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C++ Program to find minimum possible ugliness we can achieve of towers
Suppose we have an array A with n elements. Consider there are n block towers in a row. The ith tower has height A[i]. In a single day, we can perform the operation: Select two indices i and j (i != j) and move back from tower i to j. It will decrease A[i] by 1 and increase A[j] by 1. The ugliness of the buildings is max(A) − min(A). We have to find the minimum possible ugliness we can achieve.
So, if the input is like A = [1, 2, 3, 1, 5], then the output will be 1, because we can do three operations for i=2 and j=0, the new array will now be [2,2,2,1,5], then for i = 4 and j = 3, array will be [2,2,2,2,4], and for i = 4 and j = 2, the array is [2,2,3,2,3].
Steps
To solve this, we will follow these steps −
sum := 0 x := 0 n := size of A for initialize i := 0, when i < n, update (increase i by 1), do: sum := sum + A[i] if sum mod n is same as 0, then: return 0 return 1
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
int solve(vector<int> A) {
int sum = 0, x = 0;
int n = A.size();
for (int i = 0; i < n; i++)
sum += A[i];
if (sum % n == 0)
return 0;
return 1;
}
int main() {
vector<int> A = { 1, 2, 3, 1, 5 };
cout << solve(A) << endl;
}Input
{ 1, 2, 3, 1, 5 }Output
1
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