Minimum number of deletions to make a sorted sequence
Last Updated :
07 Nov, 2023
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 is in sorted order or not. The 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 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 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# 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.
<script>
// javascript 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)
{
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 deletions
function 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>
<?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.
?>
OutputMinimum number of deletions = 4
Time Complexity : O(n2)
Auxiliary Space: O(n)
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.
Approach#2: Using longest increasing subsequence
One approach to solve this problem is to find the length of the longest increasing subsequence (LIS) of the given array and subtract it from the length of the array. The difference gives us the minimum number of deletions required to make the array sorted.
Algorithm
1. Calculate the length of the longest increasing subsequence (LIS) of the array.
2. Subtract the length of the LIS from the length of the array.
3. Return the difference obtained in step 2 as the output.
C++
#include <iostream>
#include <vector>
#include <algorithm> // Required for max_element
using namespace std;
// Function to find the minimum number of deletions
int minDeletions(vector<int> arr) {
int n = arr.size();
vector<int> lis(n, 1); // Initialize LIS array with 1
// Calculate LIS values
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (arr[i] > arr[j]) {
lis[i] = max(lis[i], lis[j] + 1); // Update LIS value
}
}
}
// Find the maximum length of LIS
int maxLength = *max_element(lis.begin(), lis.end());
// Return the minimum number of deletions
return n - maxLength;
}
//Driver code
int main() {
vector<int> arr = {5, 6, 1, 7, 4};
// Call the minDeletions function and print the result
cout << minDeletions(arr) << endl;
return 0;
}
import java.util.Arrays;
public class Main {
public static int minDeletions(int[] arr) {
int n = arr.length;
int[] lis = new int[n];
Arrays.fill(lis, 1); // Initialize the LIS array with all 1's
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (arr[i] > arr[j]) {
lis[i] = Math.max(lis[i], lis[j] + 1);
}
}
}
return n - Arrays.stream(lis).max().getAsInt(); // Return the number of elements to delete
}
public static void main(String[] args) {
int[] arr = {5, 6, 1, 7, 4};
System.out.println(minDeletions(arr)); // Output: 2
}
}
def min_deletions(arr):
n = len(arr)
lis = [1] * n
for i in range(1, n):
for j in range(i):
if arr[i] > arr[j]:
lis[i] = max(lis[i], lis[j] + 1)
return n - max(lis)
arr = [5, 6, 1, 7, 4]
print(min_deletions(arr))
using System;
using System.Collections.Generic;
using System.Linq;
namespace MinDeletionsExample
{
class Program
{
static int MinDeletions(List<int> arr)
{
int n = arr.Count;
List<int> lis = Enumerable.Repeat(1, n).ToList(); // Initialize LIS array with 1
// Calculate LIS values
for (int i = 1; i < n; ++i)
{
for (int j = 0; j < i; ++j)
{
if (arr[i] > arr[j])
{
lis[i] = Math.Max(lis[i], lis[j] + 1); // Update LIS value
}
}
}
// Find the maximum length of LIS
int maxLength = lis.Max();
// Return the minimum number of deletions
return n - maxLength;
}
// Driver Code
static void Main(string[] args)
{
List<int> arr = new List<int> { 5, 6, 1, 7, 4 };
// Call the MinDeletions function and print the result
Console.WriteLine(MinDeletions(arr));
// Keep console window open until a key is pressed
Console.ReadKey();
}
}
}
function minDeletions(arr) {
let n = arr.length;
let lis = new Array(n).fill(1);
for (let i = 1; i < n; i++) {
for (let j = 0; j < i; j++) {
if (arr[i] > arr[j]) {
lis[i] = Math.max(lis[i], lis[j] + 1);
}
}
}
return n - Math.max(...lis);
}
let arr = [5, 6, 1, 7, 4];
console.log(minDeletions(arr));
Time complexity: O(n^2), where n is length of array
Space complexity: O(n), where n is length of array
Approach#3: Using binary search
This approach uses binary search to find the correct position to insert a given element into the subsequence.
Algorithm
1. Initialize a list 'sub' with the first element of the input list.
2. For each subsequent element in the input list, if it is greater than the last element in 'sub', append it to 'sub'.
3. Otherwise, use binary search to find the correct position to insert the element into 'sub'.
4. The minimum number of deletions required is equal to the length of the input list minus the length of 'sub'.
C++
#include <iostream>
#include <vector>
using namespace std;
// Function to find the minimum number of deletions to make a strictly increasing subsequence
int minDeletions(vector<int>& arr) {
int n = arr.size();
vector<int> sub; // Stores the longest increasing subsequence
sub.push_back(arr[0]); // Initialize the subsequence with the first element of the array
for (int i = 1; i < n; i++) {
if (arr[i] > sub.back()) {
// If the current element is greater than the last element of the subsequence,
// it can be added to the subsequence to make it longer.
sub.push_back(arr[i]);
} else {
int index = -1; // Initialize index to -1
int val = arr[i]; // Current element value
int l = 0, r = sub.size() - 1; // Initialize left and right pointers for binary search
// Binary search to find the index where the current element can be placed in the subsequence
while (l <= r) {
int mid = (l + r) / 2; // Calculate the middle index
if (sub[mid] >= val) {
index = mid; // Update the index if the middle element is greater or equal to the current element
r = mid - 1; // Move the right pointer to mid - 1
} else {
l = mid + 1; // Move the left pointer to mid + 1
}
}
if (index != -1) {
sub[index] = val; // Replace the element at the found index with the current element
}
}
}
// The minimum number of deletions is equal to the difference between the input array size and the size of the longest increasing subsequence
return n - sub.size();
}
int main() {
vector<int> arr = {30, 40, 2, 5, 1, 7, 45, 50, 8};
int output = minDeletions(arr);
cout << output << endl;
return 0;
}
import java.util.ArrayList;
public class Main {
// Function to find the minimum number of deletions to make a strictly increasing subsequence
static int minDeletions(ArrayList<Integer> arr) {
int n = arr.size();
ArrayList<Integer> sub = new ArrayList<>(); // Stores the longest increasing subsequence
sub.add(arr.get(0)); // Initialize the subsequence with the first element of the array
for (int i = 1; i < n; i++) {
if (arr.get(i) > sub.get(sub.size() - 1)) {
// If the current element is greater than the last element of the subsequence,
// it can be added to the subsequence to make it longer.
sub.add(arr.get(i));
} else {
int index = -1; // Initialize index to -1
int val = arr.get(i); // Current element value
int l = 0, r = sub.size() - 1; // Initialize left and right pointers for binary search
// Binary search to find the index where the current element can be placed in the subsequence
while (l <= r) {
int mid = (l + r) / 2; // Calculate the middle index
if (sub.get(mid) >= val) {
index = mid; // Update the index if the middle element is greater or equal to the current element
r = mid - 1; // Move the right pointer to mid - 1
} else {
l = mid + 1; // Move the left pointer to mid + 1
}
}
if (index != -1) {
sub.set(index, val); // Replace the element at the found index with the current element
}
}
}
// The minimum number of deletions is equal to the difference between the input array size and the size of the longest increasing subsequence
return n - sub.size();
}
public static void main(String[] args) {
ArrayList<Integer> arr = new ArrayList<>();
arr.add(30);
arr.add(40);
arr.add(2);
arr.add(5);
arr.add(1);
arr.add(7);
arr.add(45);
arr.add(50);
arr.add(8);
int output = minDeletions(arr);
System.out.println(output);
}
}
def min_deletions(arr):
def ceil_index(sub, val):
l, r = 0, len(sub)-1
while l <= r:
mid = (l + r) // 2
if sub[mid] >= val:
r = mid - 1
else:
l = mid + 1
return l
sub = [arr[0]]
for i in range(1, len(arr)):
if arr[i] > sub[-1]:
sub.append(arr[i])
else:
sub[ceil_index(sub, arr[i])] = arr[i]
return len(arr) - len(sub)
arr = [30, 40, 2, 5, 1, 7, 45, 50, 8]
output = min_deletions(arr)
print(output)
using System;
using System.Collections.Generic;
class Program
{
// Function to find the minimum number of deletions to make a strictly increasing subsequence
static int MinDeletions(List<int> arr)
{
int n = arr.Count;
List<int> sub = new List<int>(); // Stores the longest increasing subsequence
sub.Add(arr[0]); // Initialize the subsequence with the first element of the array
for (int i = 1; i < n; i++)
{
if (arr[i] > sub[sub.Count - 1])
{
// If the current element is greater than the last element of the subsequence,
// it can be added to the subsequence to make it longer.
sub.Add(arr[i]);
}
else
{
int index = -1; // Initialize index to -1
int val = arr[i]; // Current element value
int l = 0, r = sub.Count - 1; // Initialize left and right
// pointers for binary search
// Binary search to find the index where the current element
// can be placed in the subsequence
while (l <= r)
{
int mid = (l + r) / 2; // Calculate the middle index
if (sub[mid] >= val)
{
index = mid; // Update the index if the middle element is
// greater or equal to the current element
r = mid - 1; // Move the right pointer to mid - 1
}
else
{
l = mid + 1; // Move the left pointer to mid + 1
}
}
if (index != -1)
{
sub[index] = val; // Replace the element at the found index
// with the current element
}
}
}
// The minimum number of deletions is equal to the difference
// between the input list size and the size of the
// longest increasing subsequence
return n - sub.Count;
}
// Driver code
static void Main()
{
List<int> arr = new List<int> { 30, 40, 2, 5, 1, 7, 45, 50, 8 };
int output = MinDeletions(arr);
Console.WriteLine(output);
Console.ReadLine();
}
}
// Function to find the minimum number of deletions to make a strictly increasing subsequence
function minDeletions(arr) {
let n = arr.length;
let sub = []; // Stores the longest increasing subsequence
sub.push(arr[0]); // Initialize the subsequence with the first element of the array
for (let i = 1; i < n; i++) {
if (arr[i] > sub[sub.length - 1]) {
// If the current element is greater than the last element of the subsequence,
// it can be added to the subsequence to make it longer.
sub.push(arr[i]);
} else {
let index = -1; // Initialize index to -1
let val = arr[i]; // Current element value
let l = 0, r = sub.length - 1; // Initialize left and right pointers for binary search
// Binary search to find the index where the current element can be placed
// in the subsequence
while (l <= r) {
let mid = Math.floor((l + r) / 2); // Calculate the middle index
if (sub[mid] >= val) {
index = mid; // Update the index if the middle element is greater
//or equal to the current element
r = mid - 1; // Move the right pointer to mid - 1
} else {
l = mid + 1; // Move the left pointer to mid + 1
}
}
if (index !== -1) {
sub[index] = val; // Replace the element at the found index with the current element
}
}
}
// The minimum number of deletions is equal to the difference
//between the input array size and the size of the longest increasing subsequence
return n - sub.length;
}
let arr = [30, 40, 2, 5, 1, 7, 45, 50, 8];
let output = minDeletions(arr);
console.log(output);
Time Complexity: O(n log n)
Auxiliary Space: O(n)
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