每个数组元素的出现索引的绝对差之和|套装2

给定一个由N个整数组成的数组arr [] ，每个数组元素arr [i]的任务是打印| i – j |的和。对于所有可能的索引j ，使得arr [i] = arr [j] 。

例子：

Input: arr[] = {1, 3, 1, 1, 2}
Output: 5 0 3 4 0
Explanation:
For arr[0], sum = |0 – 0| + |0 – 2| + |0 – 3| = 5.
For arr[1], sum = |1 – 1| = 0.
For arr[2], sum = |2 – 0| + |2 – 2| + |2 – 3| = 3.
For arr[3], sum = |3 – 0| + |3 – 2| + |3 – 3| = 4.
For arr[4], sum = |4 – 4| = 0.
Therefore, the required output is 5 0 3 4 0.

Input: arr[] = {1, 1, 1}
Output: 3 2 3

天真的方法：请参阅本文的上一篇文章，以解决问题的最简单方法。
时间复杂度： O(N ² )
辅助空间： O(N)

基于地图的方法：请参考本文的上一篇文章，以解决使用地图的问题。
时间复杂度： O(N * L)
辅助空间： O(N)

高效方法：还可以通过在HashMap中存储先前的索引和每个元素的出现次数来优化上述方法。请按照以下步骤解决问题：

• 初始化一个HashMap，说M来存储arr [i]作为键和(count，前一个索引)作为值。
• 初始化大小为N的两个数组L []和R [] ，其中L [i]表示| i – j |的和。对于所有可能的索引j 和arr [i] = arr [j] ， R [i]表示| i – j |的和。对于所有可能的索引j> i和arr [i] = arr [j] 。
• 在[0，N – 1]范围内遍历给定数组arr [ ]并执行以下步骤：
  • 如果在映射M中存在arr [i] ，则将L [i]的值更新为0 ，将M [arr [i]]的值更新为存储对{1，i} ，其中第一个元素表示出现次数，第二个元素表示该元素的上一个索引。
  • 否则，从映射M中找到arr [i]的值，并将计数和先前索引分别存储在变量cnt和j中。
  • 将L [i]的值更新为(cnt *(i – j)+ L [j])和M中的arr [i]的值以存储对(cnt + 1，i) 。
• 重复相同的过程以更新数组R []中的值。
• 使用变量i在[0，N – 1]范围内进行迭代，并打印结果(L [i] + R [i]) 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Stores the count of occurrences
// and previous index of every element
struct pairr
{
    int count, prevIndex;
     
    // Constructor
    pairr(int countt, int prevIndexx)
    {
        count = countt;
        prevIndex = prevIndexx;
    }
};
 
// Function to calculate the sum of
// absolute differences of indices
// of occurrences of array element
void findSum(int arr[], int n)
{
     
    // Stores the count of elements
    // and their previous indices
    map mapp;
 
    // Initialize 2 arrays left[]
    // and right[] of size N
    int left[n];
    int right[n];
 
    // Traverse the given array
    for(int i = 0; i < n; i++)
    {
         
        // If arr[i] is present in the Map
        if (mapp.find(arr[i]) == mapp.end())
        {
             
            // Update left[i] to 0
            // and update the value
            // of arr[i] in map
            left[i] = 0;
            mapp[arr[i]] = new pairr(1, i);
        }
         
        // Otherwise, get the value from
        // the map and update left[i]
        else
        {
            pairr* tmp = mapp[arr[i]];
            left[i] = (tmp->count) * (i - tmp->prevIndex) +
                   left[tmp->prevIndex];
            mapp[arr[i]] = new pairr(tmp->count + 1, i);
        }
    }
 
    // Clear the map to calculate right[] array
    mapp.clear();
 
    // Traverse the array arr[] in reverse
    for(int i = n - 1; i >= 0; i--)
    {
         
        // If arr[i] is present in theMap
        if (mapp.find(arr[i]) == mapp.end())
        {
             
            // Update right[i] to 0
            // and update the value
            // of arr[i] in the Map
            right[i] = 0;
            mapp[arr[i]] = new pairr(1, i);
        }
 
        // Otherwise get the value from
        // the map and update right[i]
        else
        {
            pairr* tmp = mapp[arr[i]];
            right[i] = (tmp->count) *
                       (abs(i - tmp->prevIndex)) +
                          right[tmp->prevIndex];
 
            mapp[arr[i]] = new pairr(tmp->count + 1, i);
        }
    }
 
    // Iterate in the range [0, N-1]
    // and print the sum of left[i]
    // and right[i] as the result
    for(int i = 0; i < n; i++)
        cout << left[i] + right[i] << " ";
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 3, 1, 1, 2 };
    int N = 5;
     
    findSum(arr, N);
}
 
// This code is contributed by mohit kumar 29

Java

// Java program for the above approach
 
import java.util.*;
 
class GFG {
 
    // Stores the count of occurrences
    // and previous index of every element
    static class pair {
        int count, prevIndex;
 
        // Constructor
        pair(int count, int prevIndex)
        {
            this.count = count;
            this.prevIndex = prevIndex;
        }
    }
 
    // Function to calculate the sum of
    // absolute differences of indices
    // of occurrences of array element
    static void findSum(int[] arr, int n)
    {
        // Stores the count of elements
        // and their previous indices
        Map map = new HashMap<>();
 
        // Initialize 2 arrays left[]
        // and right[] of size N
        int[] left = new int[n];
        int[] right = new int[n];
 
        // Traverse the given array
        for (int i = 0; i < n; i++) {
 
            // If arr[i] is present in the Map
            if (!map.containsKey(arr[i])) {
 
                // Update left[i] to 0
                // and update the value
                // of arr[i] in map
                left[i] = 0;
                map.put(arr[i], new pair(1, i));
            }
 
            // Otherwise, get the value from
            // the map and update left[i]
            else {
                pair tmp = map.get(arr[i]);
                left[i] = (tmp.count)
                              * (i - tmp.prevIndex)
                          + left[tmp.prevIndex];
                map.put(
                    arr[i], new pair(
                                tmp.count + 1, i));
            }
        }
 
        // Clear the map to calculate right[] array
        map.clear();
 
        // Traverse the array arr[] in reverse
        for (int i = n - 1; i >= 0; i--) {
 
            // If arr[i] is present in theMap
            if (!map.containsKey(arr[i])) {
 
                // Update right[i] to 0
                // and update the value
                // of arr[i] in the Map
                right[i] = 0;
                map.put(arr[i], new pair(1, i));
            }
 
            // Otherwise get the value from
            // the map and update right[i]
            else {
 
                pair tmp = map.get(arr[i]);
                right[i]
                    = (tmp.count)
                          * (Math.abs(i - tmp.prevIndex))
                      + right[tmp.prevIndex];
 
                map.put(
                    arr[i], new pair(
                                tmp.count + 1, i));
            }
        }
 
        // Iterate in the range [0, N-1]
        // and print the sum of left[i]
        // and right[i] as the result
        for (int i = 0; i < n; i++)
            System.out.print(
                left[i] + right[i] + " ");
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 1, 3, 1, 1, 2 };
        int N = arr.length;
        findSum(arr, N);
    }
}

输出：

5 0 3 4 0

时间复杂度： O(N)
辅助空间： O(N)