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C++ program to find maximum of each k sized contiguous subarray
Suppose we have an array with n elements and a value k. We shall have to find maximum value for each of the contiguous subarray of size k.
So, if the input is like arr = [3,4,6,2,8], k = 3, then the output will be The contiguous subarrays of size 3 are [3,4,6], [4,6,2], [6,2,8], so the maximum elements are 6, 6 and 8.
To solve this, we will follow these steps −
- Define one deque Qi of size k
- for initialize i := 0, when i < k, update (increase i by 1), do:
- while Qi is not empty and arr[i] >= arr[last element of Qi], do:
- delete last element from Qi
- insert i at the end of Qi
- while Qi is not empty and arr[i] >= arr[last element of Qi], do:
- for i < size of arr, update (increase i by 1), do:
- display arr[first element of Qi]
- while Qi is not empty and first element of Qi <= i - k, do:
- delete front element from Qi
- while Qi is not empty and arr[i] >= arr[last element of Qi], do:
- delete last element from Qi
- insert i at the end of Qi
- display arr[first element of Qi]
Example
Let us see the following implementation to get better understanding −
#include <iostream>
#include <vector>
#include <deque>
using namespace std;
int main(){
vector<int> arr = {3,4,6,2,8};
int k = 3;
deque<int> Qi(k);
int i;
for (i = 0; i < k; ++i){
while ( (!Qi.empty()) && arr[i] >= arr[Qi.back()])
Qi.pop_back();
Qi.push_back(i);
}
for ( ; i < arr.size(); ++i){
cout << arr[Qi.front()] << " ";
while ( (!Qi.empty()) && Qi.front() <= i - k)
Qi.pop_front();
while ( (!Qi.empty()) && arr[i] >= arr[Qi.back()])
Qi.pop_back();
Qi.push_back(i);
}
cout << arr[Qi.front()] << endl;
}
Input
{3,4,6,2,8}, 3Output
6 6 8
Updated on: 07-Oct-2021
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