Equal Beauty CodeChef SnackDown 2021 Round 1A Question The beauty of an (non-empty) array of integers is defined as the difference between its largest and smallest element. For example, the beauty of the array [2,3,4,4,6] is 6−2=4. An array A is said to be good if it is possible to partition the elements of A into two non-empty arrays B1 and B2 such that B1 and B2 have the same beauty. Each element of array A should be in exactly one array: either in B1 or in B2. For example, the array [6,2,4,4,4] is good because its elements can be partitioned into two arrays B1=[6,4,4] and B2=[2,4], where both B1 and B2 have the same beauty (6−4=4−2=2). You are given an array A of length N. In one move you can: Select an index i (1≤i≤N) and either increase Ai by 1 or decrease Ai by 1. Find the minimum number of moves required to make the array A good. Input Format The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follow. Each test
Jump Game LeetCode Solution
Question
You are given an integer array nums
. You are initially positioned at the array's first index, and each element in the array represents your maximum jump length at that position.
Return true
if you can reach the last index, or false
otherwise.
Example 1:
Input: nums = [2,3,1,1,4] Output: true Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:
Input: nums = [3,2,1,0,4] Output: false Explanation: You will always arrive at index 3 no matter what. Its maximum jump length is 0, which makes it impossible to reach the last index.
Constraints:
1 <= nums.length <= 104
0 <= nums[i] <= 105
Explanation
This question is a easy question and can be solved in multiple ways using multiple concepts. This question can be solved with Dynamic Programming, Sliding Window concept, greedy approach and many more. We start moving from the first element and traverse the array. We keep a track of the maximum element after each iteration also we are checking whether the loop variable is less than the maximum element until that position, if the loop variable exceeds the maximum variable then we cannot loop to the last element so we return false. To avoid issues we also check whether the last element has been reached or not since in some cases the last element can be the element which exceeds the maximum so in that case we return true. We can do the same thing if we start from the last element and move to the first element, rest of the conditions would be same. This is the easiest approach to this problem because for beginners dynamic programming may be very complicated so with this method you can understand it easily and can code it easily too. That's it now code it. Hope you understood the explanation. I recommend first to try by yourself then if you are stuck then you can visit the solution given below in Java, C++, Python.
More information about LeetCode Daily Challenge
More Information About LeetCode Daily Challenge
LeetCode is a great coding platform which provides daily problems to solve and maintain a streak of coding. This platform helps in improvement of coding skills and attempting the daily challenge makes our practice of coding regular. With its bunch of questions available it is one of the best coding platforms where you can have self improvement. LeetCode also holds weekly and biweekly contests which is a great place to implement you knowledge and coding skills which you learnt through out the week by performing the daily challenge. It also provides badges if a coder solves all the daily problems of a month. For some lucky winners T-shirts are also given and that too for free. If you get the pro version you can get more from LeetCode like questions that are asked in interviews, detailed explanation of each chapter and each concept and much more. LeetCode challenges participants with a problem from our carefully curated collection of interview problems every 24 hours.
All users with all levels of coding background are welcome to join!
@PREMIUM USERS will have an extra carefully curated premium challenge weekly and earn extra LeetCoin rewards. The premium challenge will be available together with the first October challenge of the week and closed together with the November challenge of the week.
Starting from 2021, users who completed all Daily LeetCoding Challenge non-premium problems for the month (with or without using Time Travel Tickets) will win a badge to recognize their consistency! Your badges will be displayed on your profile page. Completing each non-premium daily challenge. (+10 LeetCoins)
Completing 25 to 30 non-premium daily challenges without using Time Travel Ticket will be eligible for an additional 25 LeetCoins. (Total = 275 to 325 LeetCoins)
Completing all 31 non-premium daily challenges without using Time Travel Ticket will be eligible for an additional 50 LeetCoins, plus a chance to win a secret prize ( Total = 385 LeetCoins + *Lucky Draw)!
Lucky Draw: Those who complete all 31 non-premium daily challenges will be automatically entered into a Lucky Draw, where LeetCode staff will randomly select 3 lucky participants to each receive one LeetCode Polo Shirt on top of their rewards! Do you have experience trying to keep a streak but failed because you missed one of the challenges?
Hoping you can travel back to the missed deadline to complete the challenge but that's not possible......or is it?
With the new Time Travel Ticket , it's possible! In order to time travel, you will need to redeem a Time Travel Ticket with your LeetCoins. You can redeem up to 3 tickets per month and the tickets are only usable through the end of the month in which they are redeemed but you can use the tickets to make up any invalid or late submission of the current year. To join, just start solving the daily problem on the calendar of the problem page. You can also find the daily problem listed on the top of the problem page. No registration is required. The daily problem will be updated on every 12 a.m. Coordinated Universal Time (UTC) and you will have 24 hours to solve that challenge. Thar's all you needed to know about Daily LeetCode Challenge.
Program Code
Java
public boolean canJump(int[] nums) {
if(nums.length==1)
return true;
int reach=0;
for(int i=0;i<nums.length;i++)
{
reach=Math.max(reach,i+nums[i]);
if(reach==i)
return false;
else if(reach>=nums.length-1)
return true;
}
return true;
}
C++
class Solution {
public:
bool canJump(vector<int>& nums)
{
int maxStepsAvailable=0;
for(int i=0;i<nums.size();i++)
{
maxStepsAvailable=max(maxStepsAvailable-1,nums[i]);
if(!maxStepsAvailable&&i!=nums.size()-1)
return false;
}
return true;
}
};
Python
class Solution:
def canJump(self, nums: List[int]) -> bool:
max_reach = nums[0]
for position in range(len(nums)):
if position > max_reach:
return False
current_reach = position + nums[position]
max_reach = max(current_reach, max_reach)
return True
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