Given an array arr[], the task is to remove at most one element and calculate the maximum length of strictly increasing subarray.
Examples:
Input: arr[] = {1, 2, 5, 3, 4}
Output: 4
After deleting 5, the resulting array will be {1, 2, 3, 4}
and the maximum length of its strictly increasing subarray is 4.Input: arr[] = {1, 2}
Output: 2
The complete array is already strictly increasing.
Approach:
- Create two arrays pre[] and pos[] of size N.
- Iterate over the input array arr[] from (0, N) to find out the contribution of the current element arr[i] in the array till now [0, i) and update the pre[] array if it contributes to the strictly increasing subarray.
- Iterate over the input array arr[] from [N – 2, 0] to find out the contribution of the current element arr[j] in the array till now (N, j) and update the pos[] array if arr[j] contributes in the longest increasing subarray.
- Calculate the maximum length of the strictly increasing subarray without removing any element.
- Iterate over the array pre[] and pos[] to find out the contribution of the current element by excluding that element.
- Maintain a variable ans to find the maximum found till now.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the maximum length of strictly// increasing subarray after removing atmost one elementint maxIncSubarr(int a[], int n){ // Create two arrays pre and pos int pre[n] = { 0 }; int pos[n] = { 0 }; pre[0] = 1; pos[n - 1] = 1; int l = 0; // Find out the contribution of the current element in // array[0, i] and update pre[i] for (int i = 1; i < n; i++) { if (a[i] > a[i - 1]) pre[i] = pre[i - 1] + 1; else pre[i] = 1; } // Find out the contribution of the current element in // array[N - 1, i] and update pos[i] l = 1; for (int i = n - 2; i >= 0; i--) { if (a[i] < a[i + 1]) pos[i] = pos[i + 1] + 1; else pos[i] = 1; } // Calculate the maximum length of the strictly // increasing subarray without removing any element int ans = 0; l = 1; for (int i = 1; i < n; i++) { if (a[i] > a[i - 1]) l++; else l = 1; ans = max(ans, l); } // Calculate the maximum length of the strictly // increasing subarray after removing the current // element for (int i = 1; i <= n - 2; i++) if (a[i - 1] < a[i + 1]) ans = max(pre[i - 1] + pos[i + 1], ans); return ans;}// Driver codeint main(){ int arr[] = { 1, 2 }; int n = sizeof(arr) / sizeof(int); cout << maxIncSubarr(arr, n); return 0;} |
C
// C implementation of the approach#include <stdio.h>// Find maximum between two numbers.int max(int num1, int num2){ return (num1 > num2) ? num1 : num2;}// Function to return the maximum length of strictly// increasing subarray after removing atmost one elementint maxIncSubarr(int a[], int n){ // Create two arrays pre and pos int pre[n]; int pos[n]; for (int i = 0; i < n; i++) pre[i] = 0; for (int i = 0; i < n; i++) pos[i] = 0; pre[0] = 1; pos[n - 1] = 1; int l = 0; // Find out the contribution of the current element in // array[0, i] and update pre[i] for (int i = 1; i < n; i++) { if (a[i] > a[i - 1]) pre[i] = pre[i - 1] + 1; else pre[i] = 1; } // Find out the contribution of the current element in // array[N - 1, i] and update pos[i] l = 1; for (int i = n - 2; i >= 0; i--) { if (a[i] < a[i + 1]) pos[i] = pos[i + 1] + 1; else pos[i] = 1; } // Calculate the maximum length of the strictly // increasing subarray without removing any element int ans = 0; l = 1; for (int i = 1; i < n; i++) { if (a[i] > a[i - 1]) l++; else l = 1; ans = max(ans, l); } // Calculate the maximum length of the strictly // increasing subarray after removing the current // element for (int i = 1; i <= n - 2; i++) if (a[i - 1] < a[i + 1]) ans = max(pre[i - 1] + pos[i + 1], ans); return ans;}// Driver codeint main(){ int arr[] = { 1, 2 }; int n = sizeof(arr) / sizeof(int); printf("%d", maxIncSubarr(arr, n)); return 0;}// This code is contributed by Sania Kumari Gupta |
Java
// Java implementation of the approach import java.io.*;class GFG { // Function to return the maximum length of // strictly increasing subarray after // removing atmost one element static int maxIncSubarr(int a[], int n) { // Create two arrays pre and pos int pre[] = new int[n] ; int pos[] = new int[n] ; pre[0] = 1; pos[n - 1] = 1; int l = 0; // Find out the contribution of the current // element in array[0, i] and update pre[i] for (int i = 1; i < n; i++) { if (a[i] > a[i - 1]) pre[i] = pre[i - 1] + 1; else pre[i] = 1; } // Find out the contribution of the current // element in array[N - 1, i] and update pos[i] l = 1; for (int i = n - 2; i >= 0; i--) { if (a[i] < a[i + 1]) pos[i] = pos[i + 1] + 1; else pos[i] = 1; } // Calculate the maximum length of the // strictly increasing subarray without // removing any element int ans = 0; l = 1; for (int i = 1; i < n; i++) { if (a[i] > a[i - 1]) l++; else l = 1; ans = Math.max(ans, l); } // Calculate the maximum length of the // strictly increasing subarray after // removing the current element for (int i = 1; i <= n - 2; i++) { if (a[i - 1] < a[i + 1]) ans = Math.max(pre[i - 1] + pos[i + 1], ans); } return ans; } // Driver code public static void main (String[] args) { int arr[] = {1, 2}; int n = arr.length; System.out.println(maxIncSubarr(arr, n)); } }// This code is contributed by AnkitRai01 |
Python3
# Python implementation of the approach # Function to return the maximum length of# strictly increasing subarray after# removing atmost one elementdef maxIncSubarr(a, n): # Create two arrays pre and pos pre = [0] * n; pos = [0] * n; pre[0] = 1; pos[n - 1] = 1; l = 0; # Find out the contribution of the current # element in array[0, i] and update pre[i] for i in range(1, n): if (a[i] > a[i - 1]): pre[i] = pre[i - 1] + 1; else: pre[i] = 1; # Find out the contribution of the current # element in array[N - 1, i] and update pos[i] l = 1; for i in range(n - 2, -1, -1): if (a[i] < a[i + 1]): pos[i] = pos[i + 1] + 1; else: pos[i] = 1; # Calculate the maximum length of the # strictly increasing subarray without # removing any element ans = 0; l = 1; for i in range(1, n): if (a[i] > a[i - 1]): l += 1; else: l = 1; ans = max(ans, l); # Calculate the maximum length of the # strictly increasing subarray after # removing the current element for i in range(1, n - 1): if (a[i - 1] < a[i + 1]): ans = max(pre[i - 1] + pos[i + 1], ans); return ans;# Driver codeif __name__ == '__main__': arr = [ 1, 2 ]; n = len(arr); print(maxIncSubarr(arr, n)); # This code is contributed by PrinciRaj1992 |
C#
// C# implementation of the approach using System;class GFG { // Function to return the maximum length of // strictly increasing subarray after // removing atmost one element static int maxIncSubarr(int []a, int n) { // Create two arrays pre and pos int []pre = new int[n] ; int []pos = new int[n] ; pre[0] = 1; pos[n - 1] = 1; int l = 0; // Find out the contribution of the current // element in array[0, i] and update pre[i] for (int i = 1; i < n; i++) { if (a[i] > a[i - 1]) pre[i] = pre[i - 1] + 1; else pre[i] = 1; } // Find out the contribution of the current // element in array[N - 1, i] and update pos[i] l = 1; for (int i = n - 2; i >= 0; i--) { if (a[i] < a[i + 1]) pos[i] = pos[i + 1] + 1; else pos[i] = 1; } // Calculate the maximum length of the // strictly increasing subarray without // removing any element int ans = 0; l = 1; for (int i = 1; i < n; i++) { if (a[i] > a[i - 1]) l++; else l = 1; ans = Math.Max(ans, l); } // Calculate the maximum length of the // strictly increasing subarray after // removing the current element for (int i = 1; i <= n - 2; i++) { if (a[i - 1] < a[i + 1]) ans = Math.Max(pre[i - 1] + pos[i + 1], ans); } return ans; } // Driver code public static void Main() { int []arr = {1, 2}; int n = arr.Length; Console.WriteLine(maxIncSubarr(arr, n)); } }// This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript implementation of the approach// Function to return the maximum length of// strictly increasing subarray after// removing atmost one elementfunction maxIncSubarr(a, n){ // Create two arrays pre and pos let pre = new Array(n); let pos = new Array(n); pre.fill(0); pos.fill(0); pre[0] = 1; pos[n - 1] = 1; let l = 0; // Find out the contribution of the current // element in array[0, i] and update pre[i] for(let i = 1; i < n; i++) { if (a[i] > a[i - 1]) pre[i] = pre[i - 1] + 1; else pre[i] = 1; } // Find out the contribution of the current // element in array[N - 1, i] and update pos[i] l = 1; for(let i = n - 2; i >= 0; i--) { if (a[i] < a[i + 1]) pos[i] = pos[i + 1] + 1; else pos[i] = 1; } // Calculate the maximum length of the // strictly increasing subarray without // removing any element let ans = 0; l = 1; for(let i = 1; i < n; i++) { if (a[i] > a[i - 1]) l++; else l = 1; ans = Math.max(ans, l); } // Calculate the maximum length of the // strictly increasing subarray after // removing the current element for(let i = 1; i <= n - 2; i++) { if (a[i - 1] < a[i + 1]) ans = Math.max(pre[i - 1] + pos[i + 1], ans); } return ans;}// Driver codelet arr = [ 1, 2 ];let n = arr.length;document.write(maxIncSubarr(arr, n));// This code is contributed by rameshtravel07</script> |
Output
2
Time Complexity: O(N)
Auxiliary Space: O(N)
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