Application Based Problems
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# Last Stone Weight

Difficulty: EASY
Avg. time to solve
27 min

Problem Statement
Suggest Edit

#### On each turn, we choose the two heaviest stones and smash them together. Suppose the stones have weights 'x' and 'y' with 'x' <= 'y'. The result of this smash will be:

``````1. If 'x' == 'y', both stones are totally destroyed;

2. If 'x' != 'y', the stone of weight 'x' is totally destroyed, and the stone of weight 'y' has a new weight equal to 'y - x'.
``````

#### In the end, there is at most 1 stone left. Return the weight of this stone or 0 if there are no stones left.

##### Input format:
``````The first line of input contains the integer 'N', representing the total number of stones.

The second line of input contains 'N' single space-separated integers, representing the weights of the stones.
``````
##### Output Format:
``````The only output line prints the weight of the last stone, if it exists, 0 otherwise.
``````
##### Note:
``````You do not need to print anything; it has already been taken care of. Just implement the given functions.
``````
##### Constraints :
``````1 <= N <= 10^5
1 <= W <= 10^6

Time Limit: 1 sec
``````
##### Sample Input 1:
``````1
10
``````
##### Sample Output 1:
``````10
``````
##### Explanation For Sample Input 1:
``````There is Only one stone so the weight of the last stone is 10
``````
##### Sample Input 2:
``````3
1 9 5
``````
##### Sample Output 2:
``````3
``````   Console