给定一个整数数组arr[] ，任务是通过以下两个操作找到使数组元素相等的最小步数 –
- 从成本为0的元素中加或减 2
- 从具有1 个成本的元素中添加或减去 1
例子：
Input: arr[] = {4, 3, 2, 1}
Output: 2
Explanation:
As in the above example all the array elements can be equal to the 4 or 3 with 2 as minimum cost
Cost of Making Array Elements equal to 4
Index 0: As index 0 is already equal to 4, the cost is 0
Index 1: 4 – 3 = 1, Adding 1 to the element will cost 1
Index 2: 4 – 2 = 2, Adding 2 to the element will cost 0
Index 3: 4 – 1 = 3, Add 2 and 1 once in the array element, which will cost 1, because 2 can be added any number of times with cost 0.
Total Cost = 1 + 1 = 2
Input: arr[] = {3, 2, 1}
Output: 1
Explanation:
As in the above example all the array elements can be made equal to 3 or 1 with 1 as minimum cost
Cost of Making Array Elements equal to 3
Index 0: As index 0 is already equal to 3, the cost is 0
Index 1: 3 – 2 = 1, Adding 1 to the element will cost 1
Index 2: 3 – 1 = 2, Adding 2 to the element will cost 0
Total Cost = 1
方法：这个想法是利用这样一个事实，即向数组的任何元素添加 2 将花费 0。第二个观察结果是，将 2 添加到任何奇数将导致奇数，类似地将 2 添加到偶数将导致只有偶数。因此，任何元素都可以在零成本下等于最接近的整数，但具有相同的属性，即如果它作为初始值是奇数，它将保持奇数，或者即使它作为初始值也将保持偶数。因此，为了找到最小成本，我们可以找到数组中奇数或偶数元素的最小数量。
算法：
- 查找数组中奇数或偶数元素的计数(例如count )。
- 找到 count 和 (length – count) 之间的最小值，其中length是数组的长度。
- 最小值将是使所有数组元素相等的所需成本。
举例说明：
Given Array be - {4, 2, 3, 1}
Count of Odd Elements - 2 ({3, 1})
Count of Even Elements - 2 ({4, 2})
Hence, the minimum cost required is 2 and
the array elements can be made equal
to any integer of even or odd value下面是上述方法的实现：
C++
// C++ implementation to make
// array elements equal with
// minimum cost
#include
using namespace std;
// Function to find the minimum
// cost required to make array
// elements equal
void makearrayequal(int arr[], int n)
{
int x = 0;
// Loop to find the
// count of odd elements
for (int i = 0; i < n; i++) {
x += arr[i] & 1;
}
cout << min(x, n - x) << endl;
}
// Driver Code
int main()
{
int arr[] = { 4, 3, 2, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
makearrayequal(arr, n);
return 0;
} Java
// Java implementation to make
// array elements equal with
// minimum cost
class GFG {
// Function to find the minimum
// cost required to make array
// elements equal
static void makearrayequal(int arr[], int n)
{
int x = 0;
// Loop to find the
// count of odd elements
for (int i = 0; i < n; i++) {
x += (arr[i] & 1);
}
System.out.println(Math.min(x, n - x));
}
// Driver Code
public static void main (String[] args)
{
int arr[] = { 4, 3, 2, 1 };
int n = arr.length;
makearrayequal(arr, n);
}
}
// This code is contributed by Yash_RPython3
# Python3 implementation to make
# array elements equal with
# minimum cost
# Function to find the minimum
# cost required to make array
# elements equal
def makearrayequal(arr, n) :
x = 0;
# Loop to find the
# count of odd elements
for i in range(n) :
x += arr[i] & 1;
print(min(x, n - x));
# Driver Code
if __name__ == "__main__" :
arr = [ 4, 3, 2, 1 ];
n = len(arr);
makearrayequal(arr, n);
# This code is contributed by Yash_RC#
// C# implementation to make
// array elements equal with
// minimum cost
using System;
class GFG {
// Function to find the minimum
// cost required to make array
// elements equal
static void makearrayequal(int []arr, int n)
{
int x = 0;
// Loop to find the
// count of odd elements
for (int i = 0; i < n; i++) {
x += (arr[i] & 1);
}
Console.WriteLine(Math.Min(x, n - x));
}
// Driver Code
public static void Main (string[] args)
{
int []arr = { 4, 3, 2, 1 };
int n = arr.Length;
makearrayequal(arr, n);
}
}
// This code is contributed by Yash_RJavascript
2性能分析：
- 时间复杂度： O(N)。
- 辅助空间： O(1)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live