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Maximum difference between first and last indexes of an element in array

Given an array of n integers. The task is to find the difference of first and last index of each distinct element so as to maximize the difference.

Examples:

Input : {2, 1, 3, 4, 2, 1, 5, 1, 7}
Output : 6
Element 1 has its first index = 1
and last index = 7
Difference = 7 - 1 = 6
Other elements have a smaller first and last
index difference

Input : {2, 2, 1, 1, 8, 8, 3, 5, 3} 
Output : 2
Maximum difference is for indexes of element 3.


A simple approach is to run two loops and find the difference for each element and accordingly update the max_diff. It has a time complexity of O(n2) and the approach also needs to keep track of the elements that have been visited so that difference for them is not calculated unnecessarily.

An efficient approach uses hashing. It has the following steps.



  1. Traverse the input array from left to right.
  2. For each distinct element map its first and last index in the hash table.
  3. Traverse the hash table and calculate the first and last index difference for each element.
  4. Accordingly update the max_diff.

In the following implementation unordered_map has been used for hashing as the range of integers is not known.

C++

// C++ implementation to find the maximum difference 
// of first and last index of array elements
#include <bits/stdc++.h>
  
using namespace std;
  
// function to find the
// maximum difference
int maxDifference(int arr[], int n)
{
    // structure to store first and last
    // index of each distinct element
    struct index
    {
        int f, l;
    };
      
    // maps each element to its
    // 'index' structure
    unordered_map<int, index> um;
      
    for (int i=0; i<n; i++)
    {
        // storing first index
        if (um.find(arr[i]) == um.end())
            um[arr[i]].f = i;
          
        // storing last index    
        um[arr[i]].l = i;    
    }
      
    int diff, max_diff = INT_MIN;
      
    unordered_map<int, index>::iterator itr;
  
    // traversing 'um'
    for (itr=um.begin(); itr != um.end(); itr++)
    {   
        // difference of last and first index
        // of each element
        diff = (itr->second).l - (itr->second).f;
          
        // update 'max_dff'
        if (max_diff < diff)
            max_diff = diff;
    }
      
    // required maximum difference
    return max_diff;
}
  
  
// Driver program to test above
int main()
{
    int arr[] = {2, 1, 3, 4, 2, 1, 5, 1, 7};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Maximum Difference = " 
         <<maxDifference(arr, n);
    return 0;     

Java

// Java implementation to find the maximum difference 
// of first and last index of array elements
import java.util.HashMap;
import java.util.Map;
  
public class MaxDiffIndexHashing {
  
    static class Element {
        int first;
        int second;
  
        public Element() {
            super();
        }
  
        public Element(int first, int second) {
            super();
            this.first = first;
            this.second = second;
        }
    }
  
    public static void main(String[] args) {
  
        int arr[]={2, 1, 3, 4, 2, 1, 5, 1, 7};
        System.out.println("Maximum Difference= "+ maxDiffIndices(arr));
    }
  
    private static int maxDiffIndices(int[] arr) {
        int n = arr.length;
        int maxDiffIndex = 0;
        Map<Integer, Element> map = new HashMap<Integer, Element>();
  
        for (int i = 0; i < n; i++) {
            if (map.containsKey(arr[i])) {
                Element e = map.get(arr[i]);
                e.second = i;
            } else {
                Element e = new Element();
                e.first = i;
                map.put(arr[i], e);
            }
  
        }
  
        for (Map.Entry<Integer, Element> entry : map.entrySet()) {
            Element e = entry.getValue();
            if ((e.second - e.first) > maxDiffIndex)
                maxDiffIndex = e.second - e.first;
        }
  
        return maxDiffIndex;
    }
  
}        


Output:

Maximum Difference = 6

Time Complexity: O(n)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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