Vpropel VIT | POD | 08/05/2020 | Point to Point | Similar to Frog Jump 404 LeetCode

point_to_point.c

#include<stdio.h>
int n,m=0,arr[80];
void check(int i,int j)
{
    if(i<n && m!=1)
    {
        int d,e,f;
        d=arr[i]-arr[i-1];
        if(d==j || d==j-1 || d==j+1)
        {
            if(i==n-1)
                {
                    m=1;
                    printf("Yes");
                    return ;
                }
            check(i+1,d);
        }
        e=arr[i+1]-arr[i-1];
        if(e==j || e==j-1 || e== j+1)
        {
            if(i==n-1)
                {
                    m=1;
                    printf("Yes");
                    return ;
                }
            check(i+2,e);
        }
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d",&arr[i]);
    check(2,1);
    if(m==0)
        printf("No");
    return 0;
}
         
//Dpk IS-19 CI0003

point_to_point.cpp

#include<iostream>
int n,m=0,arr[80];
void check(int i,int j)
{
    if(i<n && m!=1)
    {
        int d,e,f;
        d=arr[i]-arr[i-1];
        if(d==j || d==j-1 || d==j+1)
        {
            if(i==n-1)
                {
                    m=1;
                    std::cout<<"Yes";
                    exit(0);
                }
            check(i+1,d);
        }
        e=arr[i+1]-arr[i-1];
        if(e==j || e==j-1 || e== j+1)
        {
            if(i==n-1)
                {
                    m=1;
                    std::cout<<"Yes";
                    exit(0);
                }
            check(i+2,e);
        }
    }
}
int main()
{
    std::cin>>n;
    for(int i=0;i<n;i++)
        std::cin>>arr[i];
    check(2,1);
    if(m==0)
        std::cout<<"No";
    return 0;
}
//Dpk IS-19 CI0003

point_to_point.java

//wait

point_to_point.py

#wait

Question(point_to_point).txt

Point to Point
Shyam is standing in Point A. He has to reach Point B. There are m number of divisions 

between Point A and point B. 

These division may or may not have a stone. One can jump only on the stone.

Given a list of stones positions (in division) in sorted ascending order, 

determine whether Shyam can reach Point B.

Initially Shyam is at the first stone and assume the first jump is one division

If Shyam’s last jump is k divisions the he can either     k-1 or k or k+1 divisions in the next jump.

He can jump only in forward direction.

Example1:-

Let The position of stones be [0,1,3,5,6,8,12,17]

There are a total of 8 stones.

The first stone at the 0th division, second stone at the 1st division, third stone at the 3rd division, and so on... 

The last stone at the 17th division.

Output:-Yes

Explanation:-

Shyam can jump to the last stone by jumping  1 division to the 2nd stone, then 2 divisions to the 3rd stone, then  2 divisions to the 4th stone, then 3 divisions to the 6th stone, 4 divisions to the 7th stone, and 5 divisions to the 8th stone.

Example2:-

Let the positions of stones be [0,1,2,3,4,8,9,11]

Output:-No

Explanation:-

There is no way to jump to the last stone as 

the gap between the 5th and 6th stone is too large.

Input format:-

Number of stones

Next n lines –positions of n stones

Output format:-

Yes or No

Note:- The first stone is always in the 0th division