Minimum number of deletions to make a sorted sequence - GeeksforGeeks

02 Jun, 2021
Given an array of n integers. The task is to remove or delete the minimum number of elements from the array so that when the remaining elements are placed in the same sequence order to form an increasing sorted sequence.Examples :  
Input : {5, 6, 1, 7, 4}
Output : 2
Removing 1 and 4
leaves the remaining sequence order as
5 6 7 which 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:  
-->arr be the given array.
-->n number of elements in arr.
-->len be the length of longest
   increasing subsequence in arr.
-->// minimum number of deletions
   min = n - len
 
C++
Java
Python3
C#
PHP
Javascript
// 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 deletionsint 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 Codeint 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 implementation to find// minimum number of deletions// to make a sorted sequence class GFG{    /* 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);    }     // Driver Code    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 */
# Python3 implementation to find# minimum number of deletions to# make a sorted sequence # lis() returns the length# of the longest increasing# subsequence in arr[] of size ndef lis(arr, n):     result = 0    lis = [0 for i in range(n)]     # Initialize LIS values    # for all indexes    for i in range(n):        lis[i] = 1     # Compute optimized LIS values    # in bottom up manner    for i in range(1, n):        for j in range(i):            if ( arr[i] > arr[j] and                lis[i] < lis[j] + 1):                lis[i] = lis[j] + 1     # Pick resultimum    # of all LIS values    for i in range(n):        if (result < lis[i]):            result = lis[i]     return result # Function to calculate minimum# number of deletionsdef minimumNumberOfDeletions(arr, n):     # Find longest increasing    # subsequence    len = lis(arr, n)     # After removing elements    # other than the lis, we    # get sorted sequence.    return (n - len)  # Driver Codearr = [30, 40, 2, 5, 1,          7, 45, 50, 8]n = len(arr)print("Minimum number of deletions = ",      minimumNumberOfDeletions(arr, n))         # This code is contributed by Anant Agarwal.
// C# implementation to find// minimum number of deletions// to make a sorted sequenceusing System; class GfG{         /* 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);    }     // Driver Code    public static void Main (String[] args)    {        int []arr = {30, 40, 2, 5, 1,                       7, 45, 50, 8};        int n = arr.Length;        Console.Write("Minimum number of" +                          " deletions = " +         minimumNumberOfDeletions(arr, n));    }} // This code is contributed by parashar.
<?php// PHP implementation to find// minimum number of deletions// to make a sorted sequence  /* lis() returns the length of   the longest increasing subsequence   in arr[] of size n */function lis( $arr, $n ){    $result = 0;    $lis[$n] = 0;     /* Initialize LIS values       for all indexes */    for ($i = 0; $i < $n; $i++ )        $lis[$i] = 1;     /* Compute optimized LIS       values in bottom up manner */    for ($i = 1; $i < $n; $i++ )        for ($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 ($i = 0; $i < $n; $i++ )        if ($result < $lis[$i])            $result = $lis[$i];     return $result;} // function to calculate minimum// number of deletionsfunction minimumNumberOfDeletions($arr, $n){    // Find longest increasing    // subsequence    $len = lis($arr, $n);     // After removing elements    // other than the lis, we    // get sorted sequence.    return ($n - $len);} // Driver Code$arr = array(30, 40, 2, 5, 1,               7, 45, 50, 8);$n = sizeof($arr) / sizeof($arr[0]);echo "Minimum number of deletions = " ,    minimumNumberOfDeletions($arr, $n); // This code is contributed by nitin mittal.?>
<script>// javascript implementation to find// minimum number of deletions// to make a sorted sequence/* lis() returns the lengthof the longest increasingsubsequence in arr[] of size n */function lis(arr,n){    let result = 0;    let lis= new Array(n);     /* Initialize LIS values    for all indexes */    for (let i = 0; i < n; i++ )        lis[i] = 1;     /* Compute optimized LIS    values in bottom up manner */    for (let i = 1; i < n; i++ )        for (let 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 (let i = 0; i < n; i++ )        if (result < lis[i])            result = lis[i];     return result;} // function to calculate minimum// number of deletionsfunction minimumNumberOfDeletions(arr,n){     // Find longest increasing    // subsequence    let len = lis(arr,n);     // After removing elements    // other than the lis, we    // get sorted sequence.    return (n - len);}     let arr = [30, 40, 2, 5, 1,7, 45, 50, 8];    let n = arr.length;    document.write("Minimum number of deletions = "    + minimumNumberOfDeletions(arr,n)); // This code is contributed by vaibhavrabadiya117.</script>
Output :  
Minimum number of deletions = 4 
Time Complexity : O(n2)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 write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
parashar
nitin mittal
vaibhavrabadiya117
LIS
Dynamic Programming
Dynamic Programming
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