# 3107. Minimum Operations to Make Median of Array Equal to K

## Description

You are given an integer array nums and a non-negative integer k. In one operation, you can increase or decrease any element by 1.

Return the minimum number of operations needed to make the median of nums equal to k.

The median of an array is defined as the middle element of the array when it is sorted in non-decreasing order. If there are two choices for a median, the larger of the two values is taken.

Example 1:

Input: nums = [2,5,6,8,5], k = 4

Output: 2

Explanation:

We can subtract one from nums[1] and nums[4] to obtain [2, 4, 6, 8, 4]. The median of the resulting array is equal to k.

Example 2:

Input: nums = [2,5,6,8,5], k = 7

Output: 3

Explanation:

We can add one to nums[1] twice and add one to nums[2] once to obtain [2, 7, 7, 8, 5].

Example 3:

Input: nums = [1,2,3,4,5,6], k = 4

Output: 0

Explanation:

The median of the array is already equal to k.

Constraints:

• 1 <= nums.length <= 2 * 105
• 1 <= nums[i] <= 109
• 1 <= k <= 109

## Solutions

### Solution 1: Greedy + Sorting

First, we sort the array $nums$ and find the position $m$ of the median. The initial number of operations we need is $|nums[m] - k|$.

Next, we discuss in two cases:

• If $nums[m] > k$, then all elements to the right of $m$ are greater than or equal to $k$. We only need to reduce the elements greater than $k$ on the left of $m$ to $k$.
• If $nums[m] \le k$, then all elements to the left of $m$ are less than or equal to $k$. We only need to increase the elements less than $k$ on the right of $m$ to $k$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $nums$.

• class Solution {
public long minOperationsToMakeMedianK(int[] nums, int k) {
Arrays.sort(nums);
int n = nums.length;
int m = n >> 1;
long ans = Math.abs(nums[m] - k);
if (nums[m] > k) {
for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
ans += nums[i] - k;
}
} else {
for (int i = m + 1; i < n && nums[i] < k; ++i) {
ans += k - nums[i];
}
}
return ans;
}
}

• class Solution {
public:
long long minOperationsToMakeMedianK(vector<int>& nums, int k) {
sort(nums.begin(), nums.end());
int n = nums.size();
int m = n >> 1;
long long ans = abs(nums[m] - k);
if (nums[m] > k) {
for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
ans += nums[i] - k;
}
} else {
for (int i = m + 1; i < n && nums[i] < k; ++i) {
ans += k - nums[i];
}
}
return ans;
}
};

• class Solution:
def minOperationsToMakeMedianK(self, nums: List[int], k: int) -> int:
nums.sort()
n = len(nums)
m = n >> 1
ans = abs(nums[m] - k)
if nums[m] > k:
for i in range(m - 1, -1, -1):
if nums[i] <= k:
break
ans += nums[i] - k
else:
for i in range(m + 1, n):
if nums[i] >= k:
break
ans += k - nums[i]
return ans


• func minOperationsToMakeMedianK(nums []int, k int) (ans int64) {
sort.Ints(nums)
n := len(nums)
m := n >> 1
ans = int64(abs(nums[m] - k))
if nums[m] > k {
for i := m - 1; i >= 0 && nums[i] > k; i-- {
ans += int64(nums[i] - k)
}
} else {
for i := m + 1; i < n && nums[i] < k; i++ {
ans += int64(k - nums[i])
}
}
return
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• function minOperationsToMakeMedianK(nums: number[], k: number): number {
nums.sort((a, b) => a - b);
const n = nums.length;
const m = n >> 1;
let ans = Math.abs(nums[m] - k);
if (nums[m] > k) {
for (let i = m - 1; i >= 0 && nums[i] > k; --i) {
ans += nums[i] - k;
}
} else {
for (let i = m + 1; i < n && nums[i] < k; ++i) {
ans += k - nums[i];
}
}
return ans;
}