数组中每个元素右侧较大元素的计数

给定一个由N 个整数组成的数组arr[] ，任务是计算每个数组元素右侧的较大元素的数量。

例子：

Input: arr[] = {3, 7, 1, 5, 9, 2}
Output: {3, 1, 3, 1, 0, 0}
Explanation:
For arr[0], the elements greater than it on the right are {7, 5, 9}.
For arr[1], the only element greater than it on the right is {9}.
For arr[2], the elements greater than it on the right are {5, 9, 2}.
For arr[3], the only element greater than it on the right is {9}.
For arr[4] and arr[5], no greater elements exist on the right.

Input: arr[] = {5, 4, 3, 2}
Output: {0, 0, 0, 0}

朴素的方法：最简单的方法是使用两个循环迭代所有数组元素，对于每个数组元素，在其右侧计算大于它的元素数，然后打印它。
时间复杂度： O(N ² )
辅助空间： O(1)

有效的方法：可以使用降序归并排序的概念来解决问题。请按照以下步骤解决问题：

• 初始化一个数组count[] ，其中count[i]存储每个arr[i]右侧更大元素的相应计数
• 获取索引i和j ，并比较数组中的元素。
• 如果较高索引元素大于较低索引元素，则所有较高索引元素将大于该较低索引之后的所有元素。
• 由于左侧部分已经排序，因此将较低索引元素之后的元素计数添加到较低索引的count[]数组中。
• 重复以上步骤，直到整个数组排序完毕。
• 最后打印count[]数组的值。

下面是上述方法的实现：

Java

// Java program for the above approach
  
import java.util.*;
  
public class GFG {
  
    // Stores the index & value pairs
    static class Item {
  
        int val;
        int index;
  
        public Item(int val, int index)
        {
            this.val = val;
            this.index = index;
        }
    }
  
    // Function to count the number of
    // greater elements on the right
    // of every array element
    public static ArrayList
    countLarge(int[] a)
    {
        // Lemgth of the array
        int len = a.length;
  
        // Stores the index-value pairs
        Item[] items = new Item[len];
  
        for (int i = 0; i < len; i++) {
            items[i] = new Item(a[i], i);
        }
  
        // Stores the count of greater
        // elements on right
        int[] count = new int[len];
  
        // Perform MergeSort operation
        mergeSort(items, 0, len - 1,
                  count);
  
        ArrayList res
            = new ArrayList<>();
  
        for (int i : count) {
            res.add(i);
        }
  
        return res;
    }
  
    // Function to sort the array
    // using Merge Sort
    public static void mergeSort(
        Item[] items, int low int high,
        int[] count)
    {
  
        // Base Case
        if (low >= high) {
            return;
        }
  
        // Find Mid
        int mid = low + (high - low) / 2;
  
        mergeSort(items, low, mid,
                  count);
        mergeSort(items, mid + 1,
                  high, count);
  
        // Merging step
        merge(items, low, mid,
              mid + 1, high, count);
    }
  
    // Utility function to merge sorted
    // subarrays and find the count of
    // greater elements on the right
    public static void merge(
        Item[] items, int low, int lowEnd,
        int high, int highEnd, int[] count)
    {
        int m = highEnd - low + 1; // mid
  
        Item[] sorted = new Item[m];
  
        int rightCounter = 0;
        int lowInd = low, highInd = high;
        int index = 0;
  
        // Loop to store the count of
        // larger elements on right side
        // when both array have elements
        while (lowInd <= lowEnd
               && highInd <= highEnd) {
  
            if (items[lowInd].val
                < items[highInd].val) {
                rightCounter++;
                sorted[index++]
                    = items[highInd++];
            }
            else {
                count[items[lowInd].index] += rightCounter;
                sorted[index++] = items[lowInd++];
            }
        }
  
        // Loop to store the count of
        // larger elements in right side
        // when only left array have
        // some element
        while (lowInd <= lowEnd) {
  
            count[items[lowInd].index] += rightCounter;
            sorted[index++] = items[lowInd++];
        }
  
        // Loop to store the count of
        // larger elements in right side
        // when only right array have
        // some element
        while (highInd <= highEnd) {
  
            sorted[index++] = items[highInd++];
        }
  
        System.arraycopy(sorted, 0, items,
                         low, m);
    }
  
    // Utility function that prints
    // the count of greater elements
    // on the right
    public static void
    printArray(ArrayList countList)
    {
  
        for (Integer i : countList)
            System.out.print(i + " ");
  
        System.out.println();
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        // Given array
        int arr[] = { 3, 7, 1, 5, 9, 2 };
        int n = arr.length;
  
        // Function Call
        ArrayList countList
            = countLarge(arr);
  
        printArray(countList);
    }
}

输出：

3 1 3 1 0 0

时间复杂度： O(N*log N)
辅助空间： O(N)

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