查找数组中最后删除的元素的位置

给定大小数组$N$和一个整数$M$ 。在给定的数组上执行以下操作：

1. 如果a [i]> M，则将a [i] – M推到数组的末尾，否则将其从数组中删除。
2. 数组为非空时执行第一个操作。

任务是找到最后移除的元素的原始位置。

例子：

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 
 

这个想法是观察将要从数组中删除的最后一个元素。可以很容易地说，最后要删除的元素是可以减去最大次数的元素。 $M$在数组的所有元素之间。即，具有ceil(a [i] / M)最大值的元素。

因此，现在任务减少了，以找到具有ceil(a [i] / M)最大值的数组中元素的索引。

下面是上述方法的实现：

C++

// C++ program to find the position of the
// last removed element from the array
#include 
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

= 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
?>

Javascript

 
// Javascript 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(let i = 0; i < n; i++)
    {
        a[i] = (a[i] / m + (a[i] % m));
    }
 
    let ans = -1, max = -1;
    for(let 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
let a = [ 2, 5, 4 ];
let n = a.length;
let m = 2;
 
document.write(getPosition(a, n, m));
 
// This code is contributed by rameshtravel07
 

输出：

2

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