Minimum Moves To Equal Array Elements

                                            
class Solution {
public:
    int minMoves(vector<int>& nums) {
        int sum = 0;
        int minNum = INT_MAX;
        for (int num : nums) {
            sum += num;
            minNum = min(minNum, num);
        }
        return sum - minNum * nums.size();
    }
};

                                            
class Solution {
    public int minMoves(int[] nums) {
        int min = nums[0];
        int sum = nums[0];
        
        for (int i = 1; i < nums.length; i++) {
            min = Math.min(min, nums[i]);
            sum += nums[i];
        }
        
        return sum - min * nums.length;
    }
}

                                            
class Solution(object):
    def minMoves(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return sum(nums) - min(nums) * len(nums)

                                            
int minMoves(int* nums, int numsSize) {
    int min = INT_MAX;
    long long sum = 0;
    for (int i = 0; i < numsSize; i++) {
        sum += nums[i];
        min = nums[i] < min ? nums[i] : min;
    }
    return sum - min * numsSize;
}

                                            
public class Solution {
    public int MinMoves(int[] nums) {
        int minNum = nums.Min();
        int moves = 0;
        
        foreach (int num in nums) {
            moves += num - minNum;
        }
        
        return moves;
    }
}

                                            
/**
 * @param {number[]} nums
 * @return {number}
 */
const minMoves = (nums) => {
    let min = Math.min(...nums);
    let moves = 0;
    
    for (let num of nums) {
        moves += num - min;
    }
    
    return moves;
};

                                            
function minMoves(nums: number[]): number {
    const minNum = Math.min(...nums);
    let moves = 0;
    
    for (let num of nums) {
        moves += num - minNum;
    }
    
    return moves;
};

                                            
class Solution {
    /**
     * @param Integer[] $nums
     */
    function minMoves($nums) {
        $min = min($nums);
        $moves = 0;
        foreach ($nums as $num) {
            $moves += $num - $min;
        }
        return $moves;
    }
}

                                            
class Solution {
    func minMoves(_ nums: [Int]) -> Int {
        let minNum = nums.min() ?? 0
        var moves = 0
        for num in nums {
            moves += num - minNum
        }
        return moves
    }
}

                                            
class Solution {
    fun minMoves(nums: IntArray): Int {
        val minNum = nums.minOrNull() ?: 0
        var moves = 0
        for (num in nums) {
            moves += num - minNum
        }
        return moves
    }
}

                                            
class Solution {
  int minMoves(List<int> nums) {
    int sum = 0;
    int minNum = nums[0];
    for (int num in nums) {
      sum += num;
      if (num < minNum) {
        minNum = num;
      }
    }
    return sum - nums.length * minNum;
  }
}

                                            
func minMoves(nums []int) int {
    minNum := nums[0]
    sum := nums[0]
    
    for _, num := range nums[1:] {
        if num < minNum {
            minNum = num
        }
        sum += num
    }
    
    return sum - minNum * len(nums)
}

                                            
# @param {Integer[]} nums
# @return {Integer}
def min_moves(nums)
    min_num = nums.min
    moves = 0
    nums.each { |num| moves += num - min_num }
    moves
end

                                            
object Solution {
    def minMoves(nums: Array[Int]): Int = {
        val minNum = nums.min
        var moves = 0
        for (num <- nums) {
            moves += num - minNum
        }
        moves
    }
}

                                            
impl Solution {
    pub fn min_moves(nums: Vec<i32>) -> i32 {
        let min_val = nums.iter().cloned().min().unwrap();
        let mut moves = 0;
        for num in nums {
            moves += num - min_val;
        }
        moves
    }
}

                                            
(define/contract (min-moves nums)
  (-> (listof exact-integer?) exact-integer?)
  (let ((min-val (apply min nums)))
    (- (apply + nums) (* min-val (length nums)))
  )
)

                                            
-spec min_moves(Nums :: [integer()]) -> integer().
min_moves(Nums) ->
    Total = lists:sum(Nums),
    Min = lists:min(Nums),
    Length = length(Nums),
    Total - Min * Length.

                                            
defmodule Solution do
  @spec min_moves(nums :: [integer]) :: integer
  def min_moves(nums) do
    Enum.sum(nums) - length(nums) * Enum.min(nums)
  end
end

To solve this coding challenge, we will focus on finding the minimum number of moves required to make all elements of the array equal, given that in one move, we can increment \( n-1 \) elements of the array by 1. The key insight here is to recognize the concept behind the move — rather than explicitly incrementing \( n-1 \) elements, it is equivalent to reducing one element by 1 each time.

Explanation

We can achieve equal elements by n - 1 increments in each move, which effectively means decreasing the largest element in the array till all elements become the same. To elucidate, consider the array increment process:

You increment \( n-1 \) elements, which is the same as decreasing 1 element \( n-1 \) times.
The challenge can be reframed as: how many moves are required to decrease the larger elements to the smallest element in the array.

This can be thought of as a cumulative reduction problem:

Find the smallest element in the array.
For every other element in the array, determine the difference between the current element and the smallest element.
Sum these differences to get the number of moves.

Now, let’s translate this idea into pseudocode.

Detailed Steps in Pseudocode

Calculate the smallest element in the array.
Initialize a variable to store the total moves.
Iterate through each element in the array and calculate how much each element needs to be decreased to match the smallest element.
Sum these differences to get the total number of moves needed.

Here's a pseudocode representation:

                                            
# Function to find the minimum number of moves to make all array elements equal
function findMinimumMoves(array nums):
# Step 1: Calculate the smallest element in the array
smallest_element = findMinimum(nums)

# Step 2: Initialize a variable to store total moves
total_moves = 0

# Step 3: Iterate through each element and calculate the moves
for each element in nums:
# Calculate the difference from the smallest element
difference = element - smallest_element
# Add the difference to the total moves
total_moves = total_moves + difference

# Step 4: Return the total number of moves
return total_moves

# Helper function to find the smallest element in the array
function findMinimum(array nums):
# Assume the first element is the smallest
smallest = nums[0]
# Iterate through the array to find the smallest element
for each num in nums:
if num < smallest:
smallest = num
return smallest

Detailed Steps in Pseudocode

Finding the Smallest Element :

Initialize a variable (
smallest_element

) to the first element of the array.
Iterate through each element of the array and update the
smallest_element

if a smaller element is found.

Calculating Total Moves :

Initialize a
total_moves

counter to zero.
For each element in the array, compute how much to reduce it to match
smallest_element

and accumulate this in
total_moves

.

Returning the Result :

Simply return the
total_moves

variable which now holds the minimum number of moves needed to make all elements equal.

By breaking down the problem into these steps and translating it into pseudocode, we ensure a clear and understandable approach to solving the challenge, focusing on the underlying logic rather than language-specific syntax.