Given an array of n integers. Write a program to find minimum number of changes in array so that 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
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 increasingint 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 increasingint 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 increasingpublic 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 increasingdef 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 Codearr = [ 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 increasingusing 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
<?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 increasingfunction 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?> |
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> |
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.
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