Given an integer array nums
of size n
, return the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1
elements by 1.
Clarifying Questions
- Constraints on the values within the array
nums
?
- Typically, we would assume they can be any integer, but usually within the range of typical integer limits in Python.
- Can the array be empty or contain only one element?
- Generally, the array will contain at least one element.
- Do we consider negative numbers?
- Yes, the array can contain negative as well as positive numbers.
Strategy
To solve this problem, note that incrementing n-1
elements by 1 is equivalent to decrementing one element by 1 in reverse.
The crucial insight is that to equalize all elements, we need the target value that every element should match at the end:
- The smaller the overall number of operations required will be when converging all elements to the smallest element in the array.
The strategy involves:
- Calculate the sum of the array.
- Find the minimum element in the array.
- Compute the total number of moves required using the formula: ( \text{moves} = \text{sum(nums)} - n \times \text{min(nums)} )
This formula essentially is derived by considering reducing all elements to the smallest element (min(nums)
) directly.
Time Complexity
- Calculating the sum of the array: ( O(n) )
- Finding the minimum element: ( O(n) )
The overall time complexity is ( O(n) ).
Code
def minMoves(nums):
total_sum = sum(nums)
min_num = min(nums)
return total_sum - len(nums) * min_num
# Example Usage
nums = [1, 2, 3]
print(minMoves(nums)) # Output: 3
Explanation
- sum(nums): Computes the total sum of the numbers in the array.
- min(nums): Finds the minimum element in the array.
- ( \text{sum(nums)} - n \times \text{min(nums)} ): The formula we derived to calculate the minimum moves to make all elements equal.
Each operation of reducing elements to the minimum gives us the optimal number of moves required in ( O(n) ) time.
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