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Saturday, February 16, 2019

Question Name:PRIYANKA AND TOYS

import java.io.*;
import java.util.*;
public class TestClass {
  public static void main(String[] args) { 
        Scanner sc = new Scanner(System.in);
        int arraySize = sc.nextInt();
        int[]numArray = new int[arraySize];
        for(int i = 0;i<arraySize;i++){
            numArray[i] = sc.nextInt();
        }
       
        Arrays.sort(numArray);
        
        int count = 1;
        int currentToy=numArray[0];
        for(int i = 1;i<numArray.length;i++){
            if(numArray[i]>currentToy+4){
                count++;
                currentToy=numArray[i];
            }
        }
        System.out.println(count);
 }
}
  • Problem Description
    Little Priyanka visited a kids’ shop. There are N toys and their weight is represented by an array W=[w1, w2,…,wN]. Each toy costs 1 unit, and if she buys a toy with weight w’, then she can get all other toys whose weight lies between [w’,w’+4] (both inclusive) free of cost. 

    Input Format:
    The first line contains an integer N i.e. number of toys.
    Next line will contain N integers,w1,w2,…,wN , representing the weight array.

    Output Format:
    Minimum units with which Priyanka could buy all of toys.

    Constraints
    1 <= N <= 10^5
    0 <= wi <= 10^4, where i belongs to [1,N]
  • Test Case 1
    Input (stdin)5
    1 2 3 17 10
    Expected Output3
  • Test Case 2
    Input (stdin)4
    1 2 3 17
    Expected Output2

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  3. #include
    #include
    #include
    #include

    int main()
    {
    int min,n,i,x,z,j,k,a[100000]={0};
    scanf("%d",&n);
    for(i=0;ia[j] && a[j]!=-1)
    {
    min = a[j];
    }
    }
    x = min;
    //printf("%d\n",min);
    z++;
    for(j=0;j= x && a[j] <= x+4)
    {
    a[j] = -1;
    i = -1;
    }
    }
    }
    printf("%d",z);
    return 0;
    }

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  4. #include
    #include
    using namespace std;

    int w[100000];

    int main()
    {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++)
    cin >> w[i];
    sort(w, w + n);
    int ans = 1, prev = w[0];
    for (int i = 1; i < n; i++)
    if (w[i] - prev > 4)
    prev = w[i], ans++;
    cout << ans << endl;
    return 0;
    }

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Question Name:TOWER OF HANOI

#include < bits / stdc ++. h > #define lli long long using namespace std ; lli dp [ 202 ]; int main () { int t , n ; ...