通过替换由相等元素组成的最小子序列，使所有数组元素等于 0

给定一个大小为N的数组arr[] ，任务是通过用任意整数、最少次数替换相等元素的子序列的所有元素来使所有数组元素等于0 。

例子：

Input: arr[] = {3, 7, 3}, N = 3
Output: 2
Explanation:
Selecting a subsequence { 7 } and replacing all its elements by 0 modifies arr[] to { 3, 3, 3 }. 
Selecting the array { 3, 3, 3 } and replacing all its elements by 0 modifies arr[] to { 0, 0, 0 }

Input: arr[] = {1, 5, 1, 3, 2, 3, 1}, N = 7
Output: 4

方法：该问题可以使用贪心技术解决。这个想法是计算数组中存在的不等于 0 的不同元素并打印获得的计数。请按照以下步骤解决问题：

• 初始化 Set 以存储数组中存在的不同元素，该元素不等于0 。
• 遍历数组arr[]并将数组元素插入到 Set 中。
• 最后，打印 Set 的大小。

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 count of operations
// required to convert all array elements to zero
// br replacing subsequence of equal elements to 0
void minOpsToTurnArrToZero(int arr[], int N)
{
 
    // Store distinct elements
    // present in the array
    unordered_set st;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // If arr[i] is already present in
        // the Set or arr[i] is equal to 0
        if (st.find(arr[i]) != st.end()
            || arr[i] == 0) {
            continue;
        }
 
        // Otherwise, increment ans by
        // 1 and insert current element
        else {
            st.insert(arr[i]);
        }
    }
 
    cout << st.size() << endl;
}
 
// Driver Code
int main()
{
 
    // Given array
    int arr[] = { 3, 7, 3 };
 
    // Size of the given array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    minOpsToTurnArrToZero(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function to find minimum count of operations
    // required to convert all array elements to zero
    // br replacing subsequence of equal elements to 0
    static void minOpsToTurnArrToZero(int[] arr, int N)
    {
 
        // Store distinct elements
        // present in the array
        Set st = new HashSet();
        // Traverse the array
        for (int i = 0; i < N; i++) {
 
            // If arr[i] is already present in
            // the Set or arr[i] is equal to 0
            if (st.contains(arr[i]) || arr[i] == 0) {
                continue;
            }
 
            // Otherwise, increment ans by
            // 1 and insert current element
            else {
                st.add(arr[i]);
            }
        }
 
        System.out.println(st.size());
    }
 
    // Driver Code
    public static void main(String args[])
    {
        // Given array
        int arr[] = { 3, 7, 3 };
 
        // Size of the given array
        int N = arr.length;
 
        minOpsToTurnArrToZero(arr, N);
    }
}
 
// This code is contributed by 18bhupendrayadav18

Python3

# Python3 program for the above approach
 
# Function to find minimum count of
# operations required to convert all
# array elements to zero by replacing
# subsequence of equal elements to 0
def minOpsToTurnArrToZero(arr, N):
     
    # Store distinct elements
    # present in the array
    st = dict()
 
    # Traverse the array
    for i in range(N):
 
        # If arr[i] is already present in
        # the Set or arr[i] is equal to 0
        if (i in st.keys() or arr[i] == 0):
            continue
         
        # Otherwise, increment ans by
        # 1 and insert current element
        else:
            st[arr[i]] = 1
             
    print(len(st))
 
# Driver Code
 
# Given array
arr = [ 3, 7, 3 ]
 
# Size of the given array
N = len(arr)
 
minOpsToTurnArrToZero(arr, N)
 
# This code is contributed by susmitakundugoaldanga

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG {
 
  // Function to find minimum count of operations
  // required to convert all array elements to zero
  // br replacing subsequence of equal elements to 0
  static void minOpsToTurnArrToZero(int[] arr, int N)
  {
 
    // Store distinct elements
    // present in the array
    HashSet st = new HashSet();
 
    // Traverse the array
    for (int i = 0; i < N; i++)
    {
 
      // If arr[i] is already present in
      // the Set or arr[i] is equal to 0
      if (st.Contains(arr[i]) || arr[i] == 0)
      {
        continue;
      }
 
      // Otherwise, increment ans by
      // 1 and insert current element
      else
      {
        st.Add(arr[i]);
      }
    }
    Console.WriteLine(st.Count);
  }
 
  // Driver Code
  public static void Main(String []args)
  {
 
    // Given array
    int []arr = { 3, 7, 3 };
 
    // Size of the given array
    int N = arr.Length;
    minOpsToTurnArrToZero(arr, N);
  }
}
 
// This code is contributed by gauravrajput1

Javascript

 
// Javascript program for the above approach
 
// Function to find minimum count of operations
// required to convert all array elements to zero
// br replacing subsequence of equal elements to 0
function minOpsToTurnArrToZero(arr, N)
{
     
    // Store distinct elements
    // present in the array
    var st = new Set();
 
    // Traverse the array
    for(var i = 0; i < N; i++)
    {
         
        // If arr[i] is already present in
        // the Set or arr[i] is equal to 0
        if (st.has(arr[i]) || arr[i] == 0)
        {
            continue;
        }
 
        // Otherwise, increment ans by
        // 1 and insert current element
        else
        {
            st.add(arr[i]);
        }
    }
    document.write(st.size)
}
 
// Driver Code
 
// Given array
var arr = [ 3, 7, 3 ];
 
// Size of the given array
var N = arr.length;
 
minOpsToTurnArrToZero(arr, N);
 
// This code is contributed by noob2000
 

输出：

2

时间复杂度： O(N)
辅助空间： O(N)

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