Maximum distance between two occurrences of same element in array
Given an array with repeated elements, the task is to find the maximum distance two occurrences of an element.
Examples:
Input : arr[] = {3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2}
Output: 10
// maximum distance for 2 is 11-1 = 10
// maximum distance for 1 is 4-2 = 2
// maximum distance for 4 is 10-5 = 5
A simple solution for this problem is to one by one pick each element from array and find its first and last occurrence in array and take difference of first and last occurrence for maximum distance. Time complexity for this approach is O(n2).
An efficient solution for this problem is to use hashing. The idea is to traverse input array and store index of first occurrence in a hash map. For every other occurrence, find the difference between index and the first index stored in hash map. If difference is more than result so far, then update the result.
Below are implementations of the idea. The implementation uses unordered_map in .
C++
// C++ program to find maximum distance between two // same occurrences of a number. #include<bits/stdc++.h> using namespace std; // Function to find maximum distance between equal elements int maxDistance(int arr[], int n) { // Used to store element to first index mapping unordered_map<int, int> mp; // Traverse elements and find maximum distance between // same occurrences with the help of map. int max_dist = 0; for (int i=0; i<n; i++) { // If this is first occurrence of element, insert its // index in map if (mp.find(arr[i]) == mp.end()) mp[arr[i]] = i; // Else update max distance else max_dist = max(max_dist, i - mp[arr[i]]); } return max_dist; } // Driver program to run the case int main() { int arr[] = {3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2}; int n = sizeof(arr)/sizeof(arr[0]); cout << maxDistance(arr, n); return 0; } |
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Java
// Java program to find maximum distance between two // same occurrences of a number. import java.io.*; import java.util.*; class GFG { // Function to find maximum distance between equal elements static int maxDistance(int[] arr, int n) { // Used to store element to first index mapping HashMap<Integer, Integer> map = new HashMap<>(); // Traverse elements and find maximum distance between // same occurrences with the help of map. int max_dist = 0; for (int i = 0; i < n; i++) { // If this is first occurrence of element, insert its // index in map if (!map.containsKey(arr[i])) map.put(arr[i], i); // Else update max distance else max_dist = Math.max(max_dist, i - map.get(arr[i])); } return max_dist; } // Driver code public static void main(String args[]) { int[] arr = {3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2}; int n = arr.length; System.out.println(maxDistance(arr, n)); } } // This code is contributed by rachana soma |
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Python
# Python program to find maximum distance between two # same occurrences of a number. # Function to find maximum distance between equal elements def maxDistance(arr, n): # Used to store element to first index mapping mp = {} # Traverse elements and find maximum distance between # same occurrences with the help of map. maxDict = 0 for i in range(n): # If this is first occurrence of element, insert its # index in map if arr[i] not in mp.keys(): mp[arr[i]] = i # Else update max distance else: maxDict = max(maxDict, i-mp[arr[i]]) return maxDict # Driver Program if __name__=='__main__': arr = [3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2] n = len(arr) print maxDistance(arr, n) # Contributed By: Harshit Sidhwa |
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Output:
10
Time complexity : O(n) under the assumption that unordered_map’s search and insert operations take O(1) time.
This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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