给定一个由 n 个整数组成的数组。任务是从数组中移除或删除最少数量的元素，以便当剩余元素按相同的顺序排列时，形成一个递增的排序序列。
例子 ：
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 : 4一个简单的解决方案是一个一个地删除所有子序列并检查剩余的元素集是否按顺序排列。该解决方案的时间复杂度是指数级的。
一种有效的方法使用找到给定序列的最长递增子序列的长度的概念。
算法：
-->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 - lenC++
// C++ implementation to find
// minimum number of deletions
// to make a sorted sequence
#include
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
$arr[$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.
?>Javascript
输出 ：
Minimum number of deletions = 4 如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。