm/DynamicProgramming/ClimbStairs/ClimbStairs2.java

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+//this problem is very similar to climbStairs1
+//But in this problem instead of 1 or 2 steps we can Take variable steps
+//we given an array jumps where jump[i] represents from ith step we can go to i+jumps[i] ith steps;
+//we have to start from 0th step and reach the nth step using jumps
+
+public class ClimbStairs2 {
+    public static int getNoOfWays(int n, int[] jumps) {
+        int[] dp = new int[n + 1];
+        // dp[i] represents no of ways to reach nth step from ith steps
+        // we assume we can reach nth step in one way from nth step itself
+        dp[n] = 1;
+        for (int i = n - 1; i >= 0; i--) {
+            for (int j = 1; j <= jumps[i] && i + j < dp.length; j++) {
+                dp[i] += dp[i + j];
+            }
+        }
+        return dp[0];
+    }
+
+    public static void main(String[] args) {
+        int n = 6; // number of steps;
+        int[] jumps = { 2, 3, 0, 1, 2, 3 };
+        int ans = getNoOfWays(n, jumps);
+        System.out.println(ans);
+
+    }
+
+}

Java/Algorithm/DynamicProgramming/ClimbStairs/ClimbStairs3.java

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+//in ClimbStairs3 we have to find minimum no of jumps required to reach nth level
+
+import java.util.*;
+
+public class ClimbStairs3 {
+    public static int getNoOfWays(int n, int[] jumps) {
+        int[] dp = new int[n + 1];
+        Arrays.fill(dp, Integer.MAX_VALUE);
+        // dp[i] represents no of ways to reach nth step from ith steps
+        // we know that no of jumps required to reach nth step from itself is 0
+        dp[n] = 0;
+        for (int i = n - 1; i >= 0; i--) {
+            for (int j = 1; j <= jumps[i] && i + j < dp.length; j++) {
+                // find the least number of jumps requried to reach nth step
+                dp[i] = Math.min(dp[i], dp[i + j]);
+            }
+            if (dp[i] != Integer.MAX_VALUE)
+                dp[i]++; // taking a jump from ith step
+        }
+        return dp[0];
+    }
+
+    public static void main(String[] args) {
+        int n = 6; // number of steps;
+        int[] jumps = { 2, 3, 0, 1, 2, 3 };
+        int ans = getNoOfWays(n, jumps);
+        if (ans == Integer.MAX_VALUE) {
+            System.out.println("Cannot Reach");
+        } else
+            System.out.println(ans);
+
+    }
+
+}

Java/Algorithm/DynamicProgramming/CoinChange/CoinChange1.java

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+//this is dynamic programming approach for coin change problem
+//we  are given an array coins and int target
+// we have tp return minimum number of coins which can make up target equal to target
+public class CoinChange1 {
+    public static int getMinCoins(int[] coins, int target) {
+        int[] dp = new int[target + 1];
+        // dp[i] represents minimum number of coins required to make up target = i;
+
+        for (int i = 1; i < dp.length; i++) {
+            int min = Integer.MAX_VALUE;
+            for (int j = 0; j < coins.length; j++) {
+                if (coins[j] > i) // if coin's value is more than required target then we ignore it
+                    continue;
+                min = Math.min(min, dp[i - coins[j]]); // finding minimum number of coins to make up amount = i
+            }
+            if (min == Integer.MAX_VALUE || min == 100000) // if we cannnot make amount = i then we mark it as invalid
+                dp[i] = 100000;
+            else
+                dp[i] = min + 1; // else we add one coin to the min no of coins needed
+        }
+
+        return dp[target] == 100000 ? -1 : dp[target];
+    }
+
+    public static void main(String[] args) {
+        int[] coins = { 1, 2, 5 };
+        int target = 11;
+        int ans = getMinCoins(coins, target);
+        System.out.println(ans);
+    }
+}

Java/Algorithm/DynamicProgramming/CoinChange/CoinChange2.java

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+//In this problem we have to find out no of ways we can make up amount equals to the target
+
+public class CoinChange2 {
+    public static int getWays(int[] coins, int target) {
+        int[] dp = new int[target + 1]; // dp[i] represents number of ways to make up amount equals to i
+        // no of ways to make up amount equals to 0 is 1
+        dp[0] = 1;
+        for (int i = 0; i < coins.length; i++) {
+            for (int j = coins[i]; j < dp.length; j++) {
+                dp[j] += dp[j - coins[i]];
+
+            }
+        }
+        return dp[target];
+    }
+
+    public static void main(String[] args) {
+        int[] coins = { 1, 2, 5 };
+        int target = 7;
+        int ans = getWays(coins, target);
+        System.out.println(ans);
+    }
+}

Java/Algorithm/DynamicProgramming/CombinationSum/CombinationSum1.java

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+// we are given an array of number nums and an Integer target
+// we have to return if we can choose any subset of nums whose sum is equal to target
+
+public class CombinationSum1 {
+    public static boolean solve(int target, int[] nums) {
+        boolean[][] dp = new boolean[nums.length + 1][target + 1];
+        // dp[i][j] represents , is it possible to select subset from nums till index i
+        // to make sum equal to j;
+        // first row represents the empty subset;
+        // we can make sum equal to 0 with empty subset;
+        for (int i = 0; i < nums.length; i++)
+            dp[i][0] = true;
+
+        for (int i = 1; i <= nums.length; i++) {
+            for (int j = 1; j <= target; j++) {
+                if (nums[i - 1] <= j) {
+                    dp[i][j] = dp[i - 1][j] | dp[i - 1][j - nums[i - 1]]; // consideing both inclusive and exclusive
+                } else {
+                    dp[i][j] = dp[i - 1][j];
+                }
+            }
+        }
+        return dp[nums.length][target];
+
+    }
+
+    public static void main(String[] args) {
+        int[] nums = { 4, 2, 7, 1, 3 };
+        int target = 10;
+        boolean ans = solve(target, nums);
+        System.out.println(ans);
+
+    }
+}

Java/Algorithm/DynamicProgramming/CombinationSum/CombinationSum2.java

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+//In this problem repetition of numbers are allowed
+
+public class CombinationSum2 {
+    public static boolean solve(int[] nums, int target) {
+        boolean[] dp = new boolean[target + 1];
+        dp[0] = true; // we can make sum equal to 0 with an empty subset
+        for (int i = 1; i < dp.length; i++) {
+            for (int j = 0; j < nums.length; j++) {
+                if (nums[j] <= i) {
+                    dp[i] |= dp[i - nums[j]];
+                }
+            }
+        }
+
+        return dp[target];
+    }
+
+    public static void main(String[] args) {
+        int[] nums = { 4, 2, 7, 9, 3 };
+        int target = 10;
+        boolean ans = solve(nums, target);
+        System.out.println(ans);
+    }
+
+}

Java/Algorithm/DynamicProgramming/HouseRobber/HouseRobber1.java

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+// We are a Professional Robber and we have to rob maximum loot from n houses
+// along a street
+// But we cannnot steal from two adjacent houses otherwise we would get caught
+// we are given loot array where loot[i] represents ith house contains loot[i] money
+
+public class HouseRobber1 {
+    public static int getMaxLoot(int n, int[] loot) {
+        int[] dp = new int[n + 1];
+        // dp represents dp[i] represents max loot we can achieve till ith house;
+        dp[0] = 0; // loot from 0 houses is 0
+        dp[1] = loot[0]; // loot from 1st house
+        for (int i = 2; i <= n; i++) {
+            // we can either loot previous house or current house
+            // we are maximizing out loot at every house
+            dp[i] = Math.max(dp[i - 1], dp[i - 2] + loot[i - 1]);
+
+        }
+        return dp[n];
+    }
+
+    public static void main(String[] args) {
+        int n = 5; /// no of houses
+        int[] loot = { 2, 7, 9, 3, 1 }; // loots in the houses
+        int ans = getMaxLoot(n, loot);
+        System.out.println(ans);
+
+    }
+}

Java/Algorithm/DynamicProgramming/HouseRobber/HouseRobber2.java

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+//This problem is very similar to HouseRobber1
+//But in this problem house are arranged in circular manner
+//that means 0th house is adjacent to n-1th house
+public class HouseRobber2 {
+    public static int getMaxLoot(int n, int[] loot) {
+        // we will use two dp arrays inorder to solve this problem
+        // suppose we have 10 houses 1 to 10
+        // if we steal from house 1 then we cannot steal from house 10 , and for houses
+        // 2 to 9 will be same as HouseRobber1
+        // if we leave house1 then we can steal from house 10 and for houses 2 to 9 will
+        // be same as HouseRobber1
+        int[] dp1 = new int[n + 1];
+        int[] dp2 = new int[n + 1];
+
+        // stealing from house1
+        dp1[0] = 0;
+        dp1[1] = loot[0];
+        // leaving the first hosue
+        dp2[2] = 0;
+        dp2[2] = 0;
+        for (int i = 2; i <= n; i++) {
+            dp1[i] = Math.max(dp1[i - 1], dp1[i - 2] + loot[i - 1]);
+            dp2[i] = Math.max(dp2[i - 1], dp2[i - 2] + loot[i - 1]);
+        }
+
+        return Math.max(dp1[n - 1], dp2[n]);
+    }
+
+    public static void main(String[] args) {
+        int n = 5; /// no of houses
+        int[] loot = { 2, 7, 9, 3, 1 }; // loots in the houses
+        int ans = getMaxLoot(n, loot);
+        System.out.println(ans);
+    }
+}

Java/Algorithm/DynamicProgramming/KnapSack/BoundedKnapSack.java

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+public class BoundedKnapSack {
+    public static int solve(int W, int N, int[] val, int[] wt) {
+        int[][] dp = new int[N + 1][W + 1];
+        // dp[i][j] represents max value achieved with sack of capacity j and i-1 items;
+        for (int i = 1; i <= N; i++) {
+            for (int j = 1; j <= W; j++) {
+                if (wt[i - 1] <= j) {
+                    dp[i][j] = Math.max(dp[i - 1][j], val[i - 1] + dp[i - 1][j - wt[i - 1]]);
+                } else {
+                    dp[i][j] = dp[i - 1][j];
+                }
+
+            }
+        }
+
+        return dp[N][W];
+
+    }
+
+    public static void main(String[] args) {
+        int W = 4; // capacity of sack
+        int N = 3; // No of items
+        int[] val = { 1, 2, 3 }; // values of items;
+        int[] wt = { 3, 5, 1 }; // weights of items
+
+        int ans = solve(W, N, val, wt);
+        System.out.println(ans);
+    }
+}

Java/Algorithm/DynamicProgramming/KnapSack/UnboundedKS.java

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+//this is similar to KnapSack but Here we can repeat the items
+//this problem is quite similar to coinChange2
+
+public class UnboundedKS {
+    public static int solve(int W, int N, int[] val, int[] wt) {
+        int[] dp = new int[W + 1];
+        // dp[i] represents max Value achieved with sack of capacity of i
+        for (int i = 0; i <= W; i++) {
+            for (int j = 0; j < N; j++) {
+                if (wt[j] <= i) {
+                    dp[i] = Math.max(dp[i], val[j] + dp[i - wt[j]]);
+                }
+            }
+
+        }
+
+        return dp[W];
+    }
+
+    public static void main(String[] args) {
+        int W = 4; // capacity of sack
+        int N = 3; // No of items
+        int[] val = { 1, 2, 3 }; // values of items;
+        int[] wt = { 3, 5, 1 }; // weights of items
+        int ans = solve(W, N, val, wt);
+        System.out.println(ans);
+    }
+}