Given an array of n integers. The task is to remove or delete minimum number of elements from the array so that when the remaining elements are placed in the same sequence order form a sorted sequence.

Examples:

Input : {5, 6, 1, 7, 4} Output : 2 Removing1and4leaves the remaining sequence order as5 6 7which is a sorted sequence. Input : {30, 40, 2, 5, 1, 7, 45, 50, 8} Output : 4

A **simple solution** is to remove all subsequences one by one and check if remaining set of elements are in sorted order or not. Time complexity of this solution is exponential.

An **efficient approach** uses the concept of finding the length of the longest increasing subsequence of a given sequence.

**Algorithm:**

-->arrbe the given array. -->nnumber of elements inarr. -->lenbe the length of longest increasing subsequence inarr. -->// minimum number of deletionsmin= n - len

## C++

// C++ implementation to find minimum number // of deletions to make a sorted sequence #include <bits/stdc++.h> using namespace std; /* lis() returns the length of the longest increasing subsequence in arr[] of size n */ int lis( int arr[], int n ) { int result = 0; int lis[n]; /* Initialize LIS values for all indexes */ for (int i = 0; i < n; i++ ) lis[i] = 1; /* Compute optimized LIS values in bottom up manner */ for (int i = 1; i < n; i++ ) for (int j = 0; j < i; j++ ) if ( arr[i] > arr[j] && lis[i] < lis[j] + 1) lis[i] = lis[j] + 1; /* Pick resultimum of all LIS values */ for (int i = 0; i < n; i++ ) if (result < lis[i]) result = lis[i]; return result; } // function to calculate minimum // number of deletions int minimumNumberOfDeletions(int arr[], int n) { // Find longest increasing subsequence int len = lis(arr, n); // After removing elements other than // the lis, we get sorted sequence. return (n - len); } // Driver program to test above int main() { int arr[] = {30, 40, 2, 5, 1, 7, 45, 50, 8}; int n = sizeof(arr) / sizeof(arr[0]); cout << "Minimum number of deletions = " << minimumNumberOfDeletions(arr, n); return 0; }

## Java

// Java implementation to find minimum number // of deletions to make a sorted sequence class Main { /* lis() returns the length of the longest increasing subsequence in arr[] of size n */ static int lis( int arr[], int n ) { int result = 0; int[] lis = new int[n]; /* Initialize LIS values for all indexes */ for (int i = 0; i < n; i++ ) lis[i] = 1; /* Compute optimized LIS values in bottom up manner */ for (int i = 1; i < n; i++ ) for (int j = 0; j < i; j++ ) if ( arr[i] > arr[j] && lis[i] < lis[j] + 1) lis[i] = lis[j] + 1; /* Pick resultimum of all LIS values */ for (int i = 0; i < n; i++ ) if (result < lis[i]) result = lis[i]; return result; } // function to calculate minimum // number of deletions static int minimumNumberOfDeletions(int arr[], int n) { // Find longest increasing subsequence int len = lis(arr, n); // After removing elements other than // the lis, we get sorted sequence. return (n - len); } // main function public static void main (String[] args) { int arr[] = {30, 40, 2, 5, 1, 7, 45, 50, 8}; int n = arr.length; System.out.println("Minimum number of deletions = " + minimumNumberOfDeletions(arr, n)); } } /* This code is contributed by Harsh Agarwal */

Output:

Minimum number of deletions = 4

Time Complexity: O(n^{2})

Time Complexity can be decreased to O(nlogn) by finding the Longest Increasing Subsequence Size(N Log N)

This article is contributed by **Ayush Jauhari**. 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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