Maximum Increase | Editorial by khushalagarwal

This is a simple problem, with an even simpler solution. We simply transverse across the array comparing each element with the one before it, and maintaining two counters, one for the length of current increasing sequence, and other for the length of overall maxima till this point.

Even better, we don't need to store the array, we can transverse by just storing the current, and the element before it. So, the memory required is lesser.

After an analysis of this algorithm, we can say that it runs in θ(n) time.
 We just track the previous and next element and store the max continous length in a variable.
 
```
#include<bits/stdc++.h>
using namespace std;
int main() {
    int n;
    cin >> n;
    int pre, next;
    int ctr = 0, ctr1 = 0;
    long long int arr[n];
    cin >> pre;
    for (int i = 1; i < n; i++) {
        cin >> next;
        if (pre < next) {
            ctr++;
            ctr1 = max(ctr, ctr1);
        } else {
            ctr = 0;
        }
        pre = next;
    }
    cout << ctr1 + 1;
}
```

Maximum Increase | Editorial by khushalagarwal

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