Minimum Moves To Equal Array Elements Ii

                                            
class Solution {
public:
    int minMoves2(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int i = 0, j = nums.size() - 1;
        int moves = 0;
        while (i < j) {
            moves += nums[j] - nums[i];
            i++;
            j--;
        }
        return moves;
    }
};

                                            
class Solution {
    public int minMoves2(int[] nums) {
        Arrays.sort(nums);
        int median = nums[nums.length / 2];
        int moves = 0;
        for (int num : nums) {
            moves += Math.abs(num - median);
        }
        return moves;
    }
}

                                            
class Solution(object):
    def minMoves2(self, nums):
        nums.sort()
        mid = nums[len(nums)//2]
        return sum(abs(num - mid) for num in nums)

                                            
public class Solution {
    public int MinMoves2(int[] nums) {
        Array.Sort(nums);
        int median = nums[nums.Length / 2];
        int moves = 0;
        
        foreach (int num in nums) {
            moves += Math.Abs(num - median);
        }
        
        return moves;
    }
}

                                            
/**
 * @param {number[]} nums
 * @return {number}
 */
const minMoves2 = function(nums) {
    nums.sort((a, b) => a - b);
    let mid = nums[Math.floor(nums.length / 2)];
    let moves = 0;
    
    for (let num of nums) {
        moves += Math.abs(num - mid);
    }
    
    return moves;
};

                                            
function minMoves2(nums: number[]): number {
    nums.sort((a, b) => a - b);
    let mid = nums[Math.floor(nums.length / 2)];
    let moves = 0;
    for (let num of nums) {
        moves += Math.abs(num - mid);
    }
    return moves;
};

                                            
class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function minMoves2($nums) {
        sort($nums);
        $n = count($nums);
        $median = $nums[floor($n / 2)];
        
        $moves = 0;
        foreach ($nums as $num) {
            $moves += abs($num - $median);
        }
        
        return $moves;
    }
}

                                            
class Solution {
    func minMoves2(_ nums: [Int]) -> Int {
        let sortedNums = nums.sorted()
        var moves = 0
        var left = 0
        var right = sortedNums.count - 1
        
        while left < right {
            moves += sortedNums[right] - sortedNums[left]
            left += 1
            right -= 1
        }
        
        return moves
    }
}

                                            
class Solution {
    fun minMoves2(nums: IntArray): Int {
        nums.sort()
        var minMoves = 0
        var left = 0
        var right = nums.size - 1
        while (left < right) {
            minMoves += nums[right] - nums[left]
            left++
            right--
        }
        return minMoves
    }
}

                                            
class Solution {
  int minMoves2(List<int> nums) {
    nums.sort();
    
    int mid = nums[nums.length ~/ 2];
    int moves = 0;
    
    for (int num in nums) {
      moves += (num - mid).abs();
    }
    
    return moves;
  }
}

                                            
func minMoves2(nums []int) int {
    sort.Ints(nums)
    
    median := nums[len(nums)/2]
    
    moves := 0
    for _, num := range nums {
        moves += abs(num - median)
    }
    
    return moves
}

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

                                            
object Solution {
    def minMoves2(nums: Array[Int]): Int = {
        val sortedNums = nums.sorted
        val median = sortedNums(sortedNums.length / 2)
        sortedNums.map(num => math.abs(num - median)).sum
    }
}

                                            
impl Solution {
    pub fn min_moves2(nums: Vec<i32>) -> i32 {
        let mut nums = nums;
        nums.sort();

        let mut moves = 0;
        let median = nums[nums.len() / 2];

        for num in nums {
            moves += (num - median).abs();
        }

        moves
    }
}

                                            
(define/contract (min-moves2 nums)
  (-> (listof exact-integer?) exact-integer?)
  
  (define sorted-nums (sort nums <))
  (define target (list-ref sorted-nums (quotient (length sorted-nums) 2)))
  
  (for/sum ([num sorted-nums])
    (abs (- num target))
    )
  )

                                            
-spec min_moves2(Nums :: [integer()]) -> integer().
min_moves2(Nums) ->
    lists:sum(lists:map(fun(X) -> abs(X - lists:nth((length(Nums) div 2) + 1, lists:sort(Nums))) end, Nums)).

To solve this coding challenge, we need to find the minimum number of moves required to make all elements in an integer array equal. Each move consists of incrementing or decrementing an element by 1. The key to solving this challenge lies in understanding that the optimal target value for all elements — which minimizes the number of moves — is the median of the array.

Explanation

Sorting the Array : By sorting the array, the median becomes a straightforward choice for the final value of all elements. This is because the median minimizes the sum of absolute deviations.
Finding the Median : Once the array is sorted, the median can be found easily. If the array length is odd, the median is the middle element. If even, the median can be either of the two middle values (though practically, any value between these middle two values works to minimize the moves).
Calculating Moves : With the median in hand, we calculate the total number of moves required to make each element equal to the median by summing up the absolute differences between each element and the median.

Let's now break this down step-by-step in pseudocode with detailed comments explaining each part.

Step-by-Step Explanation

Step 1: Sorting the Array

Sorting the array is crucial because finding the median in an unsorted array is inefficient. Sorting allows us to directly access the median.

Step 2: Finding the Median

The median is the optimal value that minimizes the sum of absolute differences. If the array length is odd, the median is simply the middle element. If even, the average of the two middle values serves as the median (as any value between these works ideally).

Step 3: Summing Up Moves

For each element in the array, calculate the absolute difference between that element and the median. Sum these differences to get the total number of moves.

Detailed Steps in Pseudocode

                                            
# Pseudocode for the Minimum Moves to Equal Array Elements II problem

# Step 1: Sort the array
# Input: nums (array of integers)
# Output: sorted array

function sortArray(nums):
    # Use a sorting algorithm to sort nums
    # This can be any efficient sorting algorithm like quicksort or mergesort
    sort(nums)
    return nums

# Step 2: Find the median of the sorted array
# Input: sorted_nums (sorted array of integers)
# Output: median (the median value of the sorted array)
function findMedian(sorted_nums):
    # Get the length of the array
    length = length(sorted_nums)
    
    # Find the median
    if length % 2 == 1:
        # If the length is odd, return the middle element
        median = sorted_nums[length // 2]
    else:
        # If the length is even, return the average of the two middle elements
        mid1 = sorted_nums[(length // 2) - 1]
        mid2 = sorted_nums[length // 2]
        median = (mid1 + mid2) / 2  # Generally, for this problem, we can use either mid1 or mid2
    return median

# Step 3: Calculate the total number of moves to equalize the array
# Input: nums (original array of integers), median (median value from sorted array)
# Output: total_moves (minimum number of moves needed)
function calculateMoves(nums, median):
    total_moves = 0
    # Loop through each element in the array
    for each num in nums:
        # Add the absolute difference between the element and the median to total_moves
        total_moves += abs(num - median)
    return total_moves

# Main function to solve the problem
# Input: nums (array of integers)
# Output: total_moves (minimum number of moves needed to equalize the array)
function minMoves2(nums):
    # Step 1: Sort the array
    sorted_nums = sortArray(nums)
    
    # Step 2: Find the median
    median = findMedian(sorted_nums)
    
    # Step 3: Calculate the total number of moves
    total_moves = calculateMoves(nums, median)
    
    return total_moves

This pseudocode structure ensures that each step in the code is clear, efficient, and follows a logical sequence to solve the problem. By sorting the array and then using the median, we efficiently find the minimum total number of moves needed.