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Instead of storing the entire Padovan sequence in a list, you can optimize the space complexity by just storing the last three elements of the sequence. This way, you can calculate the next element based on these three elements without storing the entire sequence.

Question

Instead of storing the entire Padovan sequence in a list, you can optimize the space complexity by just storing the last three elements of the sequence. This way, you can calculate the next element based on these three elements without storing the entire sequence.

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Solution

Sure, here's how you can do it:

  1. Initialize three variables, say p0, p1, and p2 to represent the last three elements of the Padovan sequence. You can start with p0 = p1 = 1 and p2 = 0 as the first three elements of the sequence.

  2. To calculate the next element of the sequence, use the formula p_next = p0 + p2. This is based on the recurrence relation of the Padovan sequence.

  3. After calculating the next element, you need to update the last three elements. Set p0 = p1, p1 = p2, and p2 = p_next.

  4. Repeat steps 2 and 3 for as many elements as you want to generate in the sequence.

This way, you only need to store the last three elements at any point, reducing the space complexity from O(n) to O(1).

This problem has been solved

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You are given a sequence [a1,โ€ฆ,an][๐‘Ž1,โ€ฆ,๐‘Ž๐‘›] where each element ai๐‘Ž๐‘– is either 00 or 11. You can apply several (possibly zero) operations to the sequence. In each operation, you select two integers 1โ‰คlโ‰คrโ‰ค|a|1โ‰ค๐‘™โ‰ค๐‘Ÿโ‰ค|๐‘Ž| (where |a||๐‘Ž| is the current length of a๐‘Ž) and replace [al,โ€ฆ,ar][๐‘Ž๐‘™,โ€ฆ,๐‘Ž๐‘Ÿ] with a single element x๐‘ฅ, where x๐‘ฅ is the majority of [al,โ€ฆ,ar][๐‘Ž๐‘™,โ€ฆ,๐‘Ž๐‘Ÿ].Here, the majority of a sequence consisting of 00 and 11 is defined as follows: suppose there are c0๐‘0 zeros and c1๐‘1 ones in the sequence, respectively.If c0โ‰ฅc1๐‘0โ‰ฅ๐‘1, the majority is 00.If c0<c1๐‘0<๐‘1, the majority is 11.For example, suppose a=[1,0,0,0,1,1]๐‘Ž=[1,0,0,0,1,1]. If we select l=1,r=2๐‘™=1,๐‘Ÿ=2, the resulting sequence will be [0,0,0,1,1][0,0,0,1,1]. If we select l=4,r=6๐‘™=4,๐‘Ÿ=6, the resulting sequence will be [1,0,0,1][1,0,0,1].Determine if you can make a=[1]๐‘Ž=[1] with a finite number of operations.InputEach test contains multiple test cases. The first line contains the number of test cases t๐‘ก (1โ‰คtโ‰ค4โ‹…1041โ‰ค๐‘กโ‰ค4โ‹…104). Description of the test cases follows.The first line of each testcase contains one integer n๐‘› (1โ‰คnโ‰ค2โ‹…1051โ‰ค๐‘›โ‰ค2โ‹…105).The second line of each testcase contains a string consisting of 00 and 11, describing the sequence a๐‘Ž.It's guaranteed that the sum of n๐‘› over all testcases does not exceed 2โ‹…1052โ‹…105.OutputFor each testcase, if it's possible to make a=[1]๐‘Ž=[1], print YES. Otherwise, print NO. You can output the answer in any case (upper or lower). For example, the strings yEs, yes, Yes, and YES will be recognized as positive responses.ExampleinputCopy5101120191000000019000011000outputCopyNoYesNoYesNoNoteIn the fourth testcase of the example, initially a=[1,0,0,0,0,0,0,0,1]๐‘Ž=[1,0,0,0,0,0,0,0,1]. A valid sequence of operations is:Select l=2,r=8๐‘™=2,๐‘Ÿ=8 and apply the operation. a๐‘Ž becomes [1,0,1][1,0,1].Select l=1,r=3๐‘™=1,๐‘Ÿ=3 and apply the operation. a๐‘Ž becomes [1][1].

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