Given a matrix arr[][] of size N*M The task is to find the index of the row that has the maximum unique elements. If there are multiple rows possible, return the minimum indexed row.
Examples:
Input: arr[][] = { {1, 2, 3, 4, 5}, {1, 2, 2, 4, 7}, {1, 3, 1, 3, 1} }
Output: 0
Explanation: Rows 0, 1 & 2 have 5, 4 & 2 unique elements respectively.
Input: arr[][] = { {1, 2, 3, 4}, {5, 6, 7, 8}, {9, 1, 2, 3} }
Output : 0
Approach: The task can be solved using a set data structure. Iterate over the rows and store the corresponding element inside a set to keep track of the unique elements, return the minimum indexed row with maximum unique elements.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum indexed row// with maximum unique elementsint get(int n, int m, vector<vector<int> >& v){ // Stores the unique elements set<int> s; // Stores the minimum indexed row int max_ans = INT_MAX; int cnt = -1; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { s.insert(v[i][j]); } int size = (int)s.size(); if (cnt < size) { size = cnt; max_ans = min(max_ans, i); } s.clear(); } return max_ans;}// Driver Codeint main(){ vector<vector<int> > arr = { { 1, 2, 3, 4, 5 }, { 1, 2, 2, 4, 7 }, { 1, 3, 1, 3, 1 } }; int n = arr.size(); int m = arr[0].size(); cout << get(n, m, arr);} |
Java
// JAVA program for the above approachimport java.util.*;class GFG{ // Function to find the minimum indexed row // with maximum unique elements public static int get(int n, int m, ArrayList<ArrayList<Integer> > v) { // Stores the unique elements HashSet<Integer> s = new HashSet<Integer>(); // Stores the minimum indexed row int max_ans = Integer.MAX_VALUE; int cnt = -1; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { s.add(v.get(i).get(j)); } int size = (int)s.size(); if (cnt < size) { size = cnt; max_ans = Math.min(max_ans, i); } s.clear(); } return max_ans; } // Driver Code public static void main(String[] args) { ArrayList<ArrayList<Integer> > arr = new ArrayList<ArrayList<Integer> >(); ArrayList<Integer> temp1 = new ArrayList<>(Arrays.asList(1, 2, 3, 4, 5)); ArrayList<Integer> temp2 = new ArrayList<>(Arrays.asList(1, 2, 2, 4, 7)); ArrayList<Integer> temp3 = new ArrayList<>(Arrays.asList(1, 3, 1, 3, 1)); arr.add(temp1); arr.add(temp2); arr.add(temp3); int n = arr.size(); int m = arr.get(0).size(); System.out.print(get(n, m, arr)); }}// This code is contributed by Taranpreet |
Python3
# Python program for the above approach# Function to find the minimum indexed row# with maximum unique elementsdef get(n, m, v): s = set() # Stores the minimum indexed row max_ans = 1000000000 cnt = -1 for i in range(0, n): for j in range(0, m): s.add(v[i][j]) size = len(s) if cnt < size: size = cnt max_ans = min(max_ans, i) s.clear() return max_ansif __name__ == "__main__": arr = [[1, 2, 3, 4, 5], [1, 2, 2, 4, 7], [1, 3, 1, 3, 1]] n = 3 m = 5 print(get(n, m, arr))# This code is contributed by Rohit Pradhan |
C#
// C# program for the above approachusing System;using System.Linq;using System.Collections.Generic;public class GFG { // Function to find the minimum indexed row // with maximum unique elements public static int get(int n, int m, List<List<int>> v) { // Stores the unique elements HashSet<int> s = new HashSet<int>(); // Stores the minimum indexed row int max_ans = int.MaxValue; int cnt = -1; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { s.Add(v[i][j]); } int size = (int) s.Count; if (cnt < size) { size = cnt; max_ans = Math.Min(max_ans, i); } s.Clear(); } return max_ans; } // Driver Code public static void Main(String[] args) { List<List<int>> arr = new List<List<int>>(); int [] t1 = {1, 2, 3, 4, 5}; List<int> temp1 = new List<int>(); temp1 = (t1.ToList()); List<int> temp2 = new List<int>(); int []t2 = {1, 2, 2, 4, 7}; temp2 = (t2.ToList()); List<int> temp3 = new List<int>(); int[] t3 = {1, 3, 1, 3, 1}; temp3 = t3.ToList(); arr.Add(temp1); arr.Add(temp2); arr.Add(temp3); int n = arr.Count; int m = arr[0].Count; Console.Write(get(n, m, arr)); }}// This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript code for the above approach // Function to find the minimum indexed row // with maximum unique elements function get(n, m, v) { // Stores the unique elements let s = new Set(); // Stores the minimum indexed row let max_ans = Number.MAX_VALUE; let cnt = -1; for (let i = 0; i < n; i++) { for (let j = 0; j < m; j++) { s.add(v[i][j]); } let size = s.size; if (cnt < size) { size = cnt; max_ans = Math.min(max_ans, i); } s.clear(); } return max_ans; } // Driver Code let arr = [[1, 2, 3, 4, 5], [1, 2, 2, 4, 7], [1, 3, 1, 3, 1]]; let n = arr.length; let m = arr[0].length; document.write(get(n, m, arr)); // This code is contributed by Potta Lokesh </script> |
Output
0
Time Complexity: O(N*M*logM)
Auxiliary Space: O(M)
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