对给定数组中最高 2 次幂小于或等于该数的数组元素进行计数
原文:https://www . geesforgeks . org/count-array-elements-其最高 2 次方小于或等于给定数组中存在的数字/
给定一个由 N 个正整数组成的数组arr【】，任务是找出数组元素的数量，其的最高 2 次幂小于或等于数组中存在的数量。
示例:
输入: arr[] = {3，4，6，9} 输出: 2 说明: 数组中存在 2 个数组元素(4 和 6)，其 2 的最高幂小于它(即 4)。
输入: arr[] = {3，9，10，8，1 } T3】输出: 3
天真方法:给定的问题可以通过计算给定数组中的最高幂为 2 的元素来解决，并且可以通过再次遍历数组来找到这些元素。检查所有元素后，打印获得的总计数。
时间复杂度:O(N2) 辅助空间: O(1)
高效方法:上述方法也可以通过使用无序映射来保持访问元素的计数并相应地更新结果元素的计数来优化。按照以下步骤解决给定的问题:
- 初始化一个变量，比如计数，它存储数组中最高次幂 2 小于或等于次幂的元素的计数。
- 初始化一张图 M ，存储数组元素的频率。
- 遍历给定数组，对于每个元素，如果地图中存在不超过该元素的arr【I】的 2 的最高次幂的频率，则将计数的值增加 1。
- 完成上述步骤后，打印计数的值作为元素的合成计数。
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 count array elements
// whose highest power of 2 is less
// than or equal to that number is
// present in the given array
int countElement(int arr[], int N)
{
// Stores the resultant count
// of array elements
int count = 0;
// Stores frequency of
// visited array elements
unordered_map<int, int> m;
// Traverse the array
for (int i = 0; i < N; i++) {
m[arr[i]]++;
}
for (int i = 0; i < N; i++) {
// Calculate log base 2
// of the element arr[i]
int lg = log2(arr[i]);
// Highest power of 2 whose
// value is at most arr[i]
int p = pow(2, lg);
// Increment the count by 1
if (m[p]) {
count++;
}
}
// Return the resultant count
return count;
}
// Driver Code
int main()
{
int arr[] = { 3, 4, 6, 9 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << countElement(arr, N);
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.util.HashMap;
import java.io.*;
class GFG{
static int log2(int N)
{
// Calculate log2 N indirectly
// using log() method
int result = (int)(Math.log(N) /
Math.log(2));
return result;
}
// Function to count array elements
// whose highest power of 2 is less
// than or equal to that number is
// present in the given array
static int countElement(int arr[], int N)
{
// Stores the resultant count
// of array elements
int count = 0;
// Stores frequency of
// visited array elements
HashMap<Integer, Integer> m = new HashMap<>();
// Traverse the array
for(int i = 0; i < N; i++)
{
if (m.containsKey(arr[i]))
{
m.put(arr[i], m.get(arr[i]) + 1);
}
else
{
m.put(arr[i], 1);
}
}
for(int i = 0; i < N; i++)
{
// Calculate log base 2
// of the element arr[i]
int lg = log2(arr[i]);
// Highest power of 2 whose
// value is at most arr[i]
int p = (int)Math.pow(2, lg);
// Increment the count by 1
if (m.containsKey(p))
{
count++;
}
}
// Return the resultant count
return count;
}
// Driver Code
public static void main (String[] args)
{
int arr[] = { 3, 4, 6, 9 };
int N = arr.length;
System.out.println(countElement(arr, N));
}
}
// This code is contributed by Potta Lokesh
Python 3
# Python program for the above approach
from math import log2
# Function to count array elements
# whose highest power of 2 is less
# than or equal to that number is
# present in the given array
def countElement(arr, N):
# Stores the resultant count
# of array elements
count = 0
# Stores frequency of
# visited array elements
m = {}
# Traverse the array
for i in range(N):
m[arr[i]] = m.get(arr[i], 0) + 1
for i in range(N):
# Calculate log base 2
# of the element arr[i]
lg = int(log2(arr[i]))
# Highest power of 2 whose
# value is at most arr[i]
p = pow(2, lg)
# Increment the count by 1
if (p in m):
count += 1
# Return the resultant count
return count
# Driver Code
if __name__ == '__main__':
arr= [3, 4, 6, 9]
N = len(arr)
print (countElement(arr, N))
# 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 count array elements
// whose highest power of 2 is less
// than or equal to that number is
// present in the given array
static int countElement(int []arr, int N)
{
// Stores the resultant count
// of array elements
int count = 1;
// Stores frequency of
// visited array elements
Dictionary<int,int> m = new Dictionary<int,int>();
// Traverse the array
for (int i = 0; i < N; i++) {
if(m.ContainsKey(arr[i]))
m[arr[i]]++;
else
m.Add(arr[i],1);
}
for(int i = 0; i < N; i++) {
// Calculate log base 2
// of the element arr[i]
int lg = (int)Math.Log(arr[i]);
// Highest power of 2 whose
// value is at most arr[i]
int p = (int)Math.Pow(2, lg);
// Increment the count by 1
if (m.ContainsKey(p)) {
count++;
}
}
// Return the resultant count
return count;
}
// Driver Code
public static void Main()
{
int []arr = { 3, 4, 6, 9 };
int N = arr.Length;
Console.Write(countElement(arr, N));
}
}
// This code is contributed by bgangwar59.
java 描述语言
<script>
// Javascript program for the above approach
// Function to count array elements
// whose highest power of 2 is less
// than or equal to that number is
// present in the given array
function countElement(arr, N) {
// Stores the resultant count
// of array elements
let count = 0;
// Stores frequency of
// visited array elements
let m = new Map();
// Traverse the array
for (let i = 0; i < N; i++) {
if (m.has(arr[i])) {
m.set(arr[i], m.get(arr[i]) + 1)
} else {
m.set(arr[i], 1)
}
}
for (let i = 0; i < N; i++) {
// Calculate log base 2
// of the element arr[i]
let lg = Math.floor(Math.log2(arr[i]));
// Highest power of 2 whose
// value is at most arr[i]
let p = Math.pow(2, lg);
// Increment the count by 1
if (m.get(p)) {
count++;
}
}
// Return the resultant count
return count;
}
// Driver Code
let arr = [3, 4, 6, 9];
let N = arr.length
document.write(countElement(arr, N));
// This code is contributed by _saurabh_jaiswal.
</script>
Output:
2
时间复杂度:O(N) T5辅助空间:** O(N)
版权属于：月萌API www.moonapi.com，转载请注明出处
