转换为最小变化的严格递增整数数组

给定一个由 n 个整数组成的数组。编写一个程序来找出数组中变化的最小数量，使数组严格增加整数。在严格递增数组 A[i] < A[i+1] for 0 <= i < n
例子：

Input : arr[] = { 1, 2, 6, 5, 4}
Output : 2
We can change a[2] to any value 
between 2 and 5.
and a[4] to any value greater then 5. 

Input : arr[] = { 1, 2, 3, 5, 7, 11 }
Output : 0
Array is already strictly increasing.

问题是最长递增子序列的变化。已经是 LIS 一部分的数字不需要更改。因此，要更改的最小元素是数组大小和 LIS 中元素数量的差异。请注意，我们还需要确保数字是整数。所以在制作LIS时，我们不把那些不能通过在中间插入元素形成严格递增的元素作为LIS的一部分。
例子{1, 2, 5, 3, 4 }，我们认为LIS的长度是三个{1, 2, 5}，而不是{1, 2, 3, 4}，因为我们不能制作一个严格递增的整数数组这个 LIS。

C++

// CPP program to find min elements to
// change so array is strictly increasing
#include 
using namespace std;
 
// To find min elements to remove from array
// to make it strictly increasing
int minRemove(int arr[], int n)
{
    int LIS[n], len = 0;
 
    // Mark all elements of LIS as 1
    for (int i = 0; i < n; i++)
        LIS[i] = 1;
 
    // Find LIS of array
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (arr[i] > arr[j] && (i-j)<=(arr[i]-arr[j])){
                LIS[i] = max(LIS[i], LIS[j] + 1);
            }
        }
        len = max(len, LIS[i]);
    }
 
    // Return min changes for array
    // to strictly increasing
    return n - len;
}
 
// Driver program to test minRemove()
int main()
{
    int arr[] = { 1, 2, 6, 5, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << minRemove(arr, n);
 
    return 0;
}

Java

// Java program to find min elements to
// change so array is strictly increasing
public class Main {
 
    // To find min elements to remove from array
    // to make it strictly increasing
    static int minRemove(int arr[], int n)
    {
        int LIS[] = new int[n];
        int len = 0;
 
        // Mark all elements of LIS as 1
        for (int i = 0; i < n; i++)
            LIS[i] = 1;
 
        // Find LIS of array
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (arr[i] > arr[j] && (i-j)<=(arr[i]-arr[j]))
                    LIS[i] = Math.max(LIS[i],
                                 LIS[j] + 1);
            }
            len = Math.max(len, LIS[i]);
        }
 
        // Return min changes for array
        // to strictly increasing
        return n - len;
    }
 
    // Driver program to test minRemove()
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 6, 5, 4 };
        int n = arr.length;
 
        System.out.println(minRemove(arr, n));
    }
}

Python3

# Python3 program to find min elements to
# change so array is strictly increasing
 
# Find min elements to remove from array
# to make it strictly increasing
def minRemove(arr, n):
    LIS = [0 for i in range(n)]
    len = 0
 
    # Mark all elements of LIS as 1
    for i in range(n):
        LIS[i] = 1
 
    # Find LIS of array
    for i in range(1, n):
         
        for j in range(i):
            if (arr[i] > arr[j] and (i-j)<=(arr[i]-arr[j]) ):
                LIS[i] = max(LIS[i], LIS[j] + 1)
                 
        len = max(len, LIS[i])
 
    # Return min changes for array
    # to strictly increasing
    return (n - len)
 
# Driver Code
arr = [ 1, 2, 6, 5, 4 ]
n = len(arr)
print(minRemove(arr, n))
 
# This code is contributed by Azkia Anam.

C#

// C# program to find min elements to change so
// array is strictly increasing
using System;
 
class GFG
{
 
    // To find min elements to remove from array to
    // make it strictly increasing
    static int minRemove(int []arr,
                        int n)
    {
        int []LIS = new int[n];
        int len = 0;
 
        // Mark all elements
        // of LIS as 1
        for (int i = 0; i < n; i++)
            LIS[i] = 1;
 
        // Find LIS of array
        for (int i = 1; i < n; i++)
        {
            for (int j = 0; j < i; j++)
            {
                if (arr[i] > arr[j] && (i-j)<=(arr[i]-arr[j]))
                    LIS[i] = Math.Max(LIS[i],
                                LIS[j] + 1);
            }
            len = Math.Max(len, LIS[i]);
        }
 
        // Return min changes for array 
        // to strictly increasing
        return n - len;
    }
 
    // Driver Code
    public static void Main()
    {
        int []arr = {1, 2, 6, 5, 4};
        int n = arr.Length;
 
        Console.WriteLine(minRemove(arr, n));
    }
}
 
// This code is contributed
// by anuj_67.

PHP

 $arr[$j])
                $LIS[$i] = max($LIS[$i],
                            $LIS[$j] + 1);
        }
        $len = max($len, $LIS[$i]);
    }
 
    // Return min changes for array to strictly
    // increasing
    return $n - $len;
}
 
// Driver Code
$arr = array(1, 2, 6, 5, 4);
$n = count($arr);
 
echo minRemove($arr, $n);
 
// This code is contributed
// by anuj_6
?>

Javascript

 
// Javascript program to find min elements to
// change so array is strictly increasing
 
    // To find min elements to remove from array
    // to make it strictly increasing
    function minRemove(arr, n)
    {
        let LIS = new Array(n).fill(0);
        let len = 0;
   
        // Mark all elements of LIS as 1
        for (let i = 0; i < n; i++)
            LIS[i] = 1;
   
        // Find LIS of array
        for (let i = 1; i < n; i++) {
            for (let j = 0; j < i; j++) {
                if (arr[i] > arr[j] && (i-j)<=(arr[i]-arr[j]))
                    LIS[i] = Math.max(LIS[i],
                                 LIS[j] + 1);
            }
            len = Math.max(len, LIS[i]);
        }
   
        // Return min changes for array
        // to strictly increasing
        return n - len;
    }
     
// driver program
     
        let arr = [ 1, 2, 6, 5, 4 ];
        let n = arr.length;
   
        document.write(minRemove(arr, n));
 
// This code is contributed by Code_hunt.

输出：

2

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