Minimum number of deletions to make a sorted sequence

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
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++

// 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 Code
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 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

# 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 n
def 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 deletions
def 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 = [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#

// C# implementation to find
// minimum number of deletions 
// to make a sorted sequence
using 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
// 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 deletions
function 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.
?>



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 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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Improved By : parashar, nitin mittal




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