# C++ Solution to Coding Challenge 462. Minimum Moves to Equal Array Elements II

## Median - The math behind the problem

### Problem statement

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`

.

#### 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 <= 10^5`

.`-10^9 <= nums[i] <= 10^9`

.

### Solution 1: Median - The math behind the problem

You are asked to move all elements of an array to the same value `M`

. The problem can be reduced to identifying what `M`

is.

First, moving elements of an unsorted array and moving a sorted one are the same. So you can assume `nums`

is sorted in some order. Let us say it is sorted in ascending order.

Second, `M`

must be in between the minimum element and the maximum one. Apparently!

We will prove that `M`

will be the median of `nums`

, which is `nums[n/2]`

of the sorted `nums`

.

In other words, we will prove that if you choose `M`

a value different from `nums[n/2]`

then the number of moves will be increased.

In fact, if you choose `M = nums[n/2] + x`

, where `x > 0`

, then:

Each element

`nums[i]`

that is less than`M`

needs more`x`

moves, while each`nums[j]`

that is greater than`M`

can reduce`x`

moves.But the number of

`nums[i]`

is bigger than the number of`nums[j]`

.So the total number of moves is bigger.

The same arguments apply for `x < 0`

.

#### Example 3

For `nums = [0,1,2,2,10]`

. Its median is `2`

. The minimum number of moves is `2 + 1 + 0 + 0 + 8 = 11`

.

If you choose `M = 3`

(the average value, the mean), the total number of moves is `3 + 2 + 1 + 1 + 7 = 14`

.

#### Code

```
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int minMoves2(vector<int>& nums) {
sort(nums.begin(), nums.end());
const int median = nums[nums.size() / 2];
int moves = 0;
for (int& a: nums) {
moves += abs(a - median);
}
return moves;
}
int main() {
vector<int> nums{1,2,3};
cout << minMoves2(nums) << endl;
nums = {1,10,2,9};
cout << minMoves2(nums) << endl;
}
```

```
Output:
2
16
```

#### Complexity

Runtime:

`O(nlogn)`

, where`n = nums.length`

.Extra space:

`O(1)`

.

### Solution 2: Using `std::nth_element`

to compute the median

What you only need in Solution 1 is the median value. Computing the total number of moves in the `for`

loop does not require the array `nums`

to be fully sorted.

In this case, you can use `std::nth_element`

to reduce the runtime complexity.

#### Code

```
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int minMoves2(vector<int>& nums) {
const int mid = nums.size() / 2;
std::nth_element(nums.begin(), nums.begin() + mid, nums.end());
const int median = nums[mid];
int moves = 0;
for (int& a: nums) {
moves += abs(a - median);
}
return moves;
}
int main() {
vector<int> nums{1,2,3};
cout << minMoves2(nums) << endl;
nums = {1,10,2,9};
cout << minMoves2(nums) << endl;
}
```

```
Output:
2
16
```

#### Complexity

Runtime:

`O(n)`

, where`n = nums.length`

.Extra space:

`O(1)`

.

### Modern C++ notes

In the code of Solution 2, the partial sorting algorithm `std::nth_element`

will make sure for all indices `i`

and `j`

that satisfy `0 <= i <= mid <= j < nums.length`

,

```
nums[i] <= nums[mid] <= nums[j].
```

With this property, if `mid = nums.length / 2`

then the value of `nums[mid]`

is unchanged no matter how `nums`

is sorted or not.

### References

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