Suma mínima de un par al menos K distancia aparte de una array

Dada una array de enteros A[] de tamaño N , la tarea es encontrar la suma mínima que se puede obtener de cualquier par de elementos de la array que estén al menos separados por K índices entre sí.

Ejemplos:

Entrada: A[] = {1, 2, 3, 4, 5, 6}, K = 2 
Salida: 4 
Explicación: 
La suma mínima que se puede obtener es sumando 1 y 3 que están a una distancia de 2.
Entrada : A[] = {4, 2, 5, 4, 3, 2, 5}, K = 3 
Salida: 4 
Explicación: 
La suma mínima que se puede obtener es sumando 2 y 2 que están a una distancia de 4.

Enfoque ingenuo: 
el enfoque más simple para resolver el problema es iterar sobre los índices [i + K, N – 1] para cada i ^-ésimo índice y encontrar el elemento mínimo, digamos min . Compruebe si min + A[i] es menor que la suma mínima obtenida hasta el momento y actualice la suma_mínima en consecuencia. Finalmente, imprima la suma_mínima .

A continuación se muestra la implementación del enfoque anterior:

// 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 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 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# 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

<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>

4

Complejidad de Tiempo: O(N ² )  
Espacio Auxiliar: O(1)
 

Enfoque eficiente: 
el enfoque anterior se puede optimizar utilizando una array de sufijos . Siga los pasos a continuación:

• Inicialice una array de sufijos (digamos sufijo [] ), donde sufijo [i] almacena el mínimo de todos los elementos del índice N-1 a i .
• Para cualquier i ^-ésimo índice, el elemento mínimo que está separado por una distancia K se almacena en el índice i + K en la array de sufijos.
• Para i que va de 0 a N – 1 , compruebe si A[i] + sufijo[i + k] < suma_mínima o no y actualice la suma_mínima en consecuencia.
• Finalmente, imprima minimal_sum como la respuesta requerida.

A continuación se muestra la implementación del enfoque anterior:

// 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 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 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# 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

<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>

4

Complejidad temporal: O(N)  
Espacio auxiliar: O(N)