Given an array arr[] of size N, the task is to find the count of subarrays with elements in alternate increasing-decreasing order or vice-versa. A subarray {a, b, c} will be valid if and only if either (a < b > c) or (a > b < c) is satisfied.
Examples:
Input: arr[] = {9, 8, 7, 6, 5}
Output: 4
Explanation: As each element is lesser than the other,
only consider the sequence of length 2, four times: [9, 8], [8, 7], [7, 6], [6, 5]Input: arr[] = {1, 2, 1, 2, 1}
Output: 10
Explanation: All possible sequences are: [1, 2, 1, 2, 1], [1, 2, 1, 2], [2, 1, 2, 1], [1, 2, 1],
[2, 1, 2], [1, 2, 1], [1, 2], [2, 1], [1, 2], [2, 1]
Approach: The task can be solved using a sliding window approach and mathematical concepts. The solution is built using the following steps:
- Find the longest current valid subarray.
- When the current longest valid subarray breaks, take the length of the subarray which was longest and store it in an array.
- Restart the window from the point where the previous longest subarray ended and find the next longest window and repeat till the array finishes.
- The array will have the length of contiguous subarrays in it which are non-overlapping.
- For each window of size k, which is a valid subarray, all possible windows of any size are also valid subsequences.
- To find all possible subsequences in a window, if one window is k, then count of all subarrays of all sizes = k x (k-1) /2
- Do this for each window size in the array and add it to count.
- Return count as final answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the count// of valid subarraysint getSubsequenceCount(vector<int>& nums){ bool flag = false; vector<int> arr; int i = 0, j = 1; while (j < nums.size()) { if ((nums[j] > nums[j - 1] && flag != true) || (nums[j] < nums[j - 1] && flag != false)) { if (nums[j] > nums[j - 1]) flag = true; else if (nums[j] < nums[j - 1]) flag = false; j += 1; } else { if (j - i != 1) { arr.push_back(j - i); flag = false; i = j - 1; } else { i = j; j += 1; } } } if (j - i != 1) arr.push_back(j - i); int count = 0; // Number of valid subsequences for (int itm : arr) count += itm * (itm - 1) / 2; return count;}int main(){ // Driver code vector<int> nums = { 1, 2, 1, 2, 1 }; cout << getSubsequenceCount(nums) << "\n";}// This code is contributed by Taranpreet |
Java
// Java program for the above approachimport java.util.ArrayList;class GFG { // Function to find the count // of valid subarrays static int getSubsequenceCount(int[] nums) { boolean flag = false; ArrayList<Integer> arr = new ArrayList<Integer>(); int i = 0, j = 1; while (j < nums.length) { if ((nums[j] > nums[j - 1] && flag != true) || (nums[j] < nums[j - 1] && flag != false)) { if (nums[j] > nums[j - 1]) flag = true; else if (nums[j] < nums[j - 1]) flag = false; j += 1; } else { if (j - i != 1) { arr.add(j - i); flag = false; i = j - 1; } else { i = j; j += 1; } } } if (j - i != 1) arr.add(j - i); int count = 0; // Number of valid subsequences for (int itm : arr) count += itm * (itm - 1) / 2; return count; } public static void main(String args[]) { // Driver code int[] nums = { 1, 2, 1, 2, 1 }; System.out.println(getSubsequenceCount(nums)); }}// This code is contributed gfgking |
Python3
# Python program for the above approach# Function to find the count # of valid subarraysdef getSubsequenceCount(nums): flag = None length = 1 arr = [] i, j = 0, 1 while j < len(nums): if(nums[j] > nums[j-1] and flag != True) \ or (nums[j] < nums[j-1] \ and flag != False): if(nums[j] > nums[j-1]): flag = True elif(nums[j] < nums[j-1]): flag = False j += 1 else: if(j-i != 1): arr.append(j-i) flag = None i = j-1 else: i = j j += 1 if(j-i != 1): arr.append(j-i) count = 0 # Number of valid subsequences for n in arr: count += n*(n-1)//2 return count# Driver codeif __name__ == "__main__": nums = [1, 2, 1, 2, 1] print(getSubsequenceCount(nums)) |
C#
// C# program for the above approachusing System;using System.Collections.Generic;public class GFG{ // Function to find the count // of valid subarrays static int getSubsequenceCount(int[] nums) { bool flag = false; List<int> arr = new List<int>(); int i = 0, j = 1; while (j < nums.Length) { if ((nums[j] > nums[j - 1] && flag != true) || (nums[j] < nums[j - 1] && flag != false)) { if (nums[j] > nums[j - 1]) flag = true; else if (nums[j] < nums[j - 1]) flag = false; j += 1; } else { if (j - i != 1) { arr.Add(j - i); flag = false; i = j - 1; } else { i = j; j += 1; } } } if (j - i != 1) arr.Add(j - i); int count = 0; // Number of valid subsequences foreach (int itm in arr) count += itm * (itm - 1) / 2; return count; } static public void Main () { // Driver code int[] nums = { 1, 2, 1, 2, 1 }; Console.WriteLine(getSubsequenceCount(nums)); }}// This code is contributed Shubham Singh |
Javascript
<script> // JavaScript program for the above approach // Function to find the count // of valid subarrays const getSubsequenceCount = (nums) => { let flag = 0; let length = 1; let arr = []; let i = 0, j = 1; while (j < nums.length) { if ((nums[j] > nums[j - 1] && flag != true) || (nums[j] < nums[j - 1] && flag != false)) { if (nums[j] > nums[j - 1]) flag = true; else if (nums[j] < nums[j - 1]) flag = false; j += 1; } else { if (j - i != 1) { arr.push(j - i) flag = false; i = j - 1; } else { i = j; j += 1 } } } if (j - i != 1) arr.push(j - i); let count = 0; // Number of valid subsequences for (let itm in arr) count += parseInt(arr[itm] * (arr[itm] - 1) / 2) return count; } // Driver code nums = [1, 2, 1, 2, 1]; document.write(getSubsequenceCount(nums)); // This code is contributed by rakeshsahni</script> |
Output
10
Time Complexity: O(N)
Auxiliary Space: O(N)
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