Given an array of size N and an integer K. The array consists of only digits {0, 1, 2, 3, …k-1}. The task is to make array good by removing some of the elements. The array of length x is called good if x is divisible by k and one can split the given array into x/k subsequences and each of form {0, 1, 2, 3, …k-1}.
Note: An empty array is also a good array
Examples:
Input : a[] = {0, 1, 2, 3, 4, 0, 1, 0, 1, 2, 3, 4}, K = 5
Output : 2
First sequence is formed from first, second, third, fourth
and fifth element and second sequence is formed from eighth, ninth tenth, eleventh
and twelfth. so, remove last fifth and sixth elements.
Input : a[] = {0, 2, 1, 3}, k = 4
Output : 4
Remove all elements. One can’t make subsequence of the form
{0, 1, 2, 3}
Approach:
- Let cnt0 be the number of subsequences of [0], cnt1 be the number of subsequences [0, 1], cnt2 — the number of subsequences [0, 1, 2] and so on, and cntk-1 is the number of completed subsequences [0, 1, 2, 3, …k-1].
- Iterate over all elements of arr in order from left to right.If the current element in the array is zero then increase the count of cnt0 by 1.
- If the current element in the array is not zero then check if its previous element in the sequence count is greater than zero or not.
- If its previous element in the sequence is greater than zero then decrement the count of the previous element by one and increment count of the current element by one.
Below is the implementation of the above approach:
C++
// C++ program to remove minimum elements to// make the given array good#include <bits/stdc++.h>using namespace std;// Function to remove minimum elements to// make the given array goodint MinRemove(int a[], int n, int k){ // To store count of each subsequence vector<int> cnt(k, 0); for (int i = 0; i < n; i++) { // Increase the count of subsequence [0] if (a[i] == 0) cnt[0]++; // If Previous element subsequence count // is greater than zero then increment // subsequence count of current element // and decrement subsequence count of // the previous element. else if (cnt[a[i] - 1] > 0) { cnt[a[i] - 1]--; cnt[a[i]]++; } } // Return the required answer return n - (k * cnt[k - 1]);}// Driver codeint main(){ int a[] = { 0, 1, 2, 3, 4, 0, 1, 0, 1, 2, 3, 4 }, k = 5; int n = sizeof(a) / sizeof(a[0]); // Function call cout << MinRemove(a, n, k); return 0;} |
Java
// Java program to remove minimum elements to// make the given array goodimport java.util.Collections;import java.util.Vector;class GFG{ // Function to remove minimum elements to // make the given array good static int MinRemove(int[] a, int n, int k) { // To store count of each subsequence int []cnt = new int[n]; for (int i = 0; i < n; i++) { // Increase the count of subsequence [0] if (a[i] == 0) cnt[0]++; // If Previous element subsequence count // is greater than zero then increment // subsequence count of current element // and decrement subsequence count of // the previous element. else if (cnt[a[i] - 1] > 0) { cnt[a[i] - 1]--; cnt[a[i]]++; } } // Return the required answer return n - (k * cnt[k - 1]); } // Driver code public static void main(String[] args) { int a[] = { 0, 1, 2, 3, 4, 0, 1, 0, 1, 2, 3, 4 }; int k = 5; int n = a.length; // Function call System.out.println(MinRemove(a, n, k)); }}// This code contributed by Rajput-Ji |
Python3
# Python3 program to remove minimum elements to # make the given array good # Function to remove minimum elements to # make the given array good def MinRemove(a, n, k) : # To store count of each subsequence cnt = [0] * k for i in range(n) : # Increase the count of subsequence [0] if (a[i] == 0) : cnt[0] += 1; # If Previous element subsequence count # is greater than zero then increment # subsequence count of current element # and decrement subsequence count of # the previous element. elif (cnt[a[i] - 1] > 0) : cnt[a[i] - 1] -= 1; cnt[a[i]] += 1; # Return the required answer return n - (k * cnt[k - 1]); # Driver code if __name__ == "__main__" : a = [ 0, 1, 2, 3, 4, 0, 1, 0, 1, 2, 3, 4 ] k = 5; n = len(a); # Function call print(MinRemove(a, n, k)); # This code is contributed by AnkitRai01 |
C#
// C# program to remove minimum elements to// make the given array goodusing System;class GFG{ // Function to remove minimum elements to // make the given array good static int MinRemove(int[] a, int n, int k) { // To store count of each subsequence int []cnt = new int[n]; for (int i = 0; i < n; i++) { // Increase the count of subsequence [0] if (a[i] == 0) cnt[0]++; // If Previous element subsequence count // is greater than zero then increment // subsequence count of current element // and decrement subsequence count of // the previous element. else if (cnt[a[i] - 1] > 0) { cnt[a[i] - 1]--; cnt[a[i]]++; } } // Return the required answer return n - (k * cnt[k - 1]); } // Driver code public static void Main(String[] args) { int []a = { 0, 1, 2, 3, 4, 0, 1, 0, 1, 2, 3, 4 }; int k = 5; int n = a.Length; // Function call Console.WriteLine(MinRemove(a, n, k)); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to remove minimum elements to// make the given array good// Function to remove minimum elements to// make the given array goodfunction MinRemove(a, n, k){ // To store count of each subsequence let cnt = new Array(k).fill(0); for (let i = 0; i < n; i++) { // Increase the count of subsequence [0] if (a[i] == 0) cnt[0]++; // If Previous element subsequence count // is greater than zero then increment // subsequence count of current element // and decrement subsequence count of // the previous element. else if (cnt[a[i] - 1] > 0) { cnt[a[i] - 1]--; cnt[a[i]]++; } } // Return the required answer return n - (k * cnt[k - 1]);}// Driver code let a = [ 0, 1, 2, 3, 4, 0, 1, 0, 1, 2, 3, 4 ], k = 5; let n = a.length; // Function call document.write(MinRemove(a, n, k));</script> |
Output:
2
Time Complexity : O(N)
Auxiliary Space: O(k)
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