Convert to Strictly increasing integer array with minimum changes

Given an array of n integers. Write a program to find a minimum number of changes in the array so that the array is strictly increasing of integers. In strictly increasing array A[i] < A[i+1] for 0 <= i < n

Examples:

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 than 5.

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

The problem is variation of Longest Increasing Subsequence. The numbers which are already a part of LIS need not to be changed. So minimum elements to change is difference of size of array and number of elements in LIS. Note that we also need to make sure that the numbers are integers. So while making LIS, we do not consider those elements as part of LIS that cannot form strictly increasing by inserting elements in middle. 

Example { 1, 2, 5, 3, 4 }, we consider length of LIS as three {1, 2, 5}, not as {1, 2, 3, 4} because we cannot make a strictly increasing array of integers with this LIS. 

Implementation:

C++

// CPP program to find min elements to
// change so array is strictly increasing
#include <bits/stdc++.h>
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;
}

// This code is contributed by Sania Kumari Gupta

C

// C program to find min elements to
// change so array is strictly increasing
#include <stdio.h>

// Find maximum between two numbers.
int max(int num1, int num2)
{
    return (num1 > num2) ? num1 : num2;
}

// 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]);
    printf("%d", minRemove(arr, n));
    return 0;
}

// This code is contributed by Sania Kumari Gupta

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));
    }
}

// This code is contributed by Sania Kumari Gupta

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.

Javascript

<script>

// 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.
</script>

PHP

<?php
// PHP 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)
{
    $LIS = array();
    $len = 0;

    // Mark all elements
    // of LIS as 1
    for ($i = 0; $i < $n; $i++)
        $LIS[$i] = 1;

    // Find LIS of array
    for ($i = 1; $i < $n; $i++)
    {
        for ($j = 0; $j < $i; $j++)
        {
            if ($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 = array(1, 2, 6, 5, 4);
$n = count($arr);

echo minRemove($arr, $n);

// This code is contributed
// by anuj_6
?>

Output

2

Time Complexity: O(n*n), as nested loops are used
Auxiliary Space: O(n), Use of an array to store LIS values at each index.