Convert to Strictly increasing integer array with minimum changes
Last Updated :
08 Feb, 2024
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++
#include <bits/stdc++.h>
using namespace std;
int minRemove(int arr[], int n)
{
int LIS[n], len = 0;
for (int i = 0; i < n; i++)
LIS[i] = 1;
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 n - len;
}
int main()
{
int arr[] = { 1, 2, 6, 5, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << minRemove(arr, n);
return 0;
}
|
C
#include <stdio.h>
int max(int num1, int num2)
{
return (num1 > num2) ? num1 : num2;
}
int minRemove(int arr[], int n)
{
int LIS[n], len = 0;
for (int i = 0; i < n; i++)
LIS[i] = 1;
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 n - len;
}
int main()
{
int arr[] = { 1, 2, 6, 5, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d", minRemove(arr, n));
return 0;
}
|
Java
public class Main {
static int minRemove(int arr[], int n)
{
int LIS[] = new int[n];
int len = 0;
for (int i = 0; i < n; i++)
LIS[i] = 1;
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 n - len;
}
public static void main(String[] args)
{
int arr[] = { 1, 2, 6, 5, 4 };
int n = arr.length;
System.out.println(minRemove(arr, n));
}
}
|
Python3
def minRemove(arr, n):
LIS = [0 for i in range(n)]
len = 0
for i in range(n):
LIS[i] = 1
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 (n - len)
arr = [ 1, 2, 6, 5, 4 ]
n = len(arr)
print(minRemove(arr, n))
|
C#
using System;
class GFG
{
static int minRemove(int []arr,
int n)
{
int []LIS = new int[n];
int len = 0;
for (int i = 0; i < n; i++)
LIS[i] = 1;
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 n - len;
}
public static void Main()
{
int []arr = {1, 2, 6, 5, 4};
int n = arr.Length;
Console.WriteLine(minRemove(arr, n));
}
}
|
Javascript
<script>
function minRemove(arr, n)
{
let LIS = new Array(n).fill(0);
let len = 0;
for (let i = 0; i < n; i++)
LIS[i] = 1;
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 n - len;
}
let arr = [ 1, 2, 6, 5, 4 ];
let n = arr.length;
document.write(minRemove(arr, n));
</script>
|
PHP
<?php
function minRemove($arr, $n)
{
$LIS = array();
$len = 0;
for ($i = 0; $i < $n; $i++)
$LIS[$i] = 1;
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 $n - $len;
}
$arr = array(1, 2, 6, 5, 4);
$n = count($arr);
echo minRemove($arr, $n);
?>
|
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.
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