Minimum index of element with maximum multiples in Array
Last Updated :
19 Sep, 2023
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Given an array arr[] of length n, the task is to calculate the min index of the element which has maximum elements present in the form of {2 * arr[i], 3 * arr[i], 4 * arr[i], 5 * arr[i]}.
Examples:
Input: n = 4. arr[] = {2, 1, 3, 6}
Output: 1
Explanation:
- For 2, {2*2, 3*2, 4*2, 5*2} => {4, 6, 8, 10}, Here count of present elements in an array are {0, 1, 0, 0} => 0+1+0+0 = 1
- For 1, {2*1, 3*1, 4*1, 5*1} => {2, 3, 4, 5}, Here count of present elements in an array are {1, 1, 0, 0} => 1+1+0+0 =2
Input: n = 3, arr[] = {1, 4, 8}
Output: 0
Explanation:
- For 1 count is 1, For 4 count is 1, and For 8 count is 0.
- Here index 0 and 1 have similar counts. So, we have to choose the min index as an output. Therefore, output is 0.
Approach: This can be solved with the following idea:
Using the map data structure, store the occurrence of each element.
Below are the steps involved in the implementation of the code:
- In this problem, we will maintain the map which counts the occurrences of the element and stores it in the map.
- We traverse the array and store the maximum friend elements in the count variable.
- The minimum index with maximum friend elements is stored in the res variable.
- Then we return the res variable.
Below is the implementation of the above approach:
C++
// C++ code for the above approach:#include <bits/stdc++.h>using namespace std;int findMinIndex(int n, vector<int>& arr){ // Map to store the occurrences of // the arr elements unordered_map<int, int> mp; for (int i = 0; i < n; i++) { mp[arr[i]]++; } // Store the min index int res; // Maintain the maximum elements int max = 0; for (int i = 0; i < n; i++) { int count = 0; for (int j = 2; j <= 5; j++) { count += mp[j * arr[i]]; } if (count > max) { max = count; res = i; } } // Return the resultant min index return res;}// Driver codeint main(){ int n = 4; vector<int> arr = { 2, 1, 3, 6 }; // Function call cout << findMinIndex(n, arr) << endl; return 0;} |
Java
// JAVA code for the above approach:import java.util.HashMap;public class Main { static int findMinIndex(int n, int[] arr) { // Map to store the occurrences of the arr elements HashMap<Integer, Integer> mp = new HashMap<>(); for (int i = 0; i < n; i++) { mp.put(arr[i], mp.getOrDefault(arr[i], 0) + 1); } // Store the min index int res = 0; // Maintain the maximum elements int max = 0; for (int i = 0; i < n; i++) { int count = 0; for (int j = 2; j <= 5; j++) { count += mp.getOrDefault(j * arr[i], 0); } if (count > max) { max = count; res = i; } } // Return the resultant min index return res; } // Driver code public static void main(String[] args) { int n = 4; int[] arr = { 2, 1, 3, 6 }; // Function call System.out.println(findMinIndex(n, arr)); }}// This code is contributed by shivamgupta0987654321 |
Python3
def findMinIndex(n, arr): # Dictionary to store the occurrences of the arr elements mp = {} for i in range(n): mp[arr[i]] = mp.get(arr[i], 0) + 1 # Store the min index res = None # Maintain the maximum elements max_count = 0 for i in range(n): count = 0 for j in range(2, 6): count += mp.get(j * arr[i], 0) if count > max_count: max_count = count res = i # Return the resultant min index return res# Driver codeif __name__ == "__main__": n = 4 arr = [2, 1, 3, 6] # Function call print(findMinIndex(n, arr)) # This code is contributed by shivamgupta310570 |
C#
using System;using System.Collections.Generic;public class GFG { static int FindMinIndex(int n, List<int> arr) { // Dictionary to store the occurrences of the arr // elements Dictionary<int, int> mp = new Dictionary<int, int>(); for (int i = 0; i < n; i++) { if (!mp.ContainsKey(arr[i])) mp[arr[i]] = 0; mp[arr[i]]++; } // Store the min index int res = 0; // Maintain the maximum elements int max = 0; for (int i = 0; i < n; i++) { int count = 0; for (int j = 2; j <= 5; j++) { if (mp.ContainsKey(j * arr[i])) count += mp[j * arr[i]]; } if (count > max) { max = count; res = i; } } // Return the resultant min index return res; } // Driver code public static void Main(string[] args) { int n = 4; List<int> arr = new List<int>{ 2, 1, 3, 6 }; // Function call Console.WriteLine(FindMinIndex(n, arr)); }}// This code is contributed by rambabuguphka |
Javascript
function findMinIndex(n, arr) { // Map to store the occurrences of // the arr elements let mp = new Map(); for (let i = 0; i < n; i++) { if (mp.has(arr[i])) { mp.set(arr[i], mp.get(arr[i]) + 1); } else { mp.set(arr[i], 1); } } // Store the min index let res; // Maintain the maximum elements let max = 0; for (let i = 0; i < n; i++) { let count = 0; for (let j = 2; j <= 5; j++) { if (mp.has(j * arr[i])) { count += mp.get(j * arr[i]); } } if (count > max) { max = count; res = i; } } // Return the resultant min index return res;}// Driver codelet n = 4;let arr = [2, 1, 3, 6];console.log(findMinIndex(n, arr)); |
Output
1
Time Complexity: O(n)
Auxiliary Space: O(n)
