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Longest Turbulent Subarray in C++
Consider a subarray A[i], A[i+1], ..., A[j] of A is said to be turbulent when it meets these conditions −
For i <= k < j and A[k] > A[k+1] when k is odd, and A[k] < A[k+1] when k is even;
Otherwise, for i <= k < j, A[k] > A[k+1] when k is even, and A[k] < A[k+1] when k is odd.
So the subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray. Now find the length of a maximum size turbulent subarray of A. So if the input is like [9,4,2,10,7,8,8,1,9], output is 5. This is because A[1] > A[2] < A[3] > A[4] < A[5]
To solve this, we will follow these steps −
n := size of array A
prevBig := 1, prevSmall := 1, currBig := 1, currSmall := 1 and ret := 1
for i in range 1 to n – 1
if A[i] > A[i – 1], then currBig := 1 + prevSmall
if A[i] < A[i – 1], then currSmall := 1 + prevBig
ret := max of ret, currBig and currSmall
prevSmall := currSmall, prevBig := currBig, currSmall := 1, currBig := 1
return ret
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h> using namespace std; class Solution { public: int maxTurbulenceSize(vector<int>& A) { int n = A.size(); int prevBig = 1; int prevSmall = 1; int currBig = 1; int currSmall = 1; int ret = 1; for(int i = 1; i < n; i++){ if(A[i] > A[i - 1]){ currBig = 1 + prevSmall; } if(A[i] < A[i - 1]){ currSmall = 1 + prevBig; } ret = max({ret, currBig, currSmall}); prevSmall = currSmall; prevBig = currBig; currSmall = 1; currBig = 1; } return ret; } }; main(){ vector<int> v1 = {9,4,2,10,7,8,8,1,9}; Solution ob; cout << (ob.maxTurbulenceSize(v1)); }
Input
[9,4,2,10,7,8,8,1,9]
Output
5