Programa Java para número de pares con suma máxima

Dada una array arr[], cuente el número de pares arr[i], arr[j] tales que arr[i] + arr[j] es máximo e i < j.

Example :
Input  : arr[] = {1, 1, 1, 2, 2, 2}
Output : 3
Explanation: The maximum possible pair 
sum where i

Method 1 (Naive) 
Traverse a loop i from 0 to n, i.e length of the array and another loop j from i+1 to n to find all possible pairs with i

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// Java program to count pairs 
// with maximum sum. 
class GFG { 
  
// function to find the number of 
// maximum pair sums 
static int sum(int a[], int n) 
{ 
    // traverse through all the pairs 
    int maxSum = Integer.MIN_VALUE; 
    for (int i = 0; i < n; i++) 
        for (int j = i + 1; j < n; j++) 
            maxSum = Math.max(maxSum, a[i] + 
                                    a[j]); 
  
    // traverse through all pairs and 
    // keep a count of the number of 
    // maximum pairs 
    int c = 0; 
    for (int i = 0; i < n; i++) 
        for (int j = i + 1; j < n; j++) 
            if (a[i] + a[j] == maxSum) 
                c++; 
    return c; 
} 
  
// driver program to test the above function 
public static void main(String[] args) 
{ 
    int array[] = { 1, 1, 1, 2, 2, 2 }; 
    int n = array.length; 
    System.out.println(sum(array, n)); 
} 
} 
  
// This code is contributed by Prerna Saini 








Output : 
3
Time complexity:O(n^2)
Method 2 (Efficient) 
If we take a closer look, we can notice following facts. 

Maximum element is always part of solution
If maximum element appears more than once, then result is maxCount * (maxCount - 1)/2. We basically need to choose 2 elements from maxCount (maxCountC2).
If maximum element appears once, then result is equal to count of second maximum element. We can form a pair with every second max and max


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// Java program to count pairs 
// with maximum sum.
import java.io.*;
class GFG {
      
// function to find the number 
// of maximum pair sums
static int sum(int a[], int n)
{
    // Find maximum and second maximum 
    // elements. Also find their counts.
    int maxVal = a[0], maxCount = 1;
    int secondMax = Integer.MIN_VALUE,
        secondMaxCount = 0;
    for (int i = 1; i < n; i++) {
        if (a[i] == maxVal)
            maxCount++;
        else if (a[i] > maxVal) {
            secondMax = maxVal;
            secondMaxCount = maxCount;
            maxVal = a[i];
            maxCount = 1;
        }
        else if (a[i] == secondMax) {
            secondMax = a[i];
            secondMaxCount++;
        }
        else if (a[i] > secondMax) {
            secondMax = a[i];
            secondMaxCount = 1;
        }
    }
  
    // If maximum element appears
    // more than once.
    if (maxCount > 1)
        return maxCount * (maxCount - 1) / 2;
  
    // If maximum element appears
    // only once.
    return secondMaxCount;
}
  
// driver program 
public static void main(String[] args)
{
    int array[] = { 1, 1, 1, 2, 2, 2, 3 };
    int n = array.length;
    System.out.println(sum(array, n));
}
}
  
// This code is contributed by Prerna Saini










Output : 
 
3
Time complexity:O(n)
 
Please refer complete article on Number of pairs with maximum sum for more details!



Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA