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Suppose we have a sequence of numbers that is called a wiggle sequence if the differences between successive numbers strictly alternate between positive and negative. The first difference may be either positive or negative. A sequence with less than two elements is trivially a wiggle sequence. So for example, [1,7,4,9,2,5] is a wiggle sequence because if you see, the differences (6,-3,5,-7,3) are alternately positive and negative. But, [1,4,7,2,5] and [1,7,4,5,5] are not wiggle sequences, the first one because its first two differences are positive and the second one because its last difference is zero.

So we have a sequence of integers, we have to find the length of the longest subsequence that is a wiggle sequence. A subsequence is obtained by deleting some number of elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order. So if the input is like [1,7,4,9,2,5], then the output will be 6. as the entire sequence is a wiggle sequence.

To solve this, we will follow these steps −

n := size of nums

if n is 0, then return 0

set up := 1 and down := 1

for i in range 1 to n – 1

if nums[i] > nums[i – 1], then up := down + 1

otherwise when nums[i] < nums[i – 1], then down := up + 1

return max of up and down

Let us see the following implementation to get a better understanding −

#include <bits/stdc++.h> using namespace std; class Solution { public: int wiggleMaxLength(vector<int>& nums) { int n = nums.size(); if(!n) return 0; int up = 1; int down = 1; for(int i = 1; i < n; i++){ if(nums[i] > nums[i - 1]){ up = down + 1; } else if(nums[i] < nums[i - 1]){ down = up + 1; } } return max(up, down); } }; main(){ Solution ob; vector<int> v = {1,7,4,9,2,5}; cout << (ob.wiggleMaxLength(v)); }

[1,7,4,9,2,5]

6

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