Given an array arr, the task is to find the minimum distance between any two same elements in the array. If no such element is found, return -1.
Examples:
Input: arr = {1, 2, 3, 2, 1}
Output: 2
Explanation:
There are two matching pairs of values: 1 and 2 in this array.
Minimum Distance between two 1’s = 4
Minimum Distance between two 2’s = 2
Therefore, Minimum distance between any two equal elements in the Array = 2Input: arr = {3, 5, 4, 6, 5, 3}
Output: 3
Explanation:
There are two matching pairs of values: 3 and 5 in this array.
Minimum Distance between two 3’s = 5
Minimum Distance between two 5’s = 3
Therefore, Minimum distance between any two equal elements in the Array = 3
Naive Approach: The simplest approach is using two nested for loops to form each and every combination. If the elements are equal, find the minimum distance.
Time complexity: O(N2)
Efficient Approach: An efficient approach for this problem is to use map to store array elements as a key and their index as the values.
Below is the step by step algorithm:
- Traverse the array one by one.
- Check if this element is in the map or not.
- If the map does not contain this element, insert it as (element, current index) pair.
- If the array element present in the map, fetch the previous index of this element from the map.
- Find the difference between the previous index and the current index
- Compare each difference and find the minimum distance.
- If no such element found, return -1.
Below is the implementation of the above approach.
C++
// C++ program to find the minimum distance // between two occurrences of the same element #include<bits/stdc++.h> using namespace std; // Function to find the minimum // distance between the same elements int minimumDistance(int a[], int n) { // Create a HashMap to // store (key, values) pair. map<int,int> hmap; int minDistance = INT_MAX; // Initialize previousIndex // and currentIndex as 0 int previousIndex = 0, currentIndex = 0; // Traverse the array and // find the minimum distance // between the same elements with map for (int i = 0; i < n; i++) { if (hmap.find(a[i])!=hmap.end()) { currentIndex = i; // Fetch the previous index from map. previousIndex = hmap[a[i]]; // Find the minimum distance. minDistance = min((currentIndex - previousIndex),minDistance); } // Update the map. hmap[a[i]] = i; } // return minimum distance, // if no such elements found, return -1 return (minDistance == INT_MAX ? -1 : minDistance); } // Driver code int main() { // Test Case 1: int a1[] = { 1, 2, 3, 2, 1 }; int n = sizeof(a1)/sizeof(a1[0]); cout << minimumDistance(a1, n) << endl; // Test Case 2: int a2[] = { 3, 5, 4, 6, 5, 3 }; n = sizeof(a2)/sizeof(a2[0]); cout << minimumDistance(a2, n) << endl; // Test Case 3: int a3[] = { 1, 2, 1, 4, 1 }; n = sizeof(a3)/sizeof(a3[0]); cout << minimumDistance(a3, n) << endl; } // This code is contributed by Sanjit_Prasad |
Java
// Java program to find the minimum distance // between two occurrences of the same element import java.util.*; import java.math.*; class GFG { // Function to find the minimum // distance between the same elements static int minimumDistance(int[] a) { // Create a HashMap to // store (key, values) pair. HashMap<Integer, Integer> hmap = new HashMap<>(); int minDistance = Integer.MAX_VALUE; // Initialize previousIndex // and currentIndex as 0 int previousIndex = 0, currentIndex = 0; // Traverse the array and // find the minimum distance // between the same elements with map for (int i = 0; i < a.length; i++) { if (hmap.containsKey(a[i])) { currentIndex = i; // Fetch the previous index from map. previousIndex = hmap.get(a[i]); // Find the minimum distance. minDistance = Math.min( (currentIndex - previousIndex), minDistance); } // Update the map. hmap.put(a[i], i); } // return minimum distance, // if no such elements found, return -1 return ( minDistance == Integer.MAX_VALUE ? -1 : minDistance); } // Driver code public static void main(String args[]) { // Test Case 1: int a1[] = { 1, 2, 3, 2, 1 }; System.out.println(minimumDistance(a1)); // Test Case 2: int a2[] = { 3, 5, 4, 6, 5, 3 }; System.out.println(minimumDistance(a2)); // Test Case 3: int a3[] = { 1, 2, 1, 4, 1 }; System.out.println(minimumDistance(a3)); } } |
Python3
# Python3 program to find the minimum distance # between two occurrences of the same element # Function to find the minimum # distance between the same elements def minimumDistance(a): # Create a HashMap to # store (key, values) pair. hmap = dict() minDistance = 10**9 # Initialize previousIndex # and currentIndex as 0 previousIndex = 0 currentIndex = 0 # Traverse the array and # find the minimum distance # between the same elements with map for i in range(len(a)): if a[i] in hmap: currentIndex = i # Fetch the previous index from map. previousIndex = hmap[a[i]] # Find the minimum distance. minDistance = min((currentIndex - previousIndex), minDistance) # Update the map. hmap[a[i]] = i # return minimum distance, # if no such elements found, return -1 if minDistance == 10**9: return -1 return minDistance # Driver code if __name__ == '__main__': # Test Case 1: a1 = [1, 2, 3, 2, 1 ] print(minimumDistance(a1)) # Test Case 2: a2 = [3, 5, 4, 6, 5,3] print(minimumDistance(a2)) # Test Case 3: a3 = [1, 2, 1, 4, 1 ] print(minimumDistance(a3)) # This code is contributed by mohit kumar 29 |
2 3 2
Time complexity: O(N)
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