Minimum Moves to Equal Array Elements II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given an integer array nums
of size n
, return the minimum number of moves required to make all array elements equal.
In one move, you can increment or decrement an element of the array by 1
.
Test cases are designed so that the answer will fit in a 32-bit integer.
Example 1:
Input: nums = [1,2,3] Output: 2 Explanation: Only two moves are needed (remember each move increments or decrements one element): [1,2,3] => [2,2,3] => [2,2,2]
Example 2:
Input: nums = [1,10,2,9] Output: 16
Constraints:
n == nums.length
1 <= nums.length <= 105
-109 <= nums[i] <= 109
C++ Minimum Moves to Equal Array Elements II LeetCode Solution
class Solution {
public:
int minMoves2(vector<int>& nums) {
int n = nums.size(), steps = 0;
nth_element(nums.begin(), nums.begin()+(n/2), nums.end()); //Fixing ths median element
int median = nums[n/2];
for(int i=0; i<n; i++){
steps += abs(nums[i] - median);
}
return steps;
}
};
Java Minimum Moves to Equal Array Elements II LeetCode Solution
public class Solution {
public int minMoves2(int[] nums) {
Arrays.sort(nums);
int i = 0, j = nums.length-1;
int count = 0;
while(i < j){
count += nums[j]-nums[i];
i++;
j--;
}
return count;
}
}
Python 3 Minimum Moves to Equal Array Elements II LeetCode Solution
def minMoves2(self, nums):
median = sorted(nums)[len(nums) / 2]
return sum(abs(num - median) for num in nums)
def minMoves2(self, nums):
nums.sort()
return sum(nums[~i] - nums[i] for i in range(len(nums) / 2))
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