# Make all array elements equal to 0 by replacing minimum subsequences consisting of equal elements

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:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `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 0` `void` `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 Code` `int` `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 approach` `import` `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 st = ``new` `HashSet();` `        ``// 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 0` `def` `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 array` `arr ``=` `[ ``3``, ``7``, ``3` `]`   `# Size of the given array` `N ``=` `len``(arr) `   `minOpsToTurnArrToZero(arr, N)`   `# This code is contributed by susmitakundugoaldanga`

## C#

 `// C# program for the above approach` `using` `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

 ``

Output:

`2`

Time Complexity: O(N)
Auxiliary Space: O(N)

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