Suppose we have a value n, consider an array nums with n elements, where arr[i] = (2*i)+1 for all i. Now in one operation, we can choose two indices x and y where 0 <= x, y < n and subtract 1 from nums[x] and add 1 to nums[y]. We have to make all the elements of the array same. So if we have n we have to find the minimum number of operations required to make all the elements of nums same.
So, if the input is like n = 4, then the output will be 4 as for n the array is [1,3,5,7], now after first operation we can make array like [2,3,5,6], after second operation we can make [3,3,5,5],then after third [4,3,4,5] and finally [4,4,4,4]. So now all are equal.
To solve this, we will follow these steps −
if n is same as 1, then
q:= quotient of (n/2) -1
while q>=0, do
ans:= ans + (n-j)
q := q - 1
j := j + 2
Let us see the following implementation to get better understanding −
def solve(n): ans=0 if n==1: return ans q=(n//2)-1 j=1 while q>=0: ans=ans+(n-j) q-=1 j+=2 return ans n = 4 print(solve(n))