距阵列至少K距离的一对最小总和

给定大小为N的整数A []的数组，任务是找到可以由任何彼此隔开至少K个索引的数组元素对获得的最小和。

例子：

Input: A[] = {1, 2, 3, 4, 5, 6}, K = 2 
Output: 4 
Explanation: 
The minimum sum that can be obtained is by adding 1 and 3 that are at a distance of 2.
Input: A[] = {4, 2, 5, 4, 3, 2, 5}, K = 3 
Output: 4 
Explanation: 
The minimum sum that can be obtained is by adding 2 and 2 that are at a distance of 4.

天真的方法：
解决问题的最简单方法是对每个^第i^个索引迭代索引[i + K，N – 1]并找到最小元素，例如min 。检查min + A [i]是否小于到目前为止获得的最小和，并相应地更新minimum_sum 。最后，打印minimum_sum 。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include
using namespace std;
 
// Function to find the minimum
// sum of two elements that
// are atleast K distance apart
void findMinSum(int A[], int K, int n)
{
    int minimum_sum = INT_MAX;
 
    // Iterate over the array
    for(int i = 0; i < n; i++)
    {
         
        // Initialize the min value
        int mini = INT_MAX;
 
        // Iterate from i + k to N
        for(int j = i + K; j < n; j++)
 
            // Find the minimum
            mini = min(mini, A[j]);
 
        if (mini == INT_MAX)
            continue;
 
        // Update the minimum sum
        minimum_sum = min(minimum_sum,
                          A[i] + mini);
    }
 
    // Print the answer
    cout << (minimum_sum);
}
 
// Driver Code
int main()
{
    int A[] = { 4, 2, 5, 4, 3, 2, 5 };
    int K = 3;
    int n = sizeof(A) / sizeof(A[0]);
 
    findMinSum(A, K, n);
    return 0;
}
 
// This code is contributed by chitranayal

Java

// Java Program to implement
// the above approach
 
import java.util.*;
class GFG {
 
    // Function to find the minimum
    // sum of two elements that
    // are atleast K distance apart
    public static void
    findMinSum(int A[], int K)
    {
        // Length of the array
        int n = A.length;
 
        int minimum_sum
            = Integer.MAX_VALUE;
 
        // Iterate over the array
        for (int i = 0; i < n; i++) {
 
            // Initialize the min value
            int min = Integer.MAX_VALUE;
 
            // Iterate from i + k to N
            for (int j = i + K; j < n; j++)
 
                // Find the minimum
                min = Math.min(min, A[j]);
 
            if (min == Integer.MAX_VALUE)
                continue;
 
            // Update the minimum sum
            minimum_sum = Math.min(minimum_sum,
                                   A[i] + min);
        }
 
        // Print the answer
        System.out.println(minimum_sum);
    }
 
    // Driver Code
    public static void
        main(String[] args)
    {
 
        int A[] = { 4, 2, 5, 4, 3, 2, 5 };
        int K = 3;
 
        findMinSum(A, K);
    }
}

Python3

# Python3 Program to implement
# the above approach
import sys
 
# Function to find the minimum
# sum of two elements that
# are atleast K distance apart
def findMinSum(A, K):
   
    # Length of the array
    n = len(A);
 
    minimum_sum = sys.maxsize;
 
    # Iterate over the array
    for i in range(n):
 
        # Initialize the min value
        minmum = sys.maxsize;
 
        # Iterate from i + k to N
        for j in range(i + K, n, 1):
 
            # Find the minimum
            minmum = min(minmum, A[j]);
 
        if (minmum == sys.maxsize):
            continue;
 
        # Update the minimum sum
        minimum_sum = min(minimum_sum, A[i] + minmum);
 
    # Prthe answer
    print(minimum_sum);
 
# Driver Code
if __name__ == '__main__':
    A = [4, 2, 5, 4, 3, 2, 5];
    K = 3;
 
    findMinSum(A, K);
 
# This code is contributed by sapnasingh4991

C#

// C# Program to implement
// the above approach
using System;
class GFG{
 
  // Function to find the minimum
  // sum of two elements that
  // are atleast K distance apart
  public static void findMinSum(int []A,
                                int K)
  {
    // Length of the array
    int n = A.Length;
 
    int minimum_sum = int.MaxValue;
 
    // Iterate over the array
    for (int i = 0; i < n; i++)
    {
 
      // Initialize the min value
      int min = int.MaxValue;
 
      // Iterate from i + k to N
      for (int j = i + K; j < n; j++)
 
        // Find the minimum
        min = Math.Min(min, A[j]);
 
      if (min == int.MaxValue)
        continue;
 
      // Update the minimum sum
      minimum_sum = Math.Min(minimum_sum,
                             A[i] + min);
    }
 
    // Print the answer
    Console.WriteLine(minimum_sum);
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int []A = { 4, 2, 5, 4, 3, 2, 5 };
    int K = 3;
 
    findMinSum(A, K);
  }
}
 
// This code is contributed by Rohit_ranjan

Javascript

    // Javascript program to implement
      // the above approach
     
    // Function to find the minimum
    // sum of two elements that
    // are atleast K distance apart
    function findMinSum(A, K, n)
    {
        let minimum_sum = Number.MAX_VALUE;
 
        // Iterate over the array
        for(let i = 0; i < n; i++)
        {
 
            // Initialize the min value
            let mini = Number.MAX_VALUE;
 
            // Iterate from i + k to N
            for(let j = i + K; j < n; j++)
 
                // Find the minimum
                mini = Math.min(mini, A[j]);
 
            if (mini == Number.MAX_VALUE)
                continue;
 
            // Update the minimum sum
            minimum_sum = Math.min(minimum_sum,
                              A[i] + mini);
        }
 
        // Print the answer
        document.write(minimum_sum);
    }
 
 
    let A = [ 4, 2, 5, 4, 3, 2, 5 ];
    let K = 3;
    let n = A.length;
   
    findMinSum(A, K, n);
 
    // This code is conytributed by divyeshrabadiya07.

输出：

4

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：
可以使用后缀数组来优化上述方法。请按照以下步骤操作：

• 初始化一个后缀数组(例如suffix [] )，其中后缀[i]存储从索引N-1到i的所有元素的最小值。
• 对于任何^第i^个索引，相距K个距离的最小元素都存储在后缀数组中的索引i + K处。
• i范围从0到N – 1，检查是否有[I] +后缀第[i + k]的与否和更新minimum_sum相应。
• 最后，将minimum_sum打印为必需的答案。

下面是上述方法的实现：

C++

// C++ Program to implement
//the above approach
#include 
using namespace std;
 
// Function to find the minimum
// sum of two elements that
// are atleast K distance apart
void findMinSum(int A[], int K, int len)
{
 
  // Length of the array
  int n = len;
  int suffix_min[n] = {0};
 
  suffix_min[n - 1] = A[n - 1];
 
  // Find the suffix array
  for (int i = n - 2; i >= 0; i--)
    suffix_min[i] = min(suffix_min[i + 1], A[i]);
 
  int min_sum = INT_MAX;
 
  // Iterate in the array
  for (int i = 0; i < n; i++)
  {
    if (i + K < n)
 
      // Update minimum sum
      min_sum = min(min_sum, A[i] +
                    suffix_min[i + K]);
  }
 
  // Print the answer
  cout << min_sum;
}
 
 
// Driver Code
int main()
{
    int A[] = { 1, 2, 3, 4, 5, 6 };
    int K = 2;
    int n = sizeof(A) / sizeof(A[0]);
    findMinSum(A, K, n);
    return 0;
}
 
// This code is contributed by Rohit_ranjan

Java

// Java Program to implement
// the above approach
 
import java.util.*;
class GFG {
 
    // Function to find the minimum
    // sum of two elements that
    // are atleast K distance apart
    public static void
    findMinSum(int A[], int K)
    {
 
        // Length of the array
        int n = A.length;
        int suffix_min[] = new int[n];
 
        suffix_min[n - 1] = A[n - 1];
 
        // Find the suffix array
        for (int i = n - 2; i >= 0; i--)
            suffix_min[i]
                = Math.min(suffix_min[i + 1],
                           A[i]);
 
        int min_sum = Integer.MAX_VALUE;
 
        // Iterate in the array
        for (int i = 0; i < n; i++) {
 
            if (i + K < n)
 
                // Update minimum sum
                min_sum = Math.min(
                    min_sum, A[i]
                                 + suffix_min[i + K]);
        }
 
        // Print the answer
        System.out.println(min_sum);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int A[] = { 1, 2, 3, 4, 5, 6 };
        int K = 2;
 
        findMinSum(A, K);
    }
}

Python3

# Python3 program to implement
# the above approach
import sys
 
# Function to find the minimum
# sum of two elements that
# are atleast K distance apart
def findMinSum(A, K):
     
    # Length of the array
    n = len(A);
     
    suffix_min = [0] * n;
    suffix_min[n - 1] = A[n - 1];
 
    # Find the suffix array
    for i in range(n - 2, -1, -1):
        suffix_min[i] = min(suffix_min[i + 1], A[i]);
 
    min_sum = sys.maxsize;
 
    # Iterate in the array
    for i in range(n):
        if (i + K < n):
 
            # Update minimum sum
            min_sum = min(min_sum, A[i] +
                          suffix_min[i + K]);
 
    # Print the answer
    print(min_sum);
 
# Driver Code
if __name__ == '__main__':
     
    A = [ 1, 2, 3, 4, 5, 6 ];
    K = 2;
 
    findMinSum(A, K);
 
# This code is contributed by Amit Katiyar

C＃

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to find the minimum
// sum of two elements that
// are atleast K distance apart
public static void findMinSum(int []A, int K)
{
     
    // Length of the array
    int n = A.Length;
    int []suffix_min = new int[n];
 
    suffix_min[n - 1] = A[n - 1];
 
    // Find the suffix array
    for(int i = n - 2; i >= 0; i--)
        suffix_min[i] = Math.Min(suffix_min[i + 1],
                                          A[i]);
 
    int min_sum = int.MaxValue;
 
    // Iterate in the array
    for(int i = 0; i < n; i++)
    {
        if (i + K < n)
 
            // Update minimum sum
            min_sum = Math.Min(min_sum, A[i] +
                               suffix_min[i + K]);
    }
 
    // Print the answer
    Console.WriteLine(min_sum);
}
 
// Driver Code
public static void Main(String[] args)
{
    int []A = { 1, 2, 3, 4, 5, 6 };
    int K = 2;
 
    findMinSum(A, K);
}
}
 
// This code is contributed by 29AjayKumar

输出：

4

时间复杂度： O(N)
辅助空间： O(N)