Given an array of **n** elements. Make strictly increasing and strictly decreasing subsequences from the array such that each array element belongs to increasing subsequence or decreasing subsequence, but not both, or can be part of none of the subsequence. Minimize the number of elements which are not part of any of the subsequences and find the count of such elements.

Examples:

Input :arr[] = { 7, 8, 1, 2, 4, 6, 3, 5, 2, 1, 8, 7 }Output :2 Increasing sequence can be { 1, 2, 4, 5, 8 }. Decreasing sequence can be { 7, 6, 3, 2, 1 }. So, only 2 (8, 7) element is left which are not part of either of the subsequences.Input :arr[] = { 1, 4, 2, 3, 3, 2, 4, 1 }Output :0 Increasing sequence can be { 1, 2, 3, 4 }. Decreasing sequence can be { 4, 3, 2, 1 }. So, no element is left which is not part of either of the subsequences.

The idea is to make a decision on each index, starting from index 0, one by one. For each index there can be three possibilities, first, it can belong to increasing sequence, second, it can belong to decreasing sequence, third, it does not belong to any of these sequences.

So, for each index, check for the optimal answer (minimum element which is not part of any of the subsequences) by considering it once as a part of increasing subsequence or as a part of decreasing subsequence. If the optimal answer cannot be achieved by them then leave it as the element which is not part of any of the sequence.

To decrease the complexity (using Dynamic Programming), we can store the number of elements which are not part of any of the subsequences using 3D array dp[x][y][z], where x indicates the decision index, y indicates the last index of decreasing sequence, z indicates the last index of increasing sequence.

Below is C++ implementation of this approach:

// C++ program to return minimum number of elements which // are not part of increasing or decreasing subsequences. #include<bits/stdc++.h> #define MAX 102 using namespace std; // Return minimum number of elements which is not part of // any of the sequence. int countMin(int arr[], int dp[MAX][MAX][MAX], int n, int dec, int inc, int i) { // If already calculated, return value. if (dp[dec][inc][i] != -1) return dp[dec][inc][i]; // If whole array is traversed. if (i == n) return 0; // calculating by considering element as part of // decreasing sequence. if (arr[i] < arr[dec]) dp[dec][inc][i] = countMin(arr, dp, n, i, inc, i + 1); // calculating by considering element as part of // increasing sequence. if (arr[i] > arr[inc]) { // If cannot be calculated for decreasing sequence. if (dp[dec][inc][i] == -1) dp[dec][inc][i] = countMin(arr, dp, n, dec, i, i + 1); // After considering once by decreasing sequence, now try // for increasing sequence. else dp[dec][inc][i] = min(countMin(arr, dp, n, dec, i, i + 1), dp[dec][inc][i]); } // If element cannot be part of any of the sequence. if (dp[dec][inc][i] == -1) dp[dec][inc][i] = 1 + countMin(arr, dp, n, dec, inc, i + 1); // After considering element as part of increasing and // decreasing sequence trying as not part of any of the // sequence. else dp[dec][inc][i] = min(1 + countMin(arr, dp, n, dec, inc, i + 1), dp[dec][inc][i]); return dp[dec][inc][i]; } // Wrapper Function int wrapper(int arr[], int n) { // Adding two number at the end of array, so that // increasing and decreasing sequence can be made. // MAX - 2 index is assigned INT_MAX for decreasing sequence // because/ next number of sequence must be less than it. // Similarly, for Increasing sequence INT_MIN is assigned to // MAX - 1 index. arr[MAX - 2] = INT_MAX; arr[MAX - 1] = INT_MIN; int dp[MAX][MAX][MAX]; memset(dp, -1, sizeof dp); return countMin(arr, dp, n, MAX - 2, MAX - 1, 0); } // Driven Program int main() { int n = 12; int arr[MAX] = { 7, 8, 1, 2, 4, 6, 3, 5, 2, 1, 8, 7 }; cout << wrapper(arr, n) << endl; return 0; }

Output:

2

**Time Complexity :** O(n^{3})

This article is contributed by ** Anuj Chauhan**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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