与数组中小于K的最大和配对

给定大小为N的数组arr和整数K。任务是找到一对整数，使它们的总和最大但小于K
例子：

Input : arr = {30, 20, 50} , K = 70 
Output : 30, 20 
30 + 20 = 50 which is maximum possible sum which is less than K
Input : arr = {5, 20, 110, 100, 10}, K = 85 
Output : 20, 10  

方法 ：
一种有效的方法是对给定的数组进行排序并找到大于或等于K的元素。如果在索引p处找到，我们必须仅在arr [0，…，p-1]之间找到对。运行嵌套循环。一个照顾一对中的第一个元素，另一个照顾一对中的第二个元素。保持变量maxsum以及其他两个变量a和b来跟踪可能的解决方案。将maxsum初始化为0 。找出A [i] + A [j] (假设i在外循环中运行，j在内循环中运行)。如果它大于maxsum，则用该和更新maxsum并将a和b更改为该数组的第i和j个元素。
下面是上述方法的实现：

C++

// CPP program to find pair with largest
// sum which is less than K in the array
#include 
using namespace std;
 
// Function to find pair with largest
// sum which is less than K in the array
void Max_Sum(int arr[], int n, int k)
{
    // To store the break point
    int  p = n;
     
    // Sort the given array
    sort(arr, arr + n);
     
    // Find the break point
    for (int i = 0; i < n; i++)
    {
         // No need to look beyond i'th index
        if (arr[i] >= k) {
            p = i;
            break;
        }
    }
     
     
    int maxsum = 0, a, b;
     
    // Find the required pair
    for (int i = 0; i < p; i++)
    {
        for (int j = i + 1; j < p; j++)
        {
            if (arr[i] + arr[j] < k and arr[i] + arr[j] >
                                                     maxsum)
            {
                maxsum = arr[i] + arr[j];
                 
                a = arr[i];
                b = arr[j];
            }
        }
    }
     
    // Print the required answer
    cout << a << " " << b;
}
 
// Driver code
int main()
{
    int arr[] = {5, 20, 110, 100, 10}, k = 85;
     
    int n = sizeof(arr) / sizeof(arr[0]);
     
    // Function call
    Max_Sum(arr, n, k);
     
    return 0;
}

Java

// Java program to find pair with largest
// sum which is less than K in the array
import java.util.Arrays;
 
class GFG
{
 
// Function to find pair with largest
// sum which is less than K in the array
static void Max_Sum(int arr[], int n, int k)
{
    // To store the break point
    int p = n;
     
    // Sort the given array
    Arrays.sort(arr);
     
    // Find the break point
    for (int i = 0; i < n; i++)
    {
        // No need to look beyond i'th index
        if (arr[i] >= k)
        {
            p = i;
            break;
        }
    }
     
    int maxsum = 0, a = 0, b = 0;
     
    // Find the required pair
    for (int i = 0; i < p; i++)
    {
        for (int j = i + 1; j < p; j++)
        {
            if (arr[i] + arr[j] < k &&
                arr[i] + arr[j] > maxsum)
            {
                maxsum = arr[i] + arr[j];
                 
                a = arr[i];
                b = arr[j];
            }
        }
    }
     
    // Print the required answer
    System.out.print( a + " " + b);
}
 
// Driver code
public static void main (String[] args)
{
    int []arr = {5, 20, 110, 100, 10};
    int k = 85;
 
    int n = arr.length;
     
    // Function call
    Max_Sum(arr, n, k);
}
}
 
// This code is contributed by anuj_67..

Python3

# Python3 program to find pair with largest
# sum which is less than K in the array
 
# Function to find pair with largest
# sum which is less than K in the array
def Max_Sum(arr, n, k):
 
    # To store the break point
    p = n
     
    # Sort the given array
    arr.sort()
     
    # Find the break point
    for i in range(0, n):
         
        # No need to look beyond i'th index
        if (arr[i] >= k):
            p = i
            break
     
    maxsum = 0
    a = 0
    b = 0
     
    # Find the required pair
    for i in range(0, p):    
        for j in range(i + 1, p):
            if(arr[i] + arr[j] < k and
               arr[i] + arr[j] > maxsum):
                maxsum = arr[i] + arr[j]
                a = arr[i]
                b = arr[j]
                 
    # Print the required answer
    print(a, b)
 
# Driver code
arr = [5, 20, 110, 100, 10]
k = 85
 
n = len(arr)
 
# Function call
Max_Sum(arr, n, k)
 
# This code is contributed by Sanjit_Prasad

C#

// C# program to find pair with largest
// sum which is less than K in the array
using System;
 
class GFG
{
 
// Function to find pair with largest
// sum which is less than K in the array
static void Max_Sum(int []arr, int n, int k)
{
    // To store the break point
    int p = n;
     
    // Sort the given array
    Array.Sort(arr);
     
    // Find the break point
    for (int i = 0; i < n; i++)
    {
        // No need to look beyond i'th index
        if (arr[i] >= k)
        {
            p = i;
            break;
        }
    }
     
    int maxsum = 0, a = 0, b = 0;
     
    // Find the required pair
    for (int i = 0; i < p; i++)
    {
        for (int j = i + 1; j < p; j++)
        {
            if (arr[i] + arr[j] < k &&
                arr[i] + arr[j] > maxsum)
            {
                maxsum = arr[i] + arr[j];
                 
                a = arr[i];
                b = arr[j];
            }
        }
    }
     
    // Print the required answer
    Console.WriteLine( a + " " + b);
}
 
// Driver code
public static void Main ()
{
    int []arr = {5, 20, 110, 100, 10};
    int k = 85;
 
    int n = arr.Length;
     
    // Function call
    Max_Sum(arr, n, k);
}
}
 
// This code is contributed by anuj_67..

输出

10 20

时间复杂度： O(N ^ 2)

高效方法：两指针方法
排序后，我们可以在数组的左右两端放置两个指针。然后可以通过以下步骤找到所需的一对：

1. 在两个指针处找到当前值的总和。
2. 如果总和小于k：
   1. 使用先前答案的最大值和当前总和来更新答案。
   2. 增加左指针。
3. 其他减少右指针。

C++

// CPP program for the above approach
#include 
using namespace std;
 
// Function to find max sum less than k
int maxSum(vector arr, int k)
{
     
    // Sort the array
    sort(arr.begin(), arr.end());
    int n = arr.size(), l = 0, r = n - 1, ans = 0;
 
    // While l is less than r
    while (l < r) {
        if (arr[l] + arr[r] < k) {
            ans = max(ans, arr[l] + arr[r]);
            l++;
        }
        else {
            r--;
        }
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    vector A = { 20, 10, 30, 100, 110 };
    int k = 85;
 
    // Function Call
    cout << maxSum(A, k) << endl;
}

输出

50

时间复杂度：O(NlogN)

空间复杂度：O(1)