Maximum Average Subarray II in C++

Suppose we have an array with n integers, we have to find the contiguous subarray whose length is greater than or equal to k that has the maximum average value. We have to find the maximum average value.

So, if the input is like [1,12,-5,-6,50,3], k = 4, then the output will be 12.75, as when length is 5, maximum average value is 10.8, when length is 6, maximum average value is 9.16667. Thus output is 12.75.

To solve this, we will follow these steps −

• Define a function ok(), this will take x, an array nums, k,

• n := size of nums

• Define an array arr of size: n.

• for initialize i := 0, when i < n, update (increase i by 1), do −

  • arr[i] := nums[i] - x

• sum := 0, last := 0

• for initialize i := 0, when i < k, update (increase i by 1), do −

  • sum := sum + arr[i]

• if sum >= 0, then −

  • return true

• for initialize i := 0, j := k, when j < n, update (increase i by 1), (increase j by 1), do −

  • last := last + arr[i]

  • sum := sum + arr[j]

  • if last < 0, then −

    • sum := sum - last

    • last := 0

  • if sum >= 0, then −

    • return true

• return false

• From the main method do the following −

• ret := 0, low := -inf, high := inf

• while high - low > 10^-5, do

  • mid := low + (high - low) / 2

  • if ok(mid, nums, k) is true, then −

    • low := mid

    • ret := mid

  • Otherwise

    • high := mid

• return ret

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   bool ok(double x, vector <int>& nums, int k){
      int n = nums.size();
      double arr[n];
      for (int i = 0; i < n; i++) {
         arr[i] = nums[i] - x;
      }
      double sum = 0;
      double last = 0;
      for (int i = 0; i < k; i++) {
         sum += arr[i];
      }
      if (sum >= 0)
      return true;
      for (int i = 0, j = k; j < n; i++, j++) {
         last += arr[i];
         sum += arr[j];
         if (last < 0) {
            sum -= last;
            last = 0;
         }
         if (sum >= 0)
         return true;
      }
      return false;
   }
   double findMaxAverage(vector<int>& nums, int k) {
      double ret = 0;
      double low = INT_MIN;
      double high = INT_MAX;
      while (high - low > 1e-5) {
         double mid = low + (high - low) / 2;
         if (ok(mid, nums, k)) {
            low = mid;
            ret = mid;
         } else {
            high = mid;
         }
      }
      return ret;
   }
};
main(){
   Solution ob;
   vector<int> v = {1,12,-5,-6,50,3};
   cout << (ob.findMaxAverage(v, 4));
}

Input

{1,12,-5,-6,50,3},4

Output

12.75000