竞争性编程的频率测量技术
原文:https://www . geeksforgeeks . org/frequency-measurement-technologies-for-competitive-programming/
测量数组中元素的频率是一项非常方便的技能，并且需要许多竞争性的编码问题。在许多问题中，我们需要测量各种元素的频率，如数字、字母、符号等。作为我们问题的一部分。
天真法
示例:
Input : arr[] = {10, 20, 20, 10, 10, 20, 5, 20}
Output : 10 3
20 4
5 1
Input : arr[] = {10, 20, 20}
Output : 10 2
20 1
我们运行两个循环。对于每个项目计数的次数。为了避免重复打印，请跟踪已处理的项目。
C++
// C++ program to count frequencies of array items
#include <bits/stdc++.h>
using namespace std;
void countFreq(int arr[], int n)
{
// Mark all array elements as not visited
vector<int> visited(n, false);
// Traverse through array elements and
// count frequencies
for (int i = 0; i < n; i++) {
// Skip this element if already processed
if (visited[i] == true)
continue;
// Count frequency
int count = 1;
for (int j = i + 1; j < n; j++) {
if (arr[i] == arr[j]) {
visited[j] = true;
count++;
}
}
cout << arr[i] << " " << count << endl;
}
}
int main()
{
int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
int n = sizeof(arr) / sizeof(arr[0]);
countFreq(arr, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to count frequencies
// of array items
import java.util.*;
class GFG
{
static void countFreq(int arr[], int n)
{
// Mark all array elements as not visited
boolean []visited = new boolean[n];
// Traverse through array elements and
// count frequencies
for (int i = 0; i < n; i++)
{
// Skip this element if already processed
if (visited[i] == true)
continue;
// Count frequency
int count = 1;
for (int j = i + 1; j < n; j++)
{
if (arr[i] == arr[j])
{
visited[j] = true;
count++;
}
}
System.out.println(arr[i] + " " + count);
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
int n = arr.length;
countFreq(arr, n);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python program to count frequencies
# of array items
def countFreq(arr, n):
# mark all array elements as not visited
visited = [False for i in range(n)]
# Traverse through array elements
# and count frequencies
for i in range(n):
# Skip this element if already processed
if visited[i] == True:
continue
# count frequency
count = 1
for j in range(i + 1, n):
if arr[i] == arr[j]:
visited[j] = True
count += 1
print(arr[i], count)
# Driver code
a = [10, 20, 20, 10, 10, 20, 5, 20]
n = len(a)
countFreq(a, n)
# This code is contributed
# by Mohit kumar 29
C
// C# program to count frequencies
// of array items
using System;
class GFG
{
static void countFreq(int []arr, int n)
{
// Mark all array elements as not visited
Boolean []visited = new Boolean[n];
// Traverse through array elements and
// count frequencies
for (int i = 0; i < n; i++)
{
// Skip this element if already processed
if (visited[i] == true)
continue;
// Count frequency
int count = 1;
for (int j = i + 1; j < n; j++)
{
if (arr[i] == arr[j])
{
visited[j] = true;
count++;
}
}
Console.WriteLine(arr[i] + " " + count);
}
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 10, 20, 20, 10,
10, 20, 5, 20 };
int n = arr.Length;
countFreq(arr, n);
}
}
// This code is contributed by Rajput-Ji
java 描述语言
<script>
// Javascript program to count frequencies of array items
function countFreq(arr, n)
{
// Mark all array elements as not visited
let visited = new Array(n);
visited.fill(false);
// Traverse through array elements and
// count frequencies
for (let i = 0; i < n; i++) {
// Skip this element if already processed
if (visited[i] == true)
continue;
// Count frequency
let count = 1;
for (let j = i + 1; j < n; j++) {
if (arr[i] == arr[j]) {
visited[j] = true;
count++;
}
}
document.write(arr[i] + " " + count + "</br>");
}
}
let arr = [ 10, 20, 20, 10, 10, 20, 5, 20 ];
let n = arr.length;
countFreq(arr, n);
// This code is contributed by mukesh07.
</script>
Output:
10 3
20 4
5 1
优化方法:
测量元素受值限制时的频率 如果我们的输入数组有小值，我们可以使用数组元素作为计数数组中的索引，并增加计数。在下面的示例中，元素最多为 10 个。
Input : arr[] = {5, 5, 6, 6, 5, 6, 1, 2, 3, 10, 10}
limit = 10
Output : 1 1
2 1
3 1
5 3
6 3
10 2
C++
// C++ program to count frequencies of array items
// having small values.
#include <bits/stdc++.h>
using namespace std;
void countFreq(int arr[], int n, int limit)
{
// Create an array to store counts. The size
// of array is limit+1 and all values are
// initially 0
vector<int> count(limit+1, 0);
// Traverse through array elements and
// count frequencies (assuming that elements
// are limited by limit)
for (int i = 0; i < n; i++)
count[arr[i]]++;
for (int i = 0; i <= limit; i++)
if (count[i] > 0)
cout << i << " " << count[i] << endl;
}
int main()
{
int arr[] = {5, 5, 6, 6, 5, 6, 1, 2, 3, 10, 10};
int n = sizeof(arr) / sizeof(arr[0]);
int limit = 10;
countFreq(arr, n, limit);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to count frequencies of array items
// having small values.
class GFG
{
static void countFreq(int arr[], int n, int limit)
{
// Create an array to store counts. The size
// of array is limit+1 and all values are
// initially 0
int []count = new int[limit + 1];
// Traverse through array elements and
// count frequencies (assuming that elements
// are limited by limit)
for (int i = 0; i < n; i++)
count[arr[i]]++;
for (int i = 0; i <= limit; i++)
if (count[i] > 0)
System.out.println(i + " " + count[i]);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = {5, 5, 6, 6, 5, 6, 1, 2, 3, 10, 10};
int n = arr.length;
int limit = 10;
countFreq(arr, n, limit);
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python3 program to count frequencies of
# array items having small values.
def countFreq(arr, n, limit):
# Create an array to store counts.
# The size of array is limit+1 and
# all values are initially 0
count = [0 for i in range(limit + 1)]
# Traverse through array elements and
# count frequencies (assuming that
# elements are limited by limit)
for i in range(n):
count[arr[i]] += 1
for i in range(limit + 1):
if (count[i] > 0):
print(i, count[i])
# Driver Code
arr = [ 5, 5, 6, 6, 5, 6,
1, 2, 3, 10, 10 ]
n = len(arr)
limit = 10
countFreq(arr, n, limit)
# This code is contributed by avanitrachhadiya2155
C
// C# program to count frequencies of
// array items having small values.
using System;
class GFG
{
static void countFreq(int []arr,
int n, int limit)
{
// Create an array to store counts.
// The size of array is limit+1 and
// all values are initially 0
int []count = new int[limit + 1];
// Traverse through array elements and
// count frequencies (assuming that
// elements are limited by limit)
for (int i = 0; i < n; i++)
count[arr[i]]++;
for (int i = 0; i <= limit; i++)
if (count[i] > 0)
Console.WriteLine(i + " " +
count[i]);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = {5, 5, 6, 6, 5, 6,
1, 2, 3, 10, 10};
int n = arr.Length;
int limit = 10;
countFreq(arr, n, limit);
}
}
// This code is contributed
// by Princi Singh
java 描述语言
<script>
// Javascript program to count frequencies
// of array items having small values.
function countFreq(arr, n, limit)
{
// Create an array to store counts.
// The size of array is limit+1 and
// all values are initially 0
let count = new Array(limit + 1);
count.fill(0);
// Traverse through array elements and
// count frequencies (assuming that
// elements are limited by limit)
for(let i = 0; i < n; i++)
count[arr[i]]++;
for(let i = 0; i <= limit; i++)
if (count[i] > 0)
document.write(i + " " +
count[i] + "</br>");
}
// Driver code
let arr = [ 5, 5, 6, 6, 5, 6,
1, 2, 3, 10, 10 ];
let n = arr.length;
let limit = 10;
countFreq(arr, n, limit);
// This code is contributed by rameshtravel07
</script>
Output:
1 1
2 1
3 1
5 3
6 3
10 2
C++
// C++ program to count frequencies of array items
#include <bits/stdc++.h>
using namespace std;
const int limit = 255;
void countFreq(string str)
{
// Create an array to store counts. The size
// of array is limit+1 and all values are
// initially 0
vector<int> count(limit+1, 0);
// Traverse through string characters and
// count frequencies
for (int i = 0; i < str.length(); i++)
count[str[i]]++;
for (int i = 0; i <= limit; i++)
if (count[i] > 0)
cout << (char)i << " " << count[i] << endl;
}
int main()
{
string str = "GeeksforGeeks";
countFreq(str);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to count frequencies of array items
class GFG
{
static int limit = 255;
static void countFreq(String str)
{
// Create an array to store counts. The size
// of array is limit+1 and all values are
// initially 0
int []count= new int[limit + 1];
// Traverse through string characters and
// count frequencies
for (int i = 0; i < str.length(); i++)
count[str.charAt(i)]++;
for (int i = 0; i <= limit; i++)
if (count[i] > 0)
System.out.println((char)i + " " + count[i]);
}
// Driver Code
public static void main(String[] args)
{
String str = "GeeksforGeeks";
countFreq(str);
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python3 program to count frequencies of array items
limit = 255
def countFreq(Str) :
# Create an array to store counts. The size
# of array is limit+1 and all values are
# initially 0
count = [0] * (limit + 1)
# Traverse through string characters and
# count frequencies
for i in range(len(Str)) :
count[ord(Str[i])] += 1
for i in range(limit + 1) :
if (count[i] > 0) :
print(chr(i), count[i])
Str = "GeeksforGeeks"
countFreq(Str)
# This code is contributed by divyeshrabadiya07
C
// C# program to count frequencies
// of array items
using System;
class GFG
{
static int limit = 25;
static void countFreq(String str)
{
// Create an array to store counts.
// The size of array is limit+1 and
// all values are initially 0
int []count = new int[limit + 1];
// Traverse through string characters
// and count frequencies
for (int i = 0; i < str.Length; i++)
count[str[i] - 'A']++;
for (int i = 0; i <= limit; i++)
if (count[i] > 0)
Console.WriteLine((char)(i + 'A') +
" " + count[i]);
}
// Driver Code
public static void Main(String[] args)
{
String str = "GEEKSFORGEEKS";
countFreq(str);
}
}
// This code is contributed by PrinciRaj1992
java 描述语言
<script>
// Javascript program to count frequencies of array items
let limit = 255;
function countFreq(str)
{
// Create an array to store counts. The size
// of array is limit+1 and all values are
// initially 0
let count= new Array(limit + 1);
for(let i=0;i<count.length;i++)
{
count[i]=0;
}
// Traverse through string characters and
// count frequencies
for (let i = 0; i < str.length; i++)
count[str[i].charCodeAt(0)]++;
for (let i = 0; i <= limit; i++)
{ if (count[i] > 0)
document.write(String.fromCharCode(i) + " " + count[i]+"<br>");
}
}
// Driver Code
let str = "GeeksforGeeks";
countFreq(str);
// This code is contributed by unknown2108
</script>
Output:
G 2
e 4
f 1
k 2
o 1
r 1
s 2
测量元素在有限范围内的频率 例如，考虑一个只包含大写字母的字符串。字符串的元素限制在从“A”到“Z”的范围内。其思想是减去最小的元素(本例中为“A”)，得到元素的索引。
C++
// C++ program to count frequencies of array items
#include <bits/stdc++.h>
using namespace std;
const int limit = 25;
void countFreq(string str)
{
// Create an array to store counts. The size
// of array is limit+1 and all values are
// initially 0
vector<int> count(limit+1, 0);
// Traverse through string characters and
// count frequencies
for (int i = 0; i < str.length(); i++)
count[str[i] - 'A']++;
for (int i = 0; i <= limit; i++)
if (count[i] > 0)
cout << (char)(i + 'A') << " " << count[i] << endl;
}
int main()
{
string str = "GEEKSFORGEEKS";
countFreq(str);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to count frequencies
// of array items
import java.util.*;
class GFG
{
static int limit = 25;
static void countFreq(String str)
{
// Create an array to store counts.
// The size of array is limit+1 and
// all values are initially 0
int []count = new int[limit + 1];
// Traverse through string characters
// and count frequencies
for (int i = 0; i < str.length(); i++)
count[str.charAt(i) - 'A']++;
for (int i = 0; i <= limit; i++)
if (count[i] > 0)
System.out.println((char)(i + 'A') +
" " + count[i]);
}
// Driver Code
public static void main(String[] args)
{
String str = "GEEKSFORGEEKS";
countFreq(str);
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python3 program to count frequencies of array items
limit = 25;
def countFreq(str):
# Create an array to store counts. The size
# of array is limit+1 and all values are
# initially 0
count = [0 for i in range(limit + 1)]
# Traverse through string characters and
# count frequencies
for i in range(len(str)):
count[ord(str[i]) - ord('A')] += 1
for i in range(limit + 1):
if (count[i] > 0):
print(chr(i + ord('A')), count[i])
# Driver code
if __name__=='__main__':
str = "GEEKSFORGEEKS";
countFreq(str);
# This code is contributed by Pratham76
C
// C# program to count frequencies
// of array items
using System;
class GFG
{
static int limit = 25;
static void countFreq(String str)
{
// Create an array to store counts.
// The size of array is limit+1 and
// all values are initially 0
int []count = new int[limit + 1];
// Traverse through string characters
// and count frequencies
for (int i = 0; i < str.Length; i++)
count[str[i] - 'A']++;
for (int i = 0; i <= limit; i++)
if (count[i] > 0)
Console.WriteLine((char)(i + 'A') +
" " + count[i]);
}
// Driver Code
public static void Main(String[] args)
{
String str = "GEEKSFORGEEKS";
countFreq(str);
}
}
// This code contributed by PrinciRaj1992
java 描述语言
<script>
// JavaScript program to count frequencies
// of array items
let limit = 25;
function countFreq(str)
{
// Create an array to store counts.
// The size of array is limit+1 and
// all values are initially 0
let count = new Array(limit + 1);
for(let i=0;i<limit+1;i++)
{
count[i]=0;
}
// Traverse through string characters
// and count frequencies
for (let i = 0; i < str.length; i++)
count[str[i].charCodeAt(0) - 'A'.charCodeAt(0)]++;
for (let i = 0; i <= limit; i++)
if (count[i] > 0)
document.write(String.fromCharCode(i + 'A'.charCodeAt(0)) +
" " + count[i]+"<br>");
}
// Driver Code
let str = "GEEKSFORGEEKS";
countFreq(str);
// This code is contributed by rag2127
</script>
Output:
E 4
F 1
G 2
K 2
O 1
R 1
S 2
无范围无限制的情况下测量频率 想法是用哈希法(c++中的无序 _ map和HashMap【Java 中的 T6】来获取频率。
C++
// C++ program to count frequencies of array items
#include <bits/stdc++.h>
using namespace std;
void countFreq(int arr[], int n)
{
unordered_map<int, int> mp;
// Traverse through array elements and
// count frequencies
for (int i = 0; i < n; i++)
mp[arr[i]]++;
// Traverse through map and print frequencies
for (auto x : mp)
cout << x.first << " " << x.second << endl;
}
int main()
{
int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
int n = sizeof(arr) / sizeof(arr[0]);
countFreq(arr, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to count frequencies of array items
import java.util.*;
class GFG
{
static void countFreq(int arr[], int n)
{
HashMap<Integer,
Integer>mp = new HashMap<Integer,
Integer>();
// Traverse through array elements and
// count frequencies
for (int i = 0 ; i < n; i++)
{
if(mp.containsKey(arr[i]))
{
mp.put(arr[i], mp.get(arr[i]) + 1);
}
else
{
mp.put(arr[i], 1);
}
}
// Traverse through map and print frequencies
for (Map.Entry<Integer,
Integer> entry : mp.entrySet())
System.out.println(entry.getKey() + " " +
entry.getValue());
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
int n = arr.length;
countFreq(arr, n);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 program to count frequencies of array items
def countFreq(arr, n):
mp = {}
# Traverse through array elements and
# count frequencies
for i in range(n):
if arr[i] in mp:
mp[arr[i]] += 1
else:
mp[arr[i]] = 1
# Traverse through map and print frequencies
for x in sorted(mp):
print(x, mp[x])
# Driver Code
arr = [ 10, 20, 20, 10, 10, 20, 5, 20 ]
n = len(arr)
countFreq(arr, n)
# This code is contributed by divyesh072019
C
// C# program to count frequencies of array items
using System;
using System.Collections.Generic;
class GFG
{
static void countFreq(int []arr, int n)
{
Dictionary<int,
int> mp = new Dictionary<int,
int>();
// Traverse through array elements and
// count frequencies
for (int i = 0 ; i < n; i++)
{
if(mp.ContainsKey(arr[i]))
{
var val = mp[arr[i]];
mp.Remove(arr[i]);
mp.Add(arr[i], val + 1);
}
else
{
mp.Add(arr[i], 1);
}
}
// Traverse through map and print frequencies
foreach(KeyValuePair<int, int> entry in mp)
Console.WriteLine(entry.Key + " " +
entry.Value);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 10, 20, 20, 10,
10, 20, 5, 20 };
int n = arr.Length;
countFreq(arr, n);
}
}
// This code is contributed by Princi Singh
java 描述语言
<script>
// Javascript program to count frequencies of array items
function countFreq(arr, n)
{
let mp = new Map();
// Traverse through array elements and
// count frequencies
for (let i = 0 ; i < n; i++)
{
if(mp.has(arr[i]))
{
mp.set(arr[i], mp.get(arr[i]) + 1);
}
else
{
mp.set(arr[i], 1);
}
}
// Traverse through map and print frequencies
for (let [key, value] of mp.entries())
document.write(key + " " +
value+"<br>");
}
// Driver Code
let arr=[10, 20, 20, 10, 10, 20, 5, 20];
let n = arr.length;
countFreq(arr, n);
// This code is contributed by ab2127
</script>
Output:
5 1
10 3
20 4
时间复杂度: O(n) 辅助空间: O(n)
在上述高效解决方案中，如何按照元素在输入中出现的顺序打印元素？
C++
// C++ program to count frequencies of array items
#include <bits/stdc++.h>
using namespace std;
void countFreq(int arr[], int n)
{
unordered_map<int, int> mp;
// Traverse through array elements and
// count frequencies
for (int i = 0; i < n; i++)
mp[arr[i]]++;
// To print elements according to first
// occurrence, traverse array one more time
// print frequencies of elements and mark
// frequencies as -1 so that same element
// is not printed multiple times.
for (int i = 0; i < n; i++) {
if (mp[arr[i]] != -1)
{
cout << arr[i] << " " << mp[arr[i]] << endl;
mp[arr[i]] = -1;
}
}
}
int main()
{
int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
int n = sizeof(arr) / sizeof(arr[0]);
countFreq(arr, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to count frequencies of array items
import java.util.*;
class GFG
{
static void countFreq(int arr[], int n)
{
HashMap<Integer,
Integer> mp = new HashMap<Integer,
Integer>();
// Traverse through array elements and
// count frequencies
for (int i = 0 ; i < n; i++)
{
if(mp.containsKey(arr[i]))
{
mp.put(arr[i], mp.get(arr[i]) + 1);
}
else
{
mp.put(arr[i], 1);
}
}
// To print elements according to first
// occurrence, traverse array one more time
// print frequencies of elements and mark
// frequencies as -1 so that same element
// is not printed multiple times.
for (int i = 0; i < n; i++)
{
if (mp.get(arr[i]) != -1)
{
System.out.println(arr[i] + " " +
mp.get(arr[i]));
mp. put(arr[i], -1);
}
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
int n = arr.length;
countFreq(arr, n);
}
}
// This code is contributed by Princi Singh
Python 3
# Python3 program to count frequencies of array items
def countFreq(arr, n):
mp = {}
# Traverse through array elements and
# count frequencies
for i in range(n):
if arr[i] not in mp:
mp[arr[i]] = 1
else:
mp[arr[i]] += 1
# To print elements according to first
# occurrence, traverse array one more time
# print frequencies of elements and mark
# frequencies as -1 so that same element
# is not printed multiple times.
for i in range(n):
if(mp[arr[i]] != -1):
print(arr[i], mp[arr[i]])
mp[arr[i]] = -1
# Driver code
arr = [10, 20, 20, 10, 10, 20, 5, 20 ]
n = len(arr)
countFreq(arr, n)
# This code is contributed by rag2127
C
// C# program to count frequencies of array items
using System;
using System.Collections.Generic;
class GFG
{
static void countFreq(int []arr, int n)
{
Dictionary<int,
int> mp = new Dictionary<int,
int>();
// Traverse through array elements and
// count frequencies
for (int i = 0 ; i < n; i++)
{
if(mp.ContainsKey(arr[i]))
{
mp[arr[i]] = mp[arr[i]] + 1;
}
else
{
mp.Add(arr[i], 1);
}
}
// To print elements according to first
// occurrence, traverse array one more time
// print frequencies of elements and mark
// frequencies as -1 so that same element
// is not printed multiple times.
for (int i = 0; i < n; i++)
{
if (mp[arr[i]] != -1)
{
Console.WriteLine(arr[i] + " " +
mp[arr[i]]);
mp[arr[i]] = - 1;
}
}
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 10, 20, 20, 10,
10, 20, 5, 20 };
int n = arr.Length;
countFreq(arr, n);
}
}
// This code is contributed by Rajput-Ji
java 描述语言
<script>
// Javascript program to count
// frequencies of array items
function countFreq(arr, n)
{
let mp = new Map();
// Traverse through array elements and
// count frequencies
for(let i = 0 ; i < n; i++)
{
if (mp.has(arr[i]))
{
mp.set(arr[i], mp.get(arr[i]) + 1);
}
else
{
mp.set(arr[i], 1);
}
}
// To print elements according to first
// occurrence, traverse array one more time
// print frequencies of elements and mark
// frequencies as -1 so that same element
// is not printed multiple times.
for(let i = 0; i < n; i++)
{
if (mp.get(arr[i]) != -1)
{
document.write(arr[i] + " " +
mp.get(arr[i]) + "<br>");
mp.set(arr[i], -1);
}
}
}
// Driver Code
let arr = [ 10, 20, 20, 10, 10, 20, 5, 20 ];
let n = arr.length;
countFreq(arr, n);
// This code is contributed by patel2127
</script>
Output:
10 3
20 4
5 1
时间复杂度: O(n) 辅助空间: O(n)
在 Java 中，我们可以使用 LinkedHashMap 以相同的顺序获取元素。因此，我们不需要额外的循环。 很多问题都是基于频率测量的，如果我们知道如何计算给定数组中各种元素的频率，那将是一个难题。例如，尝试下面给出的基于频率测量的问题:
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