Added code and updated readme

C++/0055.cpp

@@ -0,0 +1,26 @@
+class Solution {
+public:
+ bool canJump(vector<int>& nums) {
+ int n=nums.size();
+ if(n==0) return true;
+ if(n==1) return true;
+ if(nums[0]==0) return false;
+
+ int maxreach=nums[0],step=nums[0],jump=1;
+ for(int i=1;i<n-1;i++)
+ {
+ maxreach=max(maxreach,i+nums[i]);
+ step--;
+ if(step<0) return false;
+ if(step==0)
+ {
+ jump++;
+ step=maxreach-i;
+ }
+ }
+ if(maxreach>=n-1)
+ return true;
+ return false;
+
+ }
+};

README.md

@@ -40,6 +40,7 @@ Or from the Sheet 1 of this [Google Sheet](https://bit.ly/2EUhwnw)
 | 0983 | [Minimum Cost For Tickets](https://leetcode.com/problems/minimum-cost-for-tickets/) | Here dp(i) is the cost to travel from day days[i] to the end of the plan. if days[j] < days[i] + 1 then j1=j. if days[j] < days[i] + 7 then j=j7. if days[j] < days[i] + 30 then j=j30 . dp(i)=min(dp(j1)+costs[0],dp(j7)+costs[1],dp(j30)+costs[2]) | [Java](https://github.com/sankalpdayal5/LeetCode-Solutions/blob/master/Java/0983.java) | O(n) | O(n) | Medium | Dynamic Programming | [📺](https://www.youtube.com/watch?v=2AnrAlCA578) |
 | 1347 | [Minimum Number of Steps to Make Two Strings Anagram](https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram/) | Add 1 for char in s and remove 1 for char in t | [Java](https://github.com/sankalpdayal5/LeetCode-Solutions/blob/master/Java/1347.java) [Python](https://github.com/sankalpdayal5/LeetCode-Solutions/blob/master/Python/1347.py) | O(n+m) | O(1) | Medium | Hash Table Heap | [📺](https://www.youtube.com/watch?v=xXXOpOYWtRE) |
 | 1352 | [Product of the Last K Numbers](https://leetcode.com/problems/product-of-the-last-k-numbers/) | Add new element to list by multiplying it with previous number and return arr[n-1]/arr[n-k-1] | [Python](https://github.com/sankalpdayal5/LeetCode-Solutions/blob/master/Python/1352.py) | O(1) | O(1) | Medium | Array Design | [📺](https://www.youtube.com/watch?v=8CuVduv0Kyg) |
+| 0055 | [Jump game](https://leetcode.com/problems/jump-game/) | Run a loop from 0 to n-2 then find maximum reach and update variable step and check it accordingly | [C++](https://github.com/beertocode/LeetCode-Solutions-1/blob/mybranch2/C%2B%2B/0055.cpp) | O(n) | O(1) | Medium | Dynamic Programming | [📺](https://www.youtube.com/watch?v=ymET7SJsDQc) |
 
 
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