使 Array 的所有元素小于等于 0 所需的最少操作

给定一个由N 个正数组成的数组arr[] ，任务是找到使数组的所有元素小于或等于0所需的最小操作次数。在每个操作中，必须从数组中选取最小的正元素并从该数字中减去数组的所有元素。

例子：

Input: arr[] = {1, 2, 4, 2, 2, 5, 6}
Output: 5
Explanation: The explanation is mentioned in the diagram below: 
 

 

The resulting array has met the criteria as all it’s elements are either less than or equal to 0 
 

Input: arr[] = {1, 2, 3}
Output: 3

朴素的方法：解决问题的最简单方法是遍历数组，同时数组的所有元素不小于或等于0 ，找到最小的非零正元素，然后从整个数组中减去该元素。

时间复杂度： O(N ² )
辅助空间： O(1)

有效的方法：通过观察答案将是数组中非零不同元素的数量，可以进一步优化上述方法。请按照以下步骤解决问题：

• 初始化一个哈希映射说m存储数组中存在的唯一元素。
• 使用变量i在范围[0, N-1] 中迭代并将m[arr[i]]标记为1 。
• 打印m.size()的值作为答案。

下面是上述方法的实现：

C++

// C++ Program for the above approach
#include 
using namespace std;
 
// Function to find minimum number of
// non-zero elements that has to be subtracted
// such that all the elements are less than or 0
int distinct(vector arr)
{
    // Hash map to mark elements true
    // that are present in the array
    map m;
 
    // Traverse the array
    for (int i = 0; i < arr.size(); i++) {
        // Mark arr[i] true
        m[arr[i]] = true;
    }
 
    // Finally, return the size of hasmap
    return m.size();
}
 
// Driver Code
int main()
{
 
    // Given Input
    vector arr = { 1, 2, 4, 2, 2, 5, 6 };
 
    // Function Call
    int ans = distinct(arr);
    cout << ans << endl;
    return 0;
}

Java

// Java program for the above approach
import java.util.HashMap;
 
class GFG{
     
// Function to find minimum number of
// non-zero elements that has to be subtracted
// such that all the elements are less than or 0
public static int distinct(int[] arr)
{
     
    // Hash map to mark elements true
    // that are present in the array
    HashMap m = new HashMap();
 
    // Traverse the array
    for(int i = 0; i < arr.length; i++)
    {
         
        // Mark arr[i] true
        m.put(arr[i], true);
    }
 
    // Finally, return the size of hasmap
    return m.size();
}
 
// Driver Code
public static void main(String args[])
{
     
    // Given Input
    int[] arr = { 1, 2, 4, 2, 2, 5, 6 };
 
    // Function Call
    int ans = distinct(arr);
     
    System.out.println(ans);
}
}
 
// This code is contributed by gfgking

Python3

# Python 3 Program for the above approach
 
# Function to find minimum number of
# non-zero elements that has to be subtracted
# such that all the elements are less than or 0
def distinct(arr):
 
    # Hash map to mark elements true
    # that are present in the array
    m = {}
 
    # Traverse the array
    for i in range(len(arr)):
        # Mark arr[i] true
        m[arr[i]] = True
 
    # Finally, return the size of hasmap
    return len(m)
 
 
# Driver Code
if __name__ == "__main__":
 
    # Given Input
    arr = [1, 2, 4, 2, 2, 5, 6]
 
    # Function Call
    ans = distinct(arr)
    print(ans)
 
    # This code is contributed by  ukasp.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find minimum number of
// non-zero elements that has to be subtracted
// such that all the elements are less than or 0
static int distinct(List arr)
{
     
    // Hash map to mark elements true
    // that are present in the array
    Dictionary m = new Dictionary();
 
    // Traverse the array
    for(int i = 0; i < arr.Count; i++)
    {
         
        // Mark arr[i] true
        if (m.ContainsKey(arr[i]))
            m[arr[i]] = true;
        else
            m.Add(arr[i],true);
    }
 
    // Finally, return the size of hasmap
    return m.Count;
}
 
// Driver Code
public static void Main()
{
     
    // Given Input
    List arr = new List(){ 1, 2, 4, 2, 2, 5, 6 };
 
    // Function Call
    int ans = distinct(arr);
    Console.Write(ans);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript

 
// JavaScript Program for the above approach
 
 
// Function to find minimum number of
// non-zero elements that has to be subtracted
// such that all the elements are less than or 0
function distinct(arr) {
    // Hash map to mark elements true
    // that are present in the array
    let m = new Map();
 
    // Traverse the array
    for (let i = 0; i < arr.length; i++) {
        // Mark arr[i] true
        m.set(arr[i], true);
    }
 
    // Finally, return the size of hasmap
    return m.size;
}
 
// Driver Code
 
 
// Given Input
let arr = [1, 2, 4, 2, 2, 5, 6];
 
// Function Call
let ans = distinct(arr);
document.write(ans + "<br>");
 

输出：

5

时间复杂度： O(N)
辅助空间： O(1)

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