Find the Cyclomatic Complexity of the following code segment: int average (int[ ] value, int min, int max, int N) { int i, totalValid, sum, mean; i = totalValid = sum = 0; while ( i < N || value[i] != -999 ) { if (value[i] >= min && value[i] <= max){ totalValid += 1; sum += value[i]; if (totalValid > 0) mean = sum / totalValid; else mean = -999; } i += 1; return mean; N.B. N=2466, min = 6, max = 15124 %3!

Java Programming language Please help me with this question

Format Requirement Algorithms in pseudo code MUST be placed in code blocks/fences, and use either the cpp or java syntax highlighter. Algorithms should follow the pseudo code standard described in handout 1.  Your pseudo code should be placed in a code block, for example Algorithm alg_name(a, b) { // A description of the algorithm // input: a - a positive decimal integer; b - an integer // ouput: the sum of a and b c = a + b // or, c <- a + b return c } Do NOT change the template except the answer portion.  Formulas and equations should be in math mode using Latex math symbols.  Markdown math tutorial: http://tug.ctan.org/info/undergradmath/undergradmath.pdf Ways enter into the math mode in Notion: https://www.notion.so/help/math-equations

Find error  #include<bits/stdc++.h> using namespace std; class Solution{     public:     bool isPossible(vector<int>ank,int n,int mid)    {        int count=0;        for(int i=0;i<rank.size())        {            int val= (-1 + sqrt(1+(8*mid)/rank[i]))/2;             count+=val;         }                return count>=n;     }    int findMinTime(int N, vector<int>&A, int L){        int low=*min_element(A.end()),high=1000000;        int ans=high;        while(low<=high)        {            int mid=low+(high)/2;                        if(isPossible(A,mid))            {                ans=mid;                 high=mid;             }            else            low=mid;         }        return ans;    } };   int main() {  int t;  cin>>t;  while(t--)  {      int l;      cin >> l;      vector<int>arr(l);      for(int i = 0; i < l; i++){          cin >> arr[i];      }      Solution ob;      int ans = ob.findMin(n, *arr, l);      cout…