打印所有出现至少 M 次的数组元素
原文:https://www . geesforgeks . org/print-all-array-elements-occurrent-至少-m-times/
给定一个由 N 个整数和一个正整数 M 组成的数组 arr[] ，任务是找出出现至少 M 次的数组元素的数量。
示例:
输入: arr[] = {2，3，2，2，3，5，6，3}，M = 2 输出: 2 3 说明: 在给定数组 arr[]，出现次数至少为 M 次的元素为{2，3}。
输入: arr[] = { 1，32，2，1，33，5，1，5 }，M = 2 T5】输出: 1 5
天真方法:解决问题最简单的方法如下:
- 从左到右遍历数组
- 检查一个元素是否已经出现在前面的遍历中。如果出现，检查下一个元素。否则再次从第个位置到第(n–1)个位置遍历数组。
- 如果频率是 > = M 。打印元素。
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 number of array
// elements with frequency at least M
void printElements(int arr[], int N, int M)
{
// Traverse the array
for (int i = 0; i < N; i++) {
int j;
for (j = i - 1; j >= 0; j--) {
if (arr[i] == arr[j])
break;
}
// If the element appeared before
if (j >= 0)
continue;
// Count frequency of the element
int freq = 0;
for (j = i; j < N; j++) {
if (arr[j] == arr[i])
freq++;
}
if (freq >= M)
cout << arr[i] << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 2, 3, 2, 2, 3, 5, 6, 3 };
int M = 2;
int N = sizeof(arr) / sizeof(arr[0]);
printElements(arr, N, M);
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.io.*;
class GFG{
// Function to find the number of array
// elements with frequency at least M
static void printElements(int[] arr, int N, int M)
{
// Traverse the array
for(int i = 0; i < N; i++)
{
int j;
for(j = i - 1; j >= 0; j--)
{
if (arr[i] == arr[j])
break;
}
// If the element appeared before
if (j >= 0)
continue;
// Count frequency of the element
int freq = 0;
for(j = i; j < N; j++)
{
if (arr[j] == arr[i])
freq++;
}
if (freq >= M)
System.out.print(arr[i] + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 2, 3, 2, 2, 3, 5, 6, 3 };
int M = 2;
int N = arr.length;
printElements(arr, N, M);
}
}
// This code is contributed by subhammahato348
Python 3
# Python3 program for the above approach
# Function to find the number of array
# elements with frequency at least M
def printElements(arr, N, M):
# Traverse the array
for i in range(N):
j = i - 1
while j >= 0:
if (arr[i] == arr[j]):
break
j -= 1
# If the element appeared before
if (j >= 0):
continue
# Count frequency of the element
freq = 0
for j in range(i, N):
if (arr[j] == arr[i]):
freq += 1
if (freq >= M):
print(arr[i], end = " ")
# Driver Code
arr = [ 2, 3, 2, 2, 3, 5, 6, 3 ]
M = 2
N = len(arr)
printElements(arr, N, M)
# This code is contributed by rohitsingh07052
C
// C# program for the above approach
using System;
class GFG{
// Function to find the number of array
// elements with frequency at least M
static void printElements(int[] arr, int N, int M)
{
// Traverse the array
for(int i = 0; i < N; i++)
{
int j;
for(j = i - 1; j >= 0; j--)
{
if (arr[i] == arr[j])
break;
}
// If the element appeared before
if (j >= 0)
continue;
// Count frequency of the element
int freq = 0;
for(j = i; j < N; j++)
{
if (arr[j] == arr[i])
freq++;
}
if (freq >= M)
Console.Write(arr[i] + " ");
}
}
// Driver Code
public static void Main()
{
int[] arr = { 2, 3, 2, 2, 3, 5, 6, 3 };
int M = 2;
int N = arr.Length;
printElements(arr, N, M);
}
}
// This code is contributed by subham348
java 描述语言
<script>
// Javascript program for the above approach
// Function to find the number of array
// elements with frequency at least M
function printElements(arr, N, M)
{
// Traverse the array
for (let i = 0; i < N; i++) {
let j;
for (j = i - 1; j >= 0; j--) {
if (arr[i] == arr[j])
break;
}
// If the element appeared before
if (j >= 0)
continue;
// Count frequency of the element
let freq = 0;
for (j = i; j < N; j++) {
if (arr[j] == arr[i])
freq++;
}
if (freq >= M)
document.write(arr[i] + " ");
}
}
// Driver Code
let arr = [ 2, 3, 2, 2, 3, 5, 6, 3 ];
let M = 2;
let N = arr.length;
printElements(arr, N, M);
// This code is contributed by subham348.
</script>
Output
2 3
方法:给定的问题可以通过将数组元素的频率存储在 HashMap 中，比如说 M ，将频率为的地图中所有元素至少打印 M 来解决。
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 number of array
// elements with frequency at least M
void printElements(int arr[], int N, int M)
{
// Stores the frequency of each
// array elements
unordered_map<int, int> freq;
// Traverse the array
for (int i = 0; i < N; i++) {
// Update frequency of
// current array element
freq[arr[i]]++;
}
// Traverse the map and print array
// elements occurring at least M times
for (auto it : freq) {
if (it.second >= M) {
cout << it.first << " ";
}
}
return;
}
// Driver Code
int main()
{
int arr[] = { 2, 3, 2, 2,
3, 5, 6, 3 };
int M = 2;
int N = sizeof(arr) / sizeof(arr[0]);
printElements(arr, N, M);
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.io.*;
import java.lang.*;
import java.util.*;
class GFG
{
// Function to find the number of array
// elements with frequency at least M
static void printElements(int arr[], int N, int M)
{
// Stores the frequency of each
// array elements
HashMap<Integer, Integer> freq = new HashMap<>();
// Traverse the array
for (int i = 0; i < N; i++)
{
// Update frequency of
// current array element
freq.put(arr[i],
freq.getOrDefault(arr[i], 0) + 1);
}
// Traverse the map and print array
// elements occurring at least M times
for (int key : freq.keySet())
{
if (freq.get(key) >= M) {
System.out.print(key + " ");
}
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 2, 3, 2, 2, 3, 5, 6, 3 };
int M = 2;
int N = arr.length;
printElements(arr, N, M);
}
}
// This code is contributed by Kingash.
Python 3
# Python3 program for the above approach
# Function to find the number of array
# elements with frequency at least M
def printElements(arr, N, M):
# Stores the frequency of each
# array elements
freq = {}
# Traverse the array
for i in arr:
# Update frequency of
# current array element
freq[i] = freq.get(i, 0) + 1
# Traverse the map and prarray
# elements occurring at least M times
for it in freq:
if (freq[it] >= M):
print(it, end = " ")
return
# Driver Code
if __name__ == '__main__':
arr = [2, 3, 2, 2, 3, 5, 6, 3]
M = 2
N = len(arr)
printElements(arr, N, M)
# This code is contributed by mohit kumar 29.
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the number of array
// elements with frequency at least M
static void printElements(int []arr, int N, int M)
{
// Stores the frequency of each
// array elements
Dictionary<int,
int> freq = new Dictionary<int,
int>();
// Traverse the array
for(int i = 0; i < N; i++)
{
// Update frequency of
// current array element
if (freq.ContainsKey(arr[i]))
freq[arr[i]] += 1;
else
freq[arr[i]] = 1;
}
// Traverse the map and print array
// elements occurring at least M times
foreach(var item in freq)
{
if (item.Value >= M)
{
Console.Write(item.Key + " ");
}
}
return;
}
// Driver Code
public static void Main()
{
int []arr = { 2, 3, 2, 2,
3, 5, 6, 3 };
int M = 2;
int N = arr.Length;
printElements(arr, N, M);
}
}
// This code is contributed by SURENDRA_GANGWAR
java 描述语言
<script>
// Javascript program for the above approach
// Function to find the number of array
// elements with frequency at least M
function printElements(arr, N, M) {
// Stores the frequency of each
// array elements
let freq = new Map();
// Traverse the array
for (let i = 0; i < N; i++) {
// Update frequency of
// current array element
freq[arr[i]]++;
if (freq.has(arr[i])) {
freq.set(arr[i], freq.get(arr[i]) + 1)
} else {
freq.set(arr[i], 1)
}
}
// Traverse the map and print array
// elements occurring at least M times
for (let it of freq) {
if (it[1] >= M) {
document.write(it[0] + " ");
}
}
return;
}
// Driver Code
let arr = [2, 3, 2, 2,
3, 5, 6, 3];
let M = 2;
let N = arr.length;
printElements(arr, N, M);
// This code is contributed by gfgking.
</script>
Output
2 3
时间复杂度:O(N) T5辅助空间:** O(N)
方法 2:使用内置的 python 函数:
- 使用 【计数器() 功能计算每个元素的频率
- 遍历频率数组并打印至少出现 m 次的所有元素。
下面是实现:
Python 3
# Python3 implementation
from collections import Counter
# Function to find the number of array
# elements with frequency at least M
def printElements(arr, M):
# Counting frequency of every element using Counter
mp = Counter(arr)
# Traverse the map and print all
# the elements with occurrence atleast m times
for it in mp:
if mp[it] >= M:
print(it, end=" ")
# Driver code
arr = [2, 3, 2, 2, 3, 5, 6, 3]
M = 2
printElements(arr, M)
# This code is contributed by vikkycirus
Output
2 3
时间复杂度: O(N)
辅助空间: O(N)
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