There are n types of items in a shop's inventory, where the quantity of the ith item is denoted by quantity[i].
These items are to be shipped in two consignments, where the first consignment contains items of type [1, 2, ..., j], and the second
consignment contains the remaining item types, where j can be chosen such that 1 <= j < n. Note that both consignments must be non-empty, and all items of a type must be in the same consignment.
The shopkeeper wants the item counts in each consignment to be equal. To achieve this, the shopkeeper can perform the following move any number of times: increase or decrease the quantity of any item type by 1. The quantity of each item type must remain positive throughout.
Find the minimum number of moves in which the total quantities of both consignments can be made equal if the item types are split optimally.
Function Description
Complete the function getMinimumMoves which has the following parameter:
int quantity[n]: the quantities of each item typeReturns
long_int: the minimum moves required to make the sums equal in an optimal divisionExample 1:
Input: quantity = [1, 4, 4]
Output: 1
Explanation:Considering 1-based indexing, increase quantity[3] by 1, so quantity = [1, 4, 5]. Partition using j = 2 and consignments shipped are [1, 4] and [5]. This is optimal, so return the number of operations, 1.
N/A (If you know about it, feel free to contact us! TYVM!input:
output: