Find the position of the last removed element from the array
Given an array of size and an integer
. Perform the following operations on the given array:
- If a[i] > M then push a[i] – M to end of the array, otherwise remove it from the array.
- Perform the first operation while the array is non-empty.
The task is to find the original position of the element which gets removed last.
Examples:
Input: arr[] = {4, 3}, M = 2
Output: 2
Remove 4 from the array and the array becomes {3, 2} with original positions {2, 1}
Remove 3 from the array and the array becomes {2, 1} with original positions {1, 2}
Remove 2 from the array and the array becomes {1} with original positions {2}
So, 2nd positioned element is the last to be removed from the array.Input: arr[] = {2, 5, 4}, M = 2
Output: 2
The idea is to observe the last element which will be removed from the array. It can by easily said that the element to be removed last will be the element which can be subtracted max number of times by among all elements of the array. That is, the element with maximum value of ceil(a[i] / M).
So, the task now reduces to find the index of the element in the array with maximum value of ceil(a[i] / M).
Below is the implementation of the above approach:
C++
// C++ program to find the position of the // last removed element from the array #include <bits/stdc++.h> using namespace std; // Function to find the original position // of the element which will be // removed last int getPosition( int a[], int n, int m) { // take ceil of every number for ( int i = 0; i < n; i++) { a[i] = (a[i] / m + (a[i] % m != 0)); } int ans = -1, max = -1; for ( int i = n - 1; i >= 0; i--) { if (max < a[i]) { max = a[i]; ans = i; } } // Since position is index+1 return ans + 1; } // Driver code int main() { int a[] = { 2, 5, 4 }; int n = sizeof (a) / sizeof (a[0]); int m = 2; cout << getPosition(a, n, m); return 0; } |
Java
// Java program to find the position of the // last removed element from the array import java.util.*; class solution { // Function to find the original position // of the element which will be // removed last static int getPosition( int a[], int n, int m) { // take ceil of every number for ( int i = 0 ; i < n; i++) { a[i] = (a[i] / m + (a[i] % m)); } int ans = - 1 , max = - 1 ; for ( int i = n - 1 ; i >= 0 ; i--) { if (max < a[i]) { max = a[i]; ans = i; } } // Since position is index+1 return ans + 1 ; } // Driver code public static void main(String args[]) { int a[] = { 2 , 5 , 4 }; int n = a.length; int m = 2 ; System.out.println(getPosition(a, n, m)); } } //This code is contributed by // Surendra_Gangwar |
Python3
# Python3 program to find the position of # the last removed element from the array import math as mt # Function to find the original # position of the element which # will be removed last def getPosition(a, n, m): # take ceil of every number for i in range (n): a[i] = (a[i] / / m + (a[i] % m ! = 0 )) ans, maxx = - 1 , - 1 for i in range (n - 1 , - 1 , - 1 ): if (maxx < a[i]): maxx = a[i] ans = i # Since position is index+1 return ans + 1 # Driver code a = [ 2 , 5 , 4 ] n = len (a) m = 2 print (getPosition(a, n, m)) # This is contributed by Mohit kumar 29 |
C#
// C# program to find the position of the // last removed element from the array using System; class GFG { // Function to find the original // position of the element which // will be removed last static int getPosition( int []a, int n, int m) { // take ceil of every number for ( int i = 0; i < n; i++) { a[i] = (a[i] / m + (a[i] % m)); } int ans = -1, max = -1; for ( int i = n - 1; i >= 0; i--) { if (max < a[i]) { max = a[i]; ans = i; } } // Since position is index+1 return ans + 1; } // Driver code static public void Main () { int []a = { 2, 5, 4 }; int n = a.Length; int m = 2; Console.WriteLine(getPosition(a, n, m)); } } // This code is contributed by ajit |
PHP
<?php // PHP program to find the position of the // last removed element from the array // Function to find the original // position of the element which // will be removed last function getPosition( $a , $n , $m ) { // take ceil of every number for ( $i = 0; $i < $n ; $i ++) { $a [ $i ] = ( $a [ $i ] / $m + ( $a [ $i ] % $m != 0)); } $ans = -1; $max = -1; for ( $i = $n - 1; $i >= 0; $i --) { if ( $max < $a [ $i ]) { $max = $a [ $i ]; $ans = $i ; } } // Since position is index+1 return $ans + 1; } // Driver code $a = array ( 2, 5, 4 ); $n = sizeof( $a ); $m = 2; echo getPosition( $a , $n , $m ); // This code is contributed by jit_t ?> |
2
Recommended Posts:
- Print digit's position to be removed to make a number divisible by 6
- Find the Kth position element of the given sequence
- Find the minimum number of elements that should be removed to make an array good
- Find element position in given monotonic sequence
- Find the position of the given row in a 2-D array
- Find the value at kth position in the generated array
- Find last element after deleting every second element in array of n integers
- Find an element in array such that sum of left array is equal to sum of right array
- Find the number of boxes to be removed
- Find element in array that divides all array elements
- Find if array has an element whose value is half of array sum
- Minimum number of elements that should be removed to make the array good
- Find the maximum element in the array other than Ai
- Find extra element in the second array
- Find Second largest element in an array
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.