Added few more problems.

CoinChange.java

@@ -0,0 +1,59 @@
+/*You are given an amount denoted by value. You are also given an array of coins. The array contains the denominations 
+ * of the give coins. You need to find the minimum number of coins to make the change for value using the coins of given 
+ * denominations. Also, keep in mind that you have infinite supply of the coins.
+ * */
+//Initial Template for Java
+
+import java.io.*;
+import java.util.*;
+public class CoinChange {
+	public static void main (String[] args) {
+		Scanner sc=new Scanner(System.in);
+		int testcases=sc.nextInt();
+		while(testcases-->0)
+		{
+		    int value=sc.nextInt();
+		    int numberOfCoins=sc.nextInt();
+		    int coins[]=new int[numberOfCoins];
+
+		    for(int i=0;i<numberOfCoins;i++)
+		    {
+		        int x=sc.nextInt();
+		        coins[i]=x;
+		    }
+		   Coin obj=new Coin();
+		    long answer=obj.minimumNumberOfCoins(coins,numberOfCoins,value);
+		    System.out.println(answer==-1?"Not Possible":answer);
+		}
+	}
+}
+
+
+ // } Driver Code Ends
+
+
+
+
+
+class Coin
+{
+    //Complete this function
+    public long minimumNumberOfCoins(int coins[],int numberOfCoins,int amount)
+    {
+       int dp[]=new int[amount+1];
+        Arrays.fill(dp,amount+1);
+        dp[0]=0;
+        for(int i=1;i<=amount;i++){
+            for(int j=0;j<coins.length;j++){
+                if(coins[j]<=i)
+                    dp[i]=Math.min(dp[i],dp[i-coins[j]]+1);
+            }
+}
+        return dp[amount]>amount?-1:dp[amount];
+    }
+}
+
+
+// { Driver Code Starts.
+
+  // } Driver Code Ends

CountNumberOfHops.java

@@ -0,0 +1,47 @@
+/*A frog jumps either 1, 2, or 3 steps to go to the top. In how many ways can it reach the top. As the answer will be
+ *  large find the answer modulo 1000000007.
+ */
+//Initial Template for Java
+
+import java.util.*;
+import java.io.*;
+import java.lang.*;
+
+public class CountNumberOfHops
+{
+    public static void main(String args[])
+    {
+        Scanner sc = new Scanner(System.in);
+        int t = sc.nextInt();
+
+        while(t-- > 0)
+        {
+            int n = sc.nextInt();
+            System.out.println(new Hops().countWays(n));
+
+        }
+    }
+}
+// } Driver Code Ends
+
+
+//User function Template for Java
+
+
+class Hops
+{
+    static long countWays(int n)
+    {
+        if(n==0||n==1)
+        return 1;
+        long dp[]=new long[n+1];
+        dp[0]=1;
+        dp[1]=1;
+        dp[2]=2;
+        for(int i=3;i<=n;i++)
+            dp[i]=(dp[i-1]+dp[i-2]+dp[i-3])%1000000007;
+         return dp[n];
+    }
+
+}
+

DistinctOccurrences.java

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+/*Given two strings S and T of length n and m respectively. find count of distinct occurrences of T in S as a sub-sequence. 
+ */
+import java.util.*;
+
+public class DistinctOccurrences
+{
+	public static void main(String args[])
+	{
+		Scanner sc = new Scanner(System.in);
+		int t = sc.nextInt();
+		sc.nextLine();
+		while(t>0)
+		{
+			String line = sc.nextLine();
+			String S = line.split(" ")[0];
+			String T = line.split(" ")[1];
+
+			Solution sol=new Solution();
+			System.out.println(sol.subsequenceCount(S,T));
+			t--;
+		}
+	}
+}// } Driver Code Ends
+
+
+/*You are required to complete this method*/
+class Solution
+{
+    int  subsequenceCount(String s, String t)
+    {
+	int dp[][]=new int[t.length()+1][s.length()+1];
+	for(int j=0;j<=s.length();j++)
+	dp[0][j]=1;
+	for(int i=0;i<t.length();i++){
+	    for(int j=0;j<s.length();j++){
+	        if(t.charAt(i)==s.charAt(j))
+	        dp[i+1][j+1]=(dp[i][j]+dp[i+1][j])%1000000007;
+	        else
+	        dp[i+1][j+1]=(dp[i+1][j])%1000000007;
+	    }
+	}
+	return dp[t.length()][s.length()];
+    }
+}