Brute force approach is too slow the best approach or the solution for this is the frequency array approach (C++ solution)

`int pickingNumbers(vector a) {

int maxLength = 0;

//brute force approach on n^2
/for (int i = 0; i < a.size(); i++) { int count = 0;
for (int j = 0; j < a.size(); j++) { if (a[j] == a[i] || a[j] == a[i] + 1) { count++; } } maxLength = max(maxLength, count); }/

//frequency array approach vector freq(101, 0); for (int i = 0; i < a.size(); i++) { freq[a[i]]++; // Increment the count for number a[i] }

for (int i = 1; i < 100; i++) { maxLength = max(maxLength, freq[i] + freq[i + 1]); } return maxLength;

def pickingNumbers(a): se=list(set(a)) maxi=-99 for i in se: ma=0 ma=a.count(i)+max(a.count(i-1),a.count(i+1)) maxi=max(ma,maxi) return maxi

The Picking Numbers problem typically involves finding the longest subarray where the absolute difference between any two elements is at most 1. Are you looking for an efficient way to implement this in code? 2. How can we implement code related to Cookout Barbecue Menu for some delicious options?

C++ Solution using sliding window:

int pickingNumbers(vector<int> a) {
    sort(a.begin(),a.end());
    int slow = 0;
    int fast = 0;
    int len = 1;
    while(fast <= a.size()){
       if(abs((a[fast]-a[slow]))<=1 ){
        if((fast-slow)+1>len){
            len = (fast-slow)+1;
        }
        fast++;
       }else if(slow == fast){
         fast++;
       }else{
        slow++;
       }
    }
    return len;
}

def pickingNumbers(a): se=list(set(a)) maxi=-99 for i in se: ma=0 ma=a.count(i)+max(a.count(i-1),a.count(i+1)) maxi=max(ma,maxi) return maxi