Minimum Sum of a pair at least K distance apart from an Array

Given an array of integers A[] of size N, the task is to find the minimum sum that can be obtained by any pair of array elements that are at least K indices apart from each other.

Examples:

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.

Naive Approach:
The simplest approach is to solve the problem is to iterate over the indices [i + K, N – 1] for every i^th index and find the minimum element, say min. Check if min + A[i] is less than the minimum sum obtained so far and update minimum_sum accordingly. Finally, print the minimum_sum.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include<bits/stdc++.h>
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
        minimum = sys.maxsize;

        # Iterate from i + k to N
        for j in range(i + K, n, 1):

            # Find the minimum
            minimum = min(minimum, A[j]);

        if (minimum == sys.maxsize):
            continue;

        # Update the minimum sum
        minimum_sum = min(minimum_sum, A[i] + minimum);

    # Print 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

<script>
    // 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 contributed by divyeshrabadiya07.
</script>

Output:

4

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach:
The above approach can be optimized using a Suffix Array. Follow the steps below:

• Initialize a suffix array(say suffix[]), where suffix[i] stores the minimum of all the elements from index N-1 to i.
• For any i^th index, the minimum element which is K distance apart is stored at index i + K in the suffix array.
• For i ranging from 0 to N – 1, check if A[i] + suffix[i + k] < minimum_sum or not and update minimum_sum accordingly.
• Finally, print minimum_sum as the required answer.

Below is the implementation of the above approach:

C++

// C++ Program to implement
//the above approach
#include <bits/stdc++.h>
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

Javascript

<script>
// JavaScript program for the above approach

    // Function to find the minimum
    // sum of two elements that
    // are atleast K distance apart
   function
    findMinSum(A, K)
    {

        // Length of the array
        let n = A.length;
        let suffix_min = Array.from({length: n}, (_, i) => 0);

        suffix_min[n - 1] = A[n - 1];

        // Find the suffix array
        for (let i = n - 2; i >= 0; i--)
            suffix_min[i]
                = Math.min(suffix_min[i + 1],
                           A[i]);

        let min_sum = Number.MAX_VALUE;

        // Iterate in the array
        for (let 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
        document.write(min_sum);
    }

// Driver Code

           let A = [ 1, 2, 3, 4, 5, 6 ];
        let K = 2;

        findMinSum(A, K);

</script>

Output:

4

Time Complexity: O(N)
Auxiliary Space: O(N)