Provided S - a set of n integers S = s1, s2, ..., sn. Check whether it is possible to divide S into two parts: s1, s2, ..., si and si+1, si+2, ..., sn (1 <= i < n) Remember that the first part is strictly decreasing while the second is strictly increasing one.Input FormatIn the first line you are given an integer 2<=n<=100 and in the following line n integers -100 <= si <= 100.Constraints2<=n<=100 -100 <= si <= 100Output FormatOne word Yes or No.Sample Input 05 -1 2 -1 1 -1Sample Output 0No
Question
Provided S - a set of n integers S = s1, s2, ..., sn. Check whether it is possible to divide S into two parts: s1, s2, ..., si and si+1, si+2, ..., sn (1 <= i < n) Remember that the first part is strictly decreasing while the second is strictly increasing one.Input FormatIn the first line you are given an integer 2<=n<=100 and in the following line n integers -100 <= si <= 100.Constraints2<=n<=100 -100 <= si <= 100Output FormatOne word Yes or No.Sample Input 05 -1 2 -1 1 -1Sample Output 0No
Solution
The problem is asking to determine if a given set of integers can be divided into two parts where the first part is strictly decreasing and the second part is strictly increasing.
Here are the steps to solve this problem:
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Read the input: The first line of the input gives the number of integers in the set (n). The second line of the input gives the n integers in the set (s1, s2, ..., sn).
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Initialize two pointers: One at the start of the array (i=0) and one at the end of the array (j=n-1).
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Move the first pointer forward (i.e., increment i) as long as the next number in the array is smaller than the current one. This is to find the strictly decreasing part of the array.
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Move the second pointer backward (i.e., decrement j) as long as the next number in the array is larger than the current one. This is to find the strictly increasing part of the array.
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If the two pointers meet at some point (i.e., i == j), then it is possible to divide the array into two parts as described. In this case, print "Yes".
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If the two pointers do not meet (i.e., i < j), then it is not possible to divide the array as described. In this case, print "No".
This algorithm works because it checks both the decreasing and increasing conditions at the same time, and it stops as soon as it finds a violation of either condition. It runs in O(n) time, where n is the number of integers in the set.
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