改变给定数组中的一个元素，使其成为算术级数
原文:https://www . geesforgeks . org/change-给定数组中的一个元素使其成为算术级数/
给定一个数组，它是一个改变了一个元素的原始算术级数。任务是让它再次成为算术级数。如果有许多这样的可能序列，返回其中任何一个。数组的长度总是大于 2。 例:
输入: arr = [1，3，4，7] 输出: arr = [1，3，5，7] 等差数列的公差是 2，因此得到的 等差数列是[1，3，5，7]。
输入: arr = [1，3，7] 输出: arr = [-1，3，7] 共同的区别可以是 2 或 3 或 4。 因此得到的数组可以是[1，3，5]，[-1，3，7]或[1，4，7]。 由于可以选择任意一个，[-1，3，7]作为输出返回。
进场:
算术级数的关键组成部分是初始项和公共差。所以我们的任务是找到这两个值来知道实际的算术级数。
- 如果数组的长度为 3，则取公差作为任意两个元素的差。
- 否则，尝试使用前四个元素来找到共同的区别。
- 使用公式 a[1]–a[0]= a[2]–a[1]检查前三个元素是否是算术级数。如果它们是等差数列，则公共差 d = a[1]–a[0]和初始项将是 a[0]
- 使用公式 a[2]–a[1]= a[3]–a[2]检查第二、第三和第四个元素是否在算术级数中。如果它们是等差数列，则公共差 d = a[2]–a[1]，这意味着第一个元素已被更改，因此初始项是 a[0]= a[1]–d
- 在上述两种情况下，我们已经检查了第一个元素或第四个元素是否已被更改。如果这两种情况都为假，则意味着第一个或第四个元素没有改变。所以公差带可以取为 d =(a[3]–a[0])/3，初始项= a[0]
- 使用初始术语和常见差异打印所有元素。
以下是上述方法的实现:
C/C++
// C++ program to change one element of an array such
// that the resulting array is in arithmetic progression.
#include <bits/stdc++.h>
using namespace std;
// Finds the initial term and common difference and
// prints the resulting sequence.
void makeAP(int arr[], int n)
{
int initial_term, common_difference;
if (n == 3) {
common_difference = arr[2] - arr[1];
initial_term = arr[1] - common_difference;
}
else if ((arr[1] - arr[0]) == arr[2] - arr[1]) {
// Check if the first three elements are in
// arithmetic progression
initial_term = arr[0];
common_difference = arr[1] - arr[0];
}
else if ((arr[2] - arr[1]) == (arr[3] - arr[2])) {
// Check if the first element is not
// in arithmetic progression
common_difference = arr[2] - arr[1];
initial_term = arr[1] - common_difference;
}
else {
// The first and fourth element are
// in arithmetic progression
common_difference = (arr[3] - arr[0]) / 3;
initial_term = arr[0];
}
// Print the arithmetic progression
for (int i = 0; i < n; i++)
cout << initial_term + (i * common_difference) << " ";
cout << endl;
}
// Driver Program
int main()
{
int arr[] = { 1, 3, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
makeAP(arr, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to change one element of an array such
// that the resulting array is in arithmetic progression.
import java.util.Arrays;
class AP {
static void makeAP(int arr[], int n)
{
int initial_term, common_difference;
// Finds the initial term and common difference and
// prints the resulting array.
if (n == 3) {
common_difference = arr[2] - arr[1];
initial_term = arr[1] - common_difference;
}
else if ((arr[1] - arr[0]) == arr[2] - arr[1]) {
// Check if the first three elements are in
// arithmetic progression
initial_term = arr[0];
common_difference = arr[1] - arr[0];
}
else if ((arr[2] - arr[1]) == (arr[3] - arr[2])) {
// Check if the first element is not
// in arithmetic progression
common_difference = arr[2] - arr[1];
initial_term = arr[1] - common_difference;
}
else {
// The first and fourth element are
// in arithmetic progression
common_difference = (arr[3] - arr[0]) / 3;
initial_term = arr[0];
}
// Print the arithmetic progression
for (int i = 0; i < n; i++)
System.out.print(initial_term +
(i * common_difference) + " ");
System.out.println();
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 3, 7 };
int n = arr.length;
makeAP(arr, n);
}
}
Python 3
# Python program to change one element of an array such
# that the resulting array is in arithmetic progression.
def makeAP(arr, n):
initial_term, common_difference = 0, 0
if (n == 3):
common_difference = arr[2] - arr[1]
initial_term = arr[1] - common_difference
elif((arr[1] - arr[0]) == arr[2] - arr[1]):
# Check if the first three elements are in
# arithmetic progression
initial_term = arr[0]
common_difference = arr[1] - arr[0]
elif((arr[2] - arr[1]) == (arr[3] - arr[2])):
# Check if the first element is not
# in arithmetic progression
common_difference = arr[2] - arr[1]
initial_term = arr[1] - common_difference
else:
# The first and fourth element are
# in arithmetic progression
common_difference = (arr[3] - arr[0]) / 3
initial_term = arr[0]
# Print the arithmetic progression
for i in range(n):
print(int(initial_term+
(i * common_difference)), end = " ")
print()
# Driver code
arr = [1, 3, 7]
n = len(arr)
makeAP(arr, n)
C
// C# program to change one element of an array such
// that the resulting array is in arithmetic progression.
using System;
public class AP
{
static void makeAP(int []arr, int n)
{
int initial_term, common_difference;
// Finds the initial term and common difference and
// prints the resulting array.
if (n == 3)
{
common_difference = arr[2] - arr[1];
initial_term = arr[1] - common_difference;
}
else if ((arr[1] - arr[0]) == arr[2] - arr[1])
{
// Check if the first three elements are in
// arithmetic progression
initial_term = arr[0];
common_difference = arr[1] - arr[0];
}
else if ((arr[2] - arr[1]) == (arr[3] - arr[2]))
{
// Check if the first element is not
// in arithmetic progression
common_difference = arr[2] - arr[1];
initial_term = arr[1] - common_difference;
}
else
{
// The first and fourth element are
// in arithmetic progression
common_difference = (arr[3] - arr[0]) / 3;
initial_term = arr[0];
}
// Print the arithmetic progression
for (int i = 0; i < n; i++)
Console.Write(initial_term +
(i * common_difference) + " ");
Console.WriteLine();
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 3, 7 };
int n = arr.Length;
makeAP(arr, n);
}
}
// This code contributed by Rajput-Ji
Output:
-1 3 7
时间复杂度 : O(N)，其中 N 为数组中的元素个数。
版权属于：月萌API www.moonapi.com，转载请注明出处
