Count Strictly Increasing Subarrays
Given an array of integers, count number of subarrays (of size more than one) that are strictly increasing.
Expected Time Complexity : O(n)
Expected Extra Space: O(1)
Examples:
Input: arr[] = {1, 4, 3}
Output: 1
There is only one subarray {1, 4}
Input: arr[] = {1, 2, 3, 4}
Output: 6
There are 6 subarrays {1, 2}, {1, 2, 3}, {1, 2, 3, 4}
{2, 3}, {2, 3, 4} and {3, 4}
Input: arr[] = {1, 2, 2, 4}
Output: 2
There are 2 subarrays {1, 2} and {2, 4}We strongly recommend that you click here and practice it, before moving on to the solution.
A Simple Solution is to generate all possible subarrays, and for every subarray check if subarray is strictly increasing or not. Worst case time complexity of this solution would be O(n3).
A Better Solution is to use the fact that if subarray arr[i:j] is not strictly increasing, then subarrays arr[i:j+1], arr[i:j+2], .. arr[i:n-1] cannot be strictly increasing. Below is the program based on above idea.
C++
// C++ program to count number of strictly// increasing subarrays#include<bits/stdc++.h>using namespace std;int countIncreasing(int arr[], int n){ // Initialize count of subarrays as 0 int cnt = 0; // Pick starting point for (int i=0; i<n; i++) { // Pick ending point for (int j=i+1; j<n; j++) { if (arr[j] > arr[j-1]) cnt++; // If subarray arr[i..j] is not strictly // increasing, then subarrays after it , i.e., // arr[i..j+1], arr[i..j+2], .... cannot // be strictly increasing else break; } } return cnt;}// Driver programint main(){ int arr[] = {1, 2, 2, 4}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Count of strictly increasing subarrays is " << countIncreasing(arr, n); return 0;} |
Java
// Java program to count number of strictly// increasing subarraysclass Test{ static int arr[] = new int[]{1, 2, 2, 4}; static int countIncreasing(int n) { // Initialize count of subarrays as 0 int cnt = 0; // Pick starting point for (int i=0; i<n; i++) { // Pick ending point for (int j=i+1; j<n; j++) { if (arr[j] > arr[j-1]) cnt++; // If subarray arr[i..j] is not strictly // increasing, then subarrays after it , i.e., // arr[i..j+1], arr[i..j+2], .... cannot // be strictly increasing else break; } } return cnt; } // Driver method to test the above function public static void main(String[] args) { System.out.println("Count of strictly increasing subarrays is " + countIncreasing(arr.length)); }} |
Python3
# Python3 program to count number# of strictly increasing subarraysdef countIncreasing(arr, n): # Initialize count of subarrays as 0 cnt = 0 # Pick starting point for i in range(0, n) : # Pick ending point for j in range(i + 1, n) : if arr[j] > arr[j - 1] : cnt += 1 # If subarray arr[i..j] is not strictly # increasing, then subarrays after it , i.e., # arr[i..j+1], arr[i..j+2], .... cannot # be strictly increasing else: break return cnt# Driver codearr = [1, 2, 2, 4]n = len(arr)print ("Count of strictly increasing subarrays is", countIncreasing(arr, n))# This code is contributed by Shreyanshi Arun. |
C#
// C# program to count number of// strictly increasing subarraysusing System;class Test{ static int []arr = new int[]{1, 2, 2, 4}; static int countIncreasing(int n) { // Initialize count of subarrays as 0 int cnt = 0; // Pick starting point for (int i = 0; i < n; i++) { // Pick ending point for (int j = i + 1; j < n; j++) { if (arr[j] > arr[j - 1]) cnt++; // If subarray arr[i..j] is not strictly // increasing, then subarrays after it , // i.e., arr[i..j+1], arr[i..j+2], .... // cannot be strictly increasing else break; } } return cnt; } // Driver Code public static void Main(String[] args) { Console.Write("Count of strictly increasing" + "subarrays is " + countIncreasing(arr.Length)); }}// This code is contribute by parashar. |
PHP
<?php// PHP program to count number of// strictly increasing subarraysfunction countIncreasing( $arr, $n){ // Initialize count of subarrays // as 0 $cnt = 0; // Pick starting point for ( $i = 0; $i < $n; $i++) { // Pick ending point for ( $j = $i+1; $j < $n; $j++) { if ($arr[$j] > $arr[$j-1]) $cnt++; // If subarray arr[i..j] is // not strictly increasing, // then subarrays after it, // i.e., arr[i..j+1], // arr[i..j+2], .... cannot // be strictly increasing else break; } } return $cnt;}// Driver program$arr = array(1, 2, 2, 4);$n = count($arr);echo "Count of strictly increasing ", "subarrays is ", countIncreasing($arr, $n);// This code is contribute by anuj_67.?> |
Javascript
<script>// Javascript program to count number of strictly// increasing subarraysfunction countIncreasing(arr, n){ // Initialize count of subarrays as 0 let cnt = 0; // Pick starting point for(let i = 0; i < n; i++) { // Pick ending point for (let j = i + 1; j < n; j++) { if (arr[j] > arr[j - 1]) cnt++; // If subarray arr[i..j] is not strictly // increasing, then subarrays after it , i.e., // arr[i..j+1], arr[i..j+2], .... cannot // be strictly increasing else break; } } return cnt;}// Driver codelet arr = [ 1, 2, 2, 4 ];let n = arr.length;document.write("Count of strictly " + "increasing subarrays is " + countIncreasing(arr, n)); // This code is contributed by divyesh072019</script> |
Output :
Count of strictly increasing subarrays is 2
Time complexity of the above solution is O(m) where m is number of subarrays in output
This problem and solution are contributed by Rahul Agrawal.
An Efficient Solution can count subarrays in O(n) time. The idea is based on fact that a sorted subarray of length ‘len’ adds len*(len-1)/2 to result. For example, {10, 20, 30, 40} adds 6 to the result.
C++
// C++ program to count number of strictly// increasing subarrays in O(n) time.#include<bits/stdc++.h>using namespace std;int countIncreasing(int arr[], int n){ int cnt = 0; // Initialize result // Initialize length of current increasing // subarray int len = 1; // Traverse through the array for (int i=0; i < n-1; ++i) { // If arr[i+1] is greater than arr[i], // then increment length if (arr[i + 1] > arr[i]) len++; // Else Update count and reset length else { cnt += (((len - 1) * len) / 2); len = 1; } } // If last length is more than 1 if (len > 1) cnt += (((len - 1) * len) / 2); return cnt;}// Driver programint main(){ int arr[] = {1, 2, 2, 4}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Count of strictly increasing subarrays is " << countIncreasing(arr, n); return 0;} |
Java
// Java program to count number of strictly// increasing subarraysclass Test{ static int arr[] = new int[]{1, 2, 2, 4}; static int countIncreasing(int n) { int cnt = 0; // Initialize result // Initialize length of current increasing // subarray int len = 1; // Traverse through the array for (int i=0; i < n-1; ++i) { // If arr[i+1] is greater than arr[i], // then increment length if (arr[i + 1] > arr[i]) len++; // Else Update count and reset length else { cnt += (((len - 1) * len) / 2); len = 1; } } // If last length is more than 1 if (len > 1) cnt += (((len - 1) * len) / 2); return cnt; } // Driver method to test the above function public static void main(String[] args) { System.out.println("Count of strictly increasing subarrays is " + countIncreasing(arr.length)); }} |
Python3
# Python3 program to count number of# strictlyincreasing subarrays in O(n) time.def countIncreasing(arr, n): cnt = 0 # Initialize result # Initialize length of current # increasing subarray len = 1 # Traverse through the array for i in range(0, n - 1) : # If arr[i+1] is greater than arr[i], # then increment length if arr[i + 1] > arr[i] : len += 1 # Else Update count and reset length else: cnt += (((len - 1) * len) / 2) len = 1 # If last length is more than 1 if len > 1: cnt += (((len - 1) * len) / 2) return cnt# Driver programarr = [1, 2, 2, 4]n = len(arr)print ("Count of strictly increasing subarrays is", int(countIncreasing(arr, n)))# This code is contributed by Shreyanshi Arun. |
C#
// C# program to count number of strictly// increasing subarraysusing System;class GFG { static int []arr = new int[]{1, 2, 2, 4}; static int countIncreasing(int n) { int cnt = 0; // Initialize result // Initialize length of current // increasing subarray int len = 1; // Traverse through the array for (int i = 0; i < n-1; ++i) { // If arr[i+1] is greater than // arr[i], then increment length if (arr[i + 1] > arr[i]) len++; // Else Update count and reset // length else { cnt += (((len - 1) * len) / 2); len = 1; } } // If last length is more than 1 if (len > 1) cnt += (((len - 1) * len) / 2); return cnt; } // Driver method to test the // above function public static void Main() { Console.WriteLine("Count of strictly " + "increasing subarrays is " + countIncreasing(arr.Length)); }}// This code is contribute by anuj_67. |
PHP
<?php// PHP program to count number of strictly// increasing subarrays in O(n) time.function countIncreasing( $arr, $n){ // Initialize result $cnt = 0; // Initialize length of // current increasing // subarray $len = 1; // Traverse through the array for($i = 0; $i < $n - 1; ++$i) { // If arr[i+1] is greater than arr[i], // then increment length if ($arr[$i + 1] > $arr[$i]) $len++; // Else Update count and // reset length else { $cnt += ((($len - 1) * $len) / 2); $len = 1; } } // If last length is // more than 1 if ($len > 1) $cnt += ((($len - 1) * $len) / 2); return $cnt;}// Driver Code$arr = array(1, 2, 2, 4);$n = count($arr);echo "Count of strictly increasing subarrays is " , countIncreasing($arr, $n);// This code is contribute by anuj_67?> |
Javascript
<script> // Javascript program to count number of strictly // increasing subarrays let arr = [1, 2, 2, 4]; function countIncreasing(n) { let cnt = 0; // Initialize result // Initialize length of current // increasing subarray let len = 1; // Traverse through the array for (let i = 0; i < n-1; ++i) { // If arr[i+1] is greater than // arr[i], then increment length if (arr[i + 1] > arr[i]) len++; // Else Update count and reset // length else { cnt += (((len - 1) * len) / 2); len = 1; } } // If last length is more than 1 if (len > 1) cnt += (((len - 1) * len) / 2); return cnt; } document.write("Count of strictly " + "increasing subarrays is " + countIncreasing(arr.length)); </script> |
Output :
Count of strictly increasing subarrays is 2
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