Given an array arr[] of size N, the task is to make all array elements equal to 0 by replacing all elements of a subsequences of equal elements by any integer, minimum number of times.
Examples:
Input: arr[] = {3, 7, 3}, N = 3
Output: 2
Explanation:
Selecting a subsequence { 7 } and replacing all its elements by 0 modifies arr[] to { 3, 3, 3 }.
Selecting the array { 3, 3, 3 } and replacing all its elements by 0 modifies arr[] to { 0, 0, 0 }Input: arr[] = {1, 5, 1, 3, 2, 3, 1}, N = 7
Output: 4
Approach: The problem can be solved using Greedy technique. The idea is to count the distinct elements present in the array which is not equal to 0 and print the count obtained. Follow the steps below to solve the problem:
- Initialize a Set to store the distinct elements present in the array, which is not equal to 0.
- Traverse the array arr[] and insert the array elements into the Set.
- Finally, print the size of the Set.
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 minimum count of operations// required to convert all array elements to zero// br replacing subsequence of equal elements to 0void minOpsToTurnArrToZero(int arr[], int N){ // Store distinct elements // present in the array unordered_set<int> st; // Traverse the array for (int i = 0; i < N; i++) { // If arr[i] is already present in // the Set or arr[i] is equal to 0 if (st.find(arr[i]) != st.end() || arr[i] == 0) { continue; } // Otherwise, increment ans by // 1 and insert current element else { st.insert(arr[i]); } } cout << st.size() << endl;}// Driver Codeint main(){ // Given array int arr[] = { 3, 7, 3 }; // Size of the given array int N = sizeof(arr) / sizeof(arr[0]); minOpsToTurnArrToZero(arr, N); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG { // Function to find minimum count of operations // required to convert all array elements to zero // br replacing subsequence of equal elements to 0 static void minOpsToTurnArrToZero(int[] arr, int N) { // Store distinct elements // present in the array Set<Integer> st = new HashSet<Integer>(); // Traverse the array for (int i = 0; i < N; i++) { // If arr[i] is already present in // the Set or arr[i] is equal to 0 if (st.contains(arr[i]) || arr[i] == 0) { continue; } // Otherwise, increment ans by // 1 and insert current element else { st.add(arr[i]); } } System.out.println(st.size()); } // Driver Code public static void main(String args[]) { // Given array int arr[] = { 3, 7, 3 }; // Size of the given array int N = arr.length; minOpsToTurnArrToZero(arr, N); }}// This code is contributed by 18bhupendrayadav18 |
Python3
# Python3 program for the above approach# Function to find minimum count of # operations required to convert all # array elements to zero by replacing # subsequence of equal elements to 0def minOpsToTurnArrToZero(arr, N): # Store distinct elements # present in the array st = dict() # Traverse the array for i in range(N): # If arr[i] is already present in # the Set or arr[i] is equal to 0 if (i in st.keys() or arr[i] == 0): continue # Otherwise, increment ans by # 1 and insert current element else: st[arr[i]] = 1 print(len(st))# Driver Code# Given arrayarr = [ 3, 7, 3 ]# Size of the given arrayN = len(arr) minOpsToTurnArrToZero(arr, N)# This code is contributed by susmitakundugoaldanga |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG { // Function to find minimum count of operations // required to convert all array elements to zero // br replacing subsequence of equal elements to 0 static void minOpsToTurnArrToZero(int[] arr, int N) { // Store distinct elements // present in the array HashSet<int> st = new HashSet<int>(); // Traverse the array for (int i = 0; i < N; i++) { // If arr[i] is already present in // the Set or arr[i] is equal to 0 if (st.Contains(arr[i]) || arr[i] == 0) { continue; } // Otherwise, increment ans by // 1 and insert current element else { st.Add(arr[i]); } } Console.WriteLine(st.Count); } // Driver Code public static void Main(String []args) { // Given array int []arr = { 3, 7, 3 }; // Size of the given array int N = arr.Length; minOpsToTurnArrToZero(arr, N); }}// This code is contributed by gauravrajput1 |
Javascript
<script>// Javascript program for the above approach// Function to find minimum count of operations// required to convert all array elements to zero// br replacing subsequence of equal elements to 0function minOpsToTurnArrToZero(arr, N){ // Store distinct elements // present in the array var st = new Set(); // Traverse the array for(var i = 0; i < N; i++) { // If arr[i] is already present in // the Set or arr[i] is equal to 0 if (st.has(arr[i]) || arr[i] == 0) { continue; } // Otherwise, increment ans by // 1 and insert current element else { st.add(arr[i]); } } document.write(st.size)}// Driver Code// Given arrayvar arr = [ 3, 7, 3 ];// Size of the given arrayvar N = arr.length;minOpsToTurnArrToZero(arr, N);// This code is contributed by noob2000</script> |
Output:
2
Time Complexity: O(N)
Auxiliary Space: O(N)
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