当给定原始元素的平均值时，从数组中找出删除的值
原文:https://www . geeksforgeeks . org/当给定原始元素的平均值时，从数组中查找删除的值/
给定一组长度 N + K 。也给出了数组中所有元素的平均值 avg 。如果一个恰好出现在 K 时间的元素从数组中被移除(所有出现的)并且结果数组被给出，任务是找到元素 X 。注意如果 X 不是整数，则打印 -1 。 示例:
输入: arr[] = {2，7，3}，K = 3，avg = 4 输出: 4 原始数组为{2，7，3，4，4，4} ，其中删除了 3 次出现的 4。 (2 + 7 + 3 + 4 + 4 + 4) / 6 = 4
输入: arr[] = {5，2，3}，K = 4，avg = 7； 输出: -1 所需元素为 9.75，不是整数。
进场:
- 求数组元素的和，存储在变量 sum 中。
- 既然 X 出现了 K 次那么原阵之和将是 sumOrg = sum + (X * K) 。
- 平均值为平均值，即平均值= sumOrg / (N + K) 。
- 现在， X 可以很容易地计算为 X =((平均值*(N+K))–总和)/ K
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the missing element
int findMissing(int arr[], int n, int k, int avg)
{
// Find the sum of the array elements
int sum = 0;
for (int i = 0; i < n; i++) {
sum += arr[i];
}
// The numerator and the denominator
// of the equation
int num = (avg * (n + k)) - sum;
int den = k;
// If not divisible then X is
// not an integer
// it is a floating point number
if (num % den != 0)
return -1;
// Return X
return (num / den);
}
// Driver code
int main()
{
int k = 3, avg = 4;
int arr[] = { 2, 7, 3 };
int n = sizeof(arr) / sizeof(int);
cout << findMissing(arr, n, k, avg);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to return the missing element
static int findMissing(int arr[], int n,
int k, int avg)
{
// Find the sum of the array elements
int sum = 0;
for (int i = 0; i < n; i++)
{
sum += arr[i];
}
// The numerator and the denominator
// of the equation
int num = (avg * (n + k)) - sum;
int den = k;
// If not divisible then X is
// not an integer
// it is a floating point number
if (num % den != 0)
return -1;
// Return X
return (int)(num / den);
}
// Driver code
public static void main (String[] args)
{
int k = 3, avg = 4;
int arr[] = { 2, 7, 3 };
int n = arr.length;
System.out.println(findMissing(arr, n, k, avg));
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 implementation of the approach
# Function to return the missing element
def findMissing(arr, n, k, avg):
# Find the sum of the array elements
sum = 0;
for i in range(n):
sum += arr[i];
# The numerator and the denominator
# of the equation
num = (avg * (n + k)) - sum;
den = k;
# If not divisible then X is
# not an integer
# it is a floating ponumber
if (num % den != 0):
return -1;
# Return X
return (int)(num / den);
# Driver code
k = 3; avg = 4;
arr = [2, 7, 3] ;
n = len(arr);
print(findMissing(arr, n, k, avg));
# This code is contributed by 29AjayKumar
C
// C# implementation of above approach
using System;
class GFG
{
// Function to return the missing element
static int findMissing(int []arr, int n,
int k, int avg)
{
// Find the sum of the array elements
int sum = 0;
for (int i = 0; i < n; i++)
{
sum += arr[i];
}
// The numerator and the denominator
// of the equation
int num = (avg * (n + k)) - sum;
int den = k;
// If not divisible then X is
// not an integer
// it is a floating point number
if (num % den != 0)
return -1;
// Return X
return (int)(num / den);
}
// Driver code
public static void Main (String[] args)
{
int k = 3, avg = 4;
int []arr = { 2, 7, 3 };
int n = arr.Length;
Console.WriteLine(findMissing(arr, n, k, avg));
}
}
// This code is contributed by Rajput-Ji
java 描述语言
<script>
// Javascript implementation of the
// above approach
// Function to return the missing element
function findMissing(arr, n, k, avg)
{
// Find the sum of the array elements
var sum = 0;
for (var i = 0; i < n; i++) {
sum += arr[i];
}
// The numerator and the denominator
// of the equation
var num = (avg * (n + k)) - sum;
var den = k;
// If not divisible then X is
// not an integer
// it is a floating point number
if (num % den != 0)
return -1;
// Return X
return (Math.floor(num / den));
}
// Driver code
var k = 3;
var avg = 4;
var arr = [ 2, 7, 3 ];
var n = arr.length;
document.write(findMissing(arr, n, k, avg));
// This code is contributed by ShubhamSingh10
</script>
Output:
4
时间复杂度:O(1) T3】辅助空间: O(1)
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