{"id":1300,"date":"2020-07-01T09:50:45","date_gmt":"2020-07-01T09:50:45","guid":{"rendered":"https:\/\/blog.prepbytes.com\/?p=1300"},"modified":"2022-03-25T12:07:49","modified_gmt":"2022-03-25T12:07:49","slug":"the-last-game","status":"publish","type":"post","link":"https:\/\/prepbytes.com\/blog\/the-last-game\/","title":{"rendered":"The Last Game"},"content":{"rendered":"

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Concepts Used<\/h3>\n
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Sorting<\/p>\n<\/blockquote>\n

Difficulty Level<\/h3>\n
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Easy<\/p>\n<\/blockquote>\n

Problem Statement (Simplified):<\/h3>\n

Print the last number left in the array after deleting the largest number in the array then the smallest number repeatedly.<\/p>\n

<\/strong><\/u><\/a><\/p>\n

Test Case<\/h4>\n
Input:\n1\n3\n2 1 3\n\nOutput:\n2\n\nExplanation:\nTeacher 1 wants to minimize the mark, so we remove the largest mark i.e. 3.\n\nTeacher 2 wants to maximize the mark, so we remove the smallest mark i.e. 1.\n\nNow, there is only 2 left, so we print 2 as our answer.<\/code><\/pre>\n
Solving Approach :<\/h4>\n
Bruteforce Approach<\/strong>:<\/p>\n
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1) We can find the largest and smallest element and remove them from array serial wise until the last element is left in the array.<\/p>\n
2) This would take O(N!<\/code>) in the worst time as we decrease array size by 2 in every iteration, and again iterate on the array. But this is not a very efficient approach, we can use sorting, to solve this problem more efficiently.<\/p>\n<\/blockquote>\n
Efficient Approach<\/strong>:<\/p>\n
Observation:<\/em><\/strong> We can see that size of array goes from 1 to 106<\/sup>, in worse conditions let the size of the array be  10^6<\/code> if we sort this array using any of sorting techniques i.e. Bubble Sort, Insertion Sort, Selection sort<\/em>. They all take O(N2<\/sup>) in worst cases, which results in 1012<\/sup> iterations when the size of the array is 106<\/sup>. So we, can’t choose the above three sorting algorithms. We use Merge Sort<\/em><\/strong> to sort array as it sorts array in linearithmic time complexity<\/em><\/strong> (n * log(n)<\/code>).<\/p>\n
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1) We sort the array, and can see the largest number in the array is at its very end, and the first number is the smallest number of the array.<\/p>\n
2) So if we remove both of them, our array becomes shorter by 2 elements and starts from 1st<\/code> index and ends at (last index - 1)th<\/code> index.<\/p>\n
3) Now two cases rises:<\/p>\n
    \n
  • Case-1: If the array is Odd in size<\/em><\/strong>, and we remove elements from both sides repeatedly, the last element left would be the middle element, so we print the middle element directly.<\/li>\n
  • Case-2: If the array is Odd in size<\/em><\/strong>, and we remove elements from both sides repeatedly, the last element left would be the last smaller element. As we have to remove the largest element first, the last element left would be the last smaller element, which would be (size \/ 2)th<\/sup> element in the array. So we print the (size \/ 2)th<\/code> element.<\/li>\n<\/ul>\n<\/blockquote>\n
    NOTE:<\/em><\/strong> We directly print the element, no iteration required.<\/p>\n
    Example<\/h3>\n
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    \"\"<\/p>\n<\/blockquote>\n
    Solutions<\/h3>\n
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