Given an array arr[], the task is to find all the maximal contiguous increasing subarray in a given array.
Examples:
Input:
arr[] = { 80, 50, 60, 70, 40, 50, 80, 70 }
Output:
80
50 60 70
40 50 80
70Input:
arr[] = { 10, 20, 23, 12, 5, 4, 61, 67, 87, 9 }
Output:
10 20 23
12
5
4 61 67 87
9
Approach: Iterate over the array and compare every element with its next neighboring element such that, if it is less than the next element, print it, else print it individually on the next line.
Below is the implementation of the above approach.
C++
// C++ Implementation to print all the// Maximal Increasing Sub-array of array#include <bits/stdc++.h>using namespace std;// Function to print each of maximal// contiguous increasing subarrayvoid printmaxSubseq(int arr[], int n){ int i; // Loop to iterate through the array and print // the maximal contiguous increasing subarray. for (i = 0; i < n; i++) { // Condition to check whether the element at i, is // greater than its next neighbouring element or not. if (arr[i] < arr[i + 1]) cout << arr[i] << " "; else cout << arr[i] << "\n"; }}// Driver functionint main(){ int arr[] = { 9, 8, 11, 13, 10, 15, 14, 16, 20, 5 }; int n = sizeof(arr) / sizeof(arr[0]); printmaxSubseq(arr, n); return 0;} |
Java
// Java Implementation to print all the// Maximal Increasing Sub-array of arrayimport java.util.*;class GFG{// Function to print each of maximal// contiguous increasing subarraystatic void printmaxSubseq(int arr[], int n){ int i; // Loop to iterate through the array and print // the maximal contiguous increasing subarray. for (i = 0; i < n ; i++) { // Condition to check whether the element at i, is // greater than its next neighbouring element or not. if (i + 1 < n && arr[i] < arr[i + 1]) System.out.print(arr[i] + " "); else System.out.print(arr[i] + "\n"); }}// Driver codepublic static void main(String[] args){ int arr[] = { 9, 8, 11, 13, 10, 15, 14, 16, 20, 5 }; int n = arr.length; printmaxSubseq(arr, n);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 Implementation to print all the # Maximal Increasing Sub-array of array # Function to print each of maximal # contiguous increasing subarray def printmaxSubseq(arr, n) : # Loop to iterate through the array and print # the maximal contiguous increasing subarray. for i in range(n - 1) : # Condition to check whether the element at i, is # greater than its next neighbouring element or not. if (arr[i] < arr[i + 1]) : print(arr[i], end = " "); else : print(arr[i]); print(arr[n - 1]); # Driver function if __name__ == "__main__" : arr = [ 9, 8, 11, 13, 10, 15, 14, 16, 20, 5 ]; n = len(arr); printmaxSubseq(arr, n); # This code is contributed by AnkitRai01 |
C#
// C# Implementation to print all the // Maximal Increasing Sub-array of array using System;class GFG { // Function to print each of maximal // contiguous increasing subarray static void printmaxSubseq(int []arr, int n) { int i; // Loop to iterate through the array and print // the maximal contiguous increasing subarray. for (i = 0; i < n ; i++) { // Condition to check whether the element at i, is // greater than its next neighbouring element or not. if (i + 1 < n && arr[i] < arr[i + 1]) Console.Write(arr[i] + " "); else Console.WriteLine(arr[i]); } } // Driver code public static void Main() { int []arr = { 9, 8, 11, 13, 10, 15, 14, 16, 20, 5 }; int n = arr.Length; printmaxSubseq(arr, n); } } // This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript Implementation to print all the// Maximal Increasing Sub-array of array// Function to print each of maximal// contiguous increasing subarrayfunction printmaxSubseq(arr, n){ let i; // Loop to iterate through the array and print // the maximal contiguous increasing subarray. for (i = 0; i < n; i++) { // Condition to check whether the element at i, is // greater than its next neighbouring element or not. if (arr[i] < arr[i + 1]) document.write(arr[i] + " "); else document.write(arr[i] + "<br>"); }}// Driver functionlet arr = [ 9, 8, 11, 13, 10, 15, 14, 16, 20, 5 ];let n = arr.length;printmaxSubseq(arr, n);</script> |
Output:
9 8 11 13 10 15 14 16 20 5
Time Complexity: O(n)
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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