Leetcode 2141 Maximum Running Time of N Computers

[@DevanshAgarwal-tf7jt]

You can avoid doing +1 for mid by doing this as everytime your ans will become 1 less than low:- bool check(long long time, long long n, vector batteries) { long long sum = 0; for (long long i : batteries) { sum += min(time, i); } return (sum >= (long long)time*n); } long long maxRunTime(int n, vector& batteries) { long long sum = 0; for (int i = 0; i < batteries.size(); i++) { sum += batteries[i]; } long long low = 1; long long high = sum / n; long long mid; while (low

[@JeevanKumar-code]

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[@jr_jeet]

Nice sir

[@JeevanKumar-code]

Thank you!

[@leepakshiyadav1643]

amazing, explanation :)

[@JeevanKumar-code]

Thank you!

[@ritikagrover2724]

class Solution { public boolean check(int n, int []B,long time){ long sum=0; for(int i:B){ sum+=Math.min(i,time); } return (sum>=time*n); } public long maxRunTime(int n, int[] batteries) { long sumPow=0; for(int i:batteries) sumPow+=i; long left=1, right=sumPow/n; while(left

[@saudphysics5206]

what is pool and why it is used?

[@prasidmitra6859]

The first approach is O(mlogm) m = no of batteries and the binary search approach is O(mlogk). It's given that k >> m. It seems that the frist one is better. Why use binary search?

[@devverma3168]

why are we doing +1 for calculating mid

[@JeevanKumar-code]

Hey, since we are returning left as the ans and updating l = mid. Say l = 5 and r = 6 mid = 5 and leads to infinite loop

[@geetdeveloper]

if you do not add 1, it will always pick left mid in even length or array, but here we want to pick and try check for right mid element of array, therefore added +1