added greedy problems

Greedy/ActivitySelection/ActivitySelection.java

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+/*
+Given N activities with their start and finish times. 
+Select the maximum number of activities that can be performed by a single person, 
+assuming that a person can only work on a single activity at a time.
+
+Note : The start time and end time of two activities may coincide.
+
+Input:
+The first line contains T denoting the number of testcases. 
+Then follows description of testcases. First line is N number of activities 
+then second line contains N numbers which are starting time of activies.
+Third line contains N finishing time of activities.
+ 
+Output:
+For each test case, output a single number denoting maximum activites which can be performed in new line.
+ 
+
+Constraints:
+1<=T<=50
+1<=N<=1000
+1<=A[i]<=100
+ 
+
+Example:
+Input:
+1
+6
+1 3 0 5 8 5
+2 4 6 7 9 9
+
+Output:
+4
+*/
+import java.util.*;
+import java.lang.*;
+import java.io.*;
+class Activity implements Comparable<Activity>{
+    int startTime;
+    int endTime;
+    Activity(int start,int end){
+        this.startTime=start;
+        this.endTime=end;
+    }
+    public int compareTo(Activity a){  
+          return this.endTime-a.endTime;
+    }  
+}
+class GFG
+{
+	public static void main (String[] args)
+	{
+	    //code
+	    Scanner sc=new Scanner(System.in);
+	    int t=sc.nextInt();
+	    for(int i=0;i<t;i++){
+	        ArrayList<Activity> al=new ArrayList<Activity>();
+	        int n=sc.nextInt(),temp;
+	        int[] startAr=new int[n];
+	        for(int j=0;j<n;j++){
+	            startAr[j]=sc.nextInt();
+	        }
+	        for(int j=0;j<n;j++){
+	            temp=sc.nextInt();
+	            al.add(new Activity(startAr[j],temp));
+	        }
+	        Collections.sort(al);
+	        ActivityProblem(al,n);
+	    }
+	}
+	private static void ActivityProblem(ArrayList<Activity> al,int n){
+	    Activity prev=al.get(0),curr;
+	    int count=1;
+	    for(int i=1;i<n;i++){
+	       curr=al.get(i);
+	       if(prev.endTime<=curr.startTime){
+	           count++;
+	           prev=curr;
+	       }
+	    }
+	    System.out.println(count);
+	}
+}

Greedy/MinimumCoins/MinimumNumOfCoins.java

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+/*
+Minimum number of Coins
+----------------------------------------------------------------------------------
+Given a value N, if we want to make change for N Rs,
+and we have infinite supply of each of the denominations in Indian currency,
+i.e., we have infinite supply of { 1, 2, 5, 10, 20, 50, 100, 500, 1000} valued coins/notes,
+Find the minimum number of coins and/or notes needed to make the change.
+
+Input:
+
+The first line of input contains an integer T denoting the number of test cases.
+Each test case consist of an Integer value N denoting the amount to get change for.
+
+
+Output:
+
+Print all the possible denominations needed to make the change in a separate line.
+
+
+Constraints:
+
+1 ≤ T ≤ 30
+1 ≤ N ≤ 2000
+
+
+Example:
+
+Input
+1
+43
+Output
+20 20 2 1
+*/
+
+import java.util.*;
+import java.lang.*;
+import java.io.*;
+
+class GFG {
+	public static void main (String[] args) {
+		//code
+		Scanner sc=new Scanner(System.in);
+		int t=sc.nextInt();
+		int[] array={1, 2, 5, 10, 20, 50, 100, 500, 1000};
+		for(int i=0;i<t;i++){
+		    int n=sc.nextInt();
+		    MinimumNumberOfCoins(array,n);
+		    System.out.println();
+		}
+	}
+	private static void MinimumNumberOfCoins(int[] array,int n){
+	    int length=array.length;
+	    for(int i=length-1;i>=0;i--){
+	        if(array[i]<=n){
+	            int numOfCoins=n/array[i];
+	            for(int j=0;j<numOfCoins;j++)System.out.print(array[i]+" ");
+	            n%=array[i];
+	        }
+	    }
+	}
+}