从具有和 K 的数组中计数最大可能的对
原文:https://www . geesforgeks . org/count-最大可能对-from-a-array-having-sum-k/
给定一个由 N 个整数和一个整数 K 组成的数组arr【】，任务是从给定的数组中找到最大数量的具有总和 K 的对。
注意:每个数组元素都可以是一对数组的一部分。
示例:
输入: arr[] = {1，2，3，4}，K = 5 输出: 2 说明:数组中 K 之和为(1，4)和(2，3)的对。
输入: arr[] = {3，1，3，4，3}，K = 6 输出: 1 说明:与数组中的和 K 配对为(3，3)。
两点法:思路是使用两点法。按照以下步骤解决问题:
- 将变量和初始化为 0 ，以存储最大对数和 K 。
- 按升序排列数组【arr】。
- 将两个索引变量 L 初始化为 0 ，将 R 初始化为(N–1)，在排序后的数组中查找候选元素。
- 迭代直到 L 小于 R 并执行以下操作:
- 检查arr【L】arr【R】之和是否为 K。如果发现为真，则将 ans 和 L 增加 1 ，将 R 减少 1 。
- 如果arr【L】和arr【R】之和小于 K ，则以 1 递增 L 。
- 否则，将 R 减 1 。
- 完成上述步骤后，打印和的值作为结果。
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 the maximum number
// of pairs from given array with sum K
void maxPairs(int nums[], int n, int k)
{
// Sort array in increasing order
sort(nums, nums + n);
// Stores the final result
int result = 0;
// Initialize the left and right pointers
int start = 0, end = n - 1;
// Traverse array until start < end
while (start < end) {
if (nums[start] + nums[end] > k)
// Decrement right by 1
end--;
else if (nums[start] + nums[end] < k)
// Increment left by 1
start++;
// Increment result and left
// pointer by 1 and decrement
// right pointer by 1
else
{
start++;
end--;
result++;
}
}
// Print the result
cout << result << endl;;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4 };
int n = sizeof(arr)/sizeof(arr[0]);
int K = 5;
// Function Call
maxPairs(arr, n, K);
return 0;
}
// This code is contributed by AnkThon
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.util.*;
class GFG {
// Function to count the maximum number
// of pairs from given array with sum K
public static void maxPairs(int[] nums, int k)
{
// Sort array in increasing order
Arrays.sort(nums);
// Stores the final result
int result = 0;
// Initialize the left and right pointers
int start = 0, end = nums.length - 1;
// Traverse array until start < end
while (start < end) {
if (nums[start] + nums[end] > k)
// Decrement right by 1
end--;
else if (nums[start] + nums[end] < k)
// Increment left by 1
start++;
// Increment result and left
// pointer by 1 and decrement
// right pointer by 1
else {
start++;
end--;
result++;
}
}
// Print the result
System.out.println(result);
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 2, 3, 4 };
int K = 5;
// Function Call
maxPairs(arr, K);
}
}
Python 3
# Python3 program for the above approach
# Function to count the maximum number
# of pairs from given array with sum K
def maxPairs(nums, k):
# Sort array in increasing order
nums = sorted(nums)
# Stores the final result
result = 0
# Initialize the left and right pointers
start, end = 0, len(nums) - 1
# Traverse array until start < end
while (start < end):
if (nums[start] + nums[end] > k):
# Decrement right by 1
end -= 1
elif (nums[start] + nums[end] < k):
# Increment left by 1
start += 1
# Increment result and left
# pointer by 1 and decrement
# right pointer by 1
else:
start += 1
end -= 1
result += 1
# Print the result
print(result)
# Driver Code
if __name__ == '__main__':
arr = [ 1, 2, 3, 4 ]
K = 5
# Function Call
maxPairs(arr, K)
# This code is contributed by mohit kumar 29
C
// C# program for the above approach
using System;
class GFG{
// Function to count the maximum number
// of pairs from given array with sum K
public static void maxPairs(int[] nums, int k)
{
// Sort array in increasing order
Array.Sort(nums);
// Stores the final result
int result = 0;
// Initialize the left and right pointers
int start = 0, end = nums.Length - 1;
// Traverse array until start < end
while (start < end) {
if (nums[start] + nums[end] > k)
// Decrement right by 1
end--;
else if (nums[start] + nums[end] < k)
// Increment left by 1
start++;
// Increment result and left
// pointer by 1 and decrement
// right pointer by 1
else
{
start++;
end--;
result++;
}
}
// Print the result
Console.Write(result);
}
// Driver Code
public static void Main()
{
int[] arr = { 1, 2, 3, 4 };
int K = 5;
// Function Call
maxPairs(arr, K);
}
}
// This code is contributed by susmitakundugoaldanga
java 描述语言
<script>
// JavaScript program for above approach
// Function to count the maximum number
// of pairs from given array with sum K
function maxPairs(nums, k)
{
// Sort array in increasing order
nums.sort();
// Stores the final result
let result = 0;
// Initialize the left and right pointers
let start = 0, end = nums.length - 1;
// Traverse array until start < end
while (start < end) {
if (nums[start] + nums[end] > k)
// Decrement right by 1
end--;
else if (nums[start] + nums[end] < k)
// Increment left by 1
start++;
// Increment result and left
// pointer by 1 and decrement
// right pointer by 1
else {
start++;
end--;
result++;
}
}
// Print the result
document.write(result);
}
// Driver Code
let arr = [ 1, 2, 3, 4 ];
let K = 5;
// Function Call
maxPairs(arr, K);
</script>
输出:
2
时间复杂度: O(Nlog N)* 辅助空间: O(1)
高效方法:要优化上述方法，思路是使用哈希。按照以下步骤解决问题:
- 初始化一个变量，比如说和，用总和 K 存储最大的对数。
- 初始化一个哈希表，比如 S ，以存储arr【】中元素的频率。
- 使用变量遍历数组arr【】，说出 i ，并执行以下步骤:
- 如果(K-arr[I])的频率为正，则将 ans 增加 1 ，将(K-arr[I])的频率减少 1 。
- 否则，在散列表中插入频率为 1 的arr【I】。
- 完成上述步骤后，打印和的值作为结果。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
#include <string.h>
using namespace std;
// Function to find the maximum number
// of pairs with a sum K such that
// same element can't be used twice
void maxPairs(vector<int> nums, int k)
{
// Initialize a hashm
map<int, int> m;
// Store the final result
int result = 0;
// Iterate over the array nums[]
for(auto i : nums)
{
// Decrement its frequency
// in m and increment
// the result by 1
if (m.find(i) != m.end() && m[i] > 0)
{
m[i] = m[i] - 1;
result++;
}
// Increment its frequency by 1
// if it is already present in m.
// Otherwise, set its frequency to 1
else
{
m[k - i] = m[k - i] + 1;
}
}
// Print the result
cout << result;
}
// Driver Code
int main()
{
vector<int> arr = { 1, 2, 3, 4 };
int K = 5;
// Function Call
maxPairs(arr, K);
}
// This code is contributed by grand_master
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.util.*;
class GFG {
// Function to find the maximum number
// of pairs with a sum K such that
// same element can't be used twice
public static void maxPairs(
int[] nums, int k)
{
// Initialize a hashmap
Map<Integer, Integer> map
= new HashMap<>();
// Store the final result
int result = 0;
// Iterate over the array nums[]
for (int i : nums) {
// Decrement its frequency
// in map and increment
// the result by 1
if (map.containsKey(i) &&
map.get(i) > 0)
{
map.put(i, map.get(i) - 1);
result++;
}
// Increment its frequency by 1
// if it is already present in map.
// Otherwise, set its frequency to 1
else
{
map.put(k - i,
map.getOrDefault(k - i, 0) + 1);
}
}
// Print the result
System.out.println(result);
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 2, 3, 4 };
int K = 5;
// Function Call
maxPairs(arr, K);
}
}
Python 3
# Python3 program for the above approach
# Function to find the maximum number
# of pairs with a sum K such that
# same element can't be used twice
def maxPairs(nums, k) :
# Initialize a hashm
m = {}
# Store the final result
result = 0
# Iterate over the array nums[]
for i in nums :
# Decrement its frequency
# in m and increment
# the result by 1
if ((i in m) and m[i] > 0) :
m[i] = m[i] - 1
result += 1
# Increment its frequency by 1
# if it is already present in m.
# Otherwise, set its frequency to 1
else :
if k - i in m :
m[k - i] += 1
else :
m[k - i] = 1
# Print the result
print(result)
# Driver code
arr = [ 1, 2, 3, 4 ]
K = 5
# Function Call
maxPairs(arr, K)
# This code is contributed by divyesh072019
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the maximum number
// of pairs with a sum K such that
// same element can't be used twice
public static void maxPairs(
int[] nums, int k)
{
// Initialize a hashmap
Dictionary<int, int> map
= new Dictionary<int, int>();
// Store the readonly result
int result = 0;
// Iterate over the array nums[]
foreach (int i in nums)
{
// Decrement its frequency
// in map and increment
// the result by 1
if (map.ContainsKey(i) &&
map[i] > 0)
{
map[i] = map[i] - 1;
result++;
}
// Increment its frequency by 1
// if it is already present in map.
// Otherwise, set its frequency to 1
else
{
if (!map.ContainsKey(k - i))
map.Add(k - i, 1);
else
map[i] = map[i] + 1;
}
}
// Print the result
Console.WriteLine(result);
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = {1, 2, 3, 4};
int K = 5;
// Function Call
maxPairs(arr, K);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript program for the above approach
// Function to find the maximum number
// of pairs with a sum K such that
// same element can't be used twice
function maxPairs(nums, k)
{
// Initialize a hashm
var m = new Map();
// Store the final result
var result = 0;
// Iterate over the array nums[]
nums.forEach(i => {
// Decrement its frequency
// in m and increment
// the result by 1
if (m.has(i) && m.get(i) > 0)
{
m.set(i, m.get(i)-1);
result++;
}
// Increment its frequency by 1
// if it is already present in m.
// Otherwise, set its frequency to 1
else
{
if(m.has(k-i))
m.set(k-i, m.get(k-i)+1)
else
m.set(k-i, 1)
}
});
// Print the result
document.write( result);
}
// Driver Code
var arr = [1, 2, 3, 4];
var K = 5;
// Function Call
maxPairs(arr, K);
</script>
Output:
2
时间复杂度:O(N) T5辅助空间:** O(N)
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