Aggressive cows

#1

Hey , why is end = stalls[n-1]-stalls[0] it should [n-1] only

#2

hello @soumyabhardwaj75
the first stall is present at stall[0]
the last stall is present at stall[n-1]
then what is maximum sepration possible between two cows?
it should be possible when we put one cow at stall[0] and othercow at stall[n-1]
and distance will be stall[n-1]-stall[0]

#3

we have used int min_space but we haven’t provided any value to it

#4

#include using namespace std; bool canPlaceCows(int stalls[], int n, int cows , int min_sep) { int last_cow = stalls[0]; // plpace last cow in the first stall int cnt = 1; for (int i = 0; i < n; i++) { if (stalls[i] - last_cow >= min_sep) { last_cow = stalls[i]; cnt++; if (cnt == cows) { return true; } } } return false; } int main () { int stalls[] = {1, 2, 4, 8, 9}; int n = 5; int cows = 3; // binary search algorithm int s = 0; int e = stalls[n - 1] - stalls[0]; int ans = 0; while (s <= e) { int mid = (s + e) / 2; bool cowRakhpaye = canPlaceCows(stalls, n, mid, cows); if (cowRakhpaye) { // agar cow rakh paye to usse bade space me try krenge ans = mid; s = mid + 1; } else { e = mid - 1; } } cout << ans << endl; return 0; }

#5

min_step is argument of the function, we dont have to initialise /assign any value to it.

when we call the function while passing value then at that time min_step will get initialised by passed value

#6

https://ide.codingblocks.com/s/466831 can u tell what’s the problem with this code ? it is not working properly

#7

check now->

#8

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