从数组元素中减去的值的最小和，使所有数组元素相等
原文:https://www . geeksforgeeks . org/最小值-从数组元素中减去-使所有数组元素相等/
给定一个由正整数 N 组成的数组arr【】，任务是找出需要从每个数组元素中减去的所有数组元素的和，使得剩余的数组元素都相等。
示例:
输入: arr[] = {1，2} 输出: 1 解释:从 arr[1]中减去 1 将 arr[]修改为{1，1}。因此，所需的总和为 1。
输入: arr[] = {1，2，3} 输出: 3 解释:从 arr[1]和 arr[2]中减去 1 和 2 将 arr[]修改为{1，1，1}。因此，所需的总和= 1 + 2 = 3。
方法:想法是将所有数组元素减少到数组中存在的最小元素。按照以下步骤解决问题:
- 初始化一个变量，比如和，存储所有减去的值的和。
- 使用 min_element() 找到数组中存在的最小元素，说最小。
- 遍历数组，对于每个数组元素，说出 arr[i] ，将(arr[I]–最小值)加到所需的和上。
- 完成数组遍历后，打印得到的和。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the sum of values
// removed to make all array elements equal
int minValue(int arr[], int n)
{
// Stores the minimum of the array
int minimum = *min_element(
arr, arr + n);
// Stores required sum
int sum = 0;
// Traverse the array
for (int i = 0; i < n; i++) {
// Add the value subtracted
// from the current element
sum = sum + (arr[i] - minimum);
}
// Return the total sum
return sum;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << minValue(arr, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.Arrays;
class GFG
{
// Function to find the sum of values
// removed to make all array elements equal
static int minValue(int []arr, int n)
{
Arrays.sort(arr);
// Stores the minimum of the array
int minimum = arr[0];
// Stores required sum
int sum = 0;
// Traverse the array
for(int i = 0; i < n; i++)
{
// Add the value subtracted
// from the current element
sum = sum + (arr[i] - minimum);
}
// Return the total sum
return sum;
}
// Driver Code
static public void main(String args[])
{
int []arr = { 1, 2, 3 };
int N = arr.length;
// Function Call
System.out.println(minValue(arr, N));
}
}
// This code is contributed by AnkThon
Python 3
# Python3 program for the above approach
# Function to find the sum of values
# removed to make all array elements equal
def minValue(arr, n):
# Stores the minimum of the array
minimum = min(arr)
# Stores required sum
sum = 0
# Traverse the array
for i in range(n):
# Add the value subtracted
# from the current element
sum = sum + (arr[i] - minimum)
# Return the total sum
return sum
# Driver Code
if __name__ == '__main__':
arr = [ 1, 2, 3 ]
N = len(arr)
# Function Call
print(minValue(arr, N))
# This code is contributed by mohit kumar 29
C
// C# program for the above approach
using System;
class GFG{
// Function to find the sum of values
// removed to make all array elements equal
static int minValue(int []arr, int n)
{
Array.Sort(arr);
// Stores the minimum of the array
int minimum = arr[0];
// Stores required sum
int sum = 0;
// Traverse the array
for(int i = 0; i < n; i++)
{
// Add the value subtracted
// from the current element
sum = sum + (arr[i] - minimum);
}
// Return the total sum
return sum;
}
// Driver Code
static public void Main ()
{
int []arr = { 1, 2, 3 };
int N = arr.Length;
// Function Call
Console.WriteLine(minValue(arr, N));
}
}
// This code is contributed by AnkThon
java 描述语言
<script>
// Javascript program for the above approach
// Function to find the sum of values
// removed to make all array elements equal
function minValue(arr, n)
{
// Stores the minimum of the array
var minimum = Math.min.apply(Math,arr);
// Stores required sum
var sum = 0;
var i;
// Traverse the array
for (i = 0; i < n; i++) {
// Add the value subtracted
// from the current element
sum = sum + (arr[i] - minimum);
}
// Return the total sum
return sum;
}
// Driver Code
var arr = [1, 2, 3];
var N = arr.length;
// Function Call
document.write(minValue(arr, N));
</script>
Output:
3
时间复杂度:O(N) T5辅助空间:** O(1)
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