Coin change Geeks For Geeks

-

 class Solution

{

    public long count(int arr[], int m, int n) 

    { 

       long t[][]=new long[m+1][n+1];

       for(int i=0;i<m+1;i++){

           for(int j=0;j<n+1;j++){

               if(i==0){

                   t[i][j]=0;

               }

               if(j==0){

                   t[i][j]=1;

               }

           }

           

       }

       for(int i=1;i<m+1;i++){

           for(int j=1;j<n+1;j++){

               if(arr[i-1]>j){

                   t[i][j]=t[i-1][j];

               }

               else{

                   t[i][j]=(t[i-1][j] + t[i][j-arr[i-1]]);

               }

           }

           

       }

       return t[m][n];

       

    } 

}


Approach: Dynamic Programming

Application of Subset Sum problem with just one modification of traversing one element several times(as it is unbounded knapsack).

Time: O(Array Size* Sum)[GFG Time: 0.6/1.5]

Space:O(Sum)


Thanks For Reading.😇

Comments

Post a Comment

Popular posts from this blog

Solutions Of Practice Questions Dated 01-06-2022

-
Question 1: Transmit the Message: Solution: import java.util.*; class Solution{     public static void main(String arg[]){         Scanner sc=new Scanner(System.in);         String s=sc.next();                          ArrayList<Integer>aa=new ArrayList<Integer>();         ArrayList<Character>ab=new ArrayList<Character>();                  char flag=s.charAt(0);        int index=0;        int count=1;         for(int i=1;i<s.length();i++){             if(flag==s.charAt(i)){                 count++;             }             else{                 aa.add(index,count);                 ab.add(index,flag);                 count=1;                 flag=s.charAt(i);                 index++;             }         }         aa.add(index,count);         ab.add(index,flag);                  int max=Collections.max(aa);                  for(int i=aa.size()-1;i>0;i--){             if(aa.get(i)==max){                 flag=ab.get(i);                 break;             }
Read more

CODEFORCES SPY DETECTED ROUND 713

-
 import java.util.*; public class CodeForcesSpyDetected{ public static void main(String arg[]){ Scanner sc=new Scanner(System.in); int t=sc.nextInt(); ArrayList <Integer> aa=new ArrayList<Integer>();  while(t -->0){ int n=sc.nextInt(); int arr2[]=new int[n]; for(int i=0;i<n;i++){ arr2[i]=sc.nextInt(); } for(int i=0;i<n;i++){ aa.add(arr2[i]); } for(int i=0;i<aa.size();i++){ //System.out.println(Collections.frequency(aa,aa.get(i))); if(Collections.frequency(aa,aa.get(i))==1){ System.out.println((i+1)); aa.clear(); break; } } } } }
Read more

Maximum Winning Score Geeks For Geeks

-
Problem Link: https://practice.geeksforgeeks.org/problems/4ead9c3991a3822f578309e2232bc5415ac35cb9/1# It is also the problem of the day for 5 march. Solution: class Solution {     public static Long findMaxScore(Node root)     {         Node temp=root;         long sum=0;         while(temp!=null){             if(temp.right==null && temp.left==null){                 return (long) temp.data;             }             else{                 return (long) temp.data*Long.max(findMaxScore(temp.left),findMaxScore(temp.right));             }         }         return sum;     } } Time Complexity:[GFG Time:0.7/2.9] Space Complexity: O(N)[Number of nodes given in tree] Auxiliary Space: O(1) Approach Used: Here we used an approach that we will traverse recursively. And when we travel recursive we follow the bottom to the top approach of the tree.  It means we will see the leaf node and then move up to the tree.  For recursion, we should have a base condition and that is when node. right or
Read more