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Number of non-decreasing sub-arrays of length K

 3 years ago
source link: https://www.geeksforgeeks.org/number-of-non-decreasing-sub-arrays-of-length-k/
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Number of non-decreasing sub-arrays of length K
  • Last Updated : 06 Nov, 2019

Given an array arr[] of length N, the task is to find the number of non-decreasing sub-arrays of length K.

Examples:

Input: arr[] = {1, 2, 3, 2, 5}, K = 2
Output: 3
{1, 2}, {2, 3} and {2, 5} are the increasing
subarrays of length 2.

Input: arr[] = {1, 2, 3, 2, 5}, K = 1
Output: 5

Naive approach Generate all the sub-arrays of length K and then check whether the sub-array satisfies the condition. Thus, the time complexity of the approach will be O(N * K).

Better approach: A better approach will be using two-pointer technique. Let’s say the current index is i.

  • Find the largest index j, such that the sub-array arr[i…j] is non-decreasing. This can be achieved by simply incrementing the value of j starting from i + 1 and checking whether arr[j] is greater than arr[j – 1].
  • Let’s say the length of the sub-array found in the previous step is L. The number of sub-arrays of length K contained in it will be max(L – K + 1, 0).
  • Now, update i = j and repeat the above steps while i is in the index range.

Below is the implementation of the above approach:

  • Python3

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of
// increasing subarrays of length k
int cntSubArrays(int* arr, int n, int k)
{
// To store the final result
int res = 0;
int i = 0;
// Two pointer loop
while (i < n) {
// Initialising j
int j = i + 1;
// Looping till the subarray increases
while (j < n and arr[j] >= arr[j - 1])
j++;
// Updating the required count
res += max(j - i - k + 1, 0);
// Updating i
i = j;
}
// Returning res
return res;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 2, 5 };
int n = sizeof(arr) / sizeof(int);
int k = 2;
cout << cntSubArrays(arr, n, k);
return 0;
}
Output: 
3

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