找到具有共同差D的Array的前缀和后缀和的位置形成AP
给定一个大小为N的数组arr[]和一个正整数D ，任务是找到一个元素在arr[]中的位置i ，使得前缀和 arr[i] 和后缀和在算术级数中差 D，即 arr[i] – sum(arr[0, . . ., i-1]) = sum(arr[i+1, . . ., N-1]) – arr[i] = D。如果不存在这样的位置，则返回-1。
例子：
Input: arr[] = { 4, 6, 20, 10, 15, 5 }, D = 10
Output: 3
Explanation: Sum till 3rd position is 4+6 = 10.
Element at 3rd position is 20.
Suffix sum is 10 + 15 + 5 = 30.
So 10, 20, 30 is forming an AP whose common difference is 10.
Input: arr[] ={ 1, 3, 5 }, D = 7
Output: -1
方法：给定的问题可以通过使用基于以下观察的线性搜索方法来解决：
- 如果数组的大小小于3 ，则不可能有序列，因此只需返回-1 。
- 计算数组arr[]的总和并将其存储在Sum中。
- 如果Sum % 3 != 0 ，则返回0 。
- 初始化一个变量说Mid = Sum / 3来存储 AP 系列的中间元素。
- 从索引i = 1迭代arr[]到N – 2 ：
- 计算直到i-1的前缀总和（比如 temp）
- 如果temp等于Mid – D则后缀 sum 为Mid + D 。所以返回位置i+1 。
- 否则返回-1。
- 如果循环终止并且arr[]中没有元素等于 mid 然后简单地返回 -1。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if there is
// an element forming A.P. series
// having common difference d
int checkArray(int arr[], int N, int d)
{
// If size of array is less than
// three then return -1
if (N < 3)
return -1;
// Initialize the variables
int i, Sum = 0, temp = 0;
// Calculate total sum of array
for (i = 0; i < N; i++)
Sum += arr[i];
if (Sum % 3 != 0)
return 0;
// Calculate Middle element of A.P. series
int Mid = Sum / 3;
// Iterate over the range
for (i = 1; i < N - 1; i++) {
// Store the first element of A.P.
// series in the variable temp
temp += arr[i - 1];
if (arr[i] == Mid) {
// Return position of middle element
// of the A.P. series if the first
// element is in A.P. with middle element
// having common difference d
if (temp == Mid - d)
return i + 1;
// Else return 0
else
return 0;
}
}
// If middle element is not found in arr[]
return 0;
}
// Driver Code
int main()
{
int arr[] = { 4, 6, 20, 10, 15, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
int D = 10;
// Function call
cout << checkArray(arr, N, D) << endl;
return 0;
} Java
// JAVA program for the above approach
import java.util.*;
class GFG
{
// Function to check if there is
// an element forming A.P. series
// having common difference d
public static int checkArray(int arr[], int N, int d)
{
// If size of array is less than
// three then return -1
if (N < 3)
return -1;
// Initialize the variables
int i, Sum = 0, temp = 0;
// Calculate total sum of array
for (i = 0; i < N; i++)
Sum += arr[i];
if (Sum % 3 != 0)
return 0;
// Calculate Middle element of A.P. series
int Mid = Sum / 3;
// Iterate over the range
for (i = 1; i < N - 1; i++) {
// Store the first element of A.P.
// series in the variable temp
temp += arr[i - 1];
if (arr[i] == Mid) {
// Return position of middle element
// of the A.P. series if the first
// element is in A.P. with middle element
// having common difference d
if (temp == Mid - d)
return i + 1;
// Else return 0
else
return 0;
}
}
// If middle element is not found in arr[]
return 0;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 4, 6, 20, 10, 15, 5 };
int N = arr.length;
int D = 10;
// Function call
System.out.println(checkArray(arr, N, D));
}
}
// This code is contributed by TaranpreetPython3
#Python program for the above approach
# Function to check if there is
# an element forming A.P. series
# having common difference d
def checkArray(arr, N, d):
# If size of array is less than
# three then return -1
if (N < 3):
return -1
# Initialize the variables
i = 0
Sum = 0
temp = 0
# Calculate total sum of array
for i in range (N):
Sum += arr[i]
if (Sum % 3 != 0):
return 0
# Calculate Middle element of A.P. series
Mid = Sum / 3
# Iterate over the range
for i in range(1, N - 1):
# Store the first element of A.P.
# series in the variable temp
temp += arr[i - 1]
if (arr[i] == Mid):
# Return position of middle element
# of the A.P. series if the first
# element is in A.P. with middle element
# having common difference d
if (temp == Mid - d):
return i + 1
# Else return 0
else:
return 0
# If middle element is not found in arr[]
return 0
# Driver Code
if __name__ == "__main__":
arr = [ 4, 6, 20, 10, 15, 5 ]
N = len(arr)
D = 10
# Function call
print(checkArray(arr, N, D))
# This code is contributed by hrithikgarg03188.C#
// C# program for the above approach
using System;
class GFG
{
// Function to check if there is
// an element forming A.P. series
// having common difference d
static int checkArray(int []arr, int N, int d)
{
// If size of array is less than
// three then return -1
if (N < 3)
return -1;
// Initialize the variables
int i, Sum = 0, temp = 0;
// Calculate total sum of array
for (i = 0; i < N; i++)
Sum += arr[i];
if (Sum % 3 != 0)
return 0;
// Calculate Middle element of A.P. series
int Mid = Sum / 3;
// Iterate over the range
for (i = 1; i < N - 1; i++) {
// Store the first element of A.P.
// series in the variable temp
temp += arr[i - 1];
if (arr[i] == Mid) {
// Return position of middle element
// of the A.P. series if the first
// element is in A.P. with middle element
// having common difference d
if (temp == Mid - d)
return i + 1;
// Else return 0
else
return 0;
}
}
// If middle element is not found in arr[]
return 0;
}
// Driver Code
public static void Main()
{
int []arr = { 4, 6, 20, 10, 15, 5 };
int N = arr.Length;
int D = 10;
// Function call
Console.WriteLine(checkArray(arr, N, D));
}
}
// This code is contributed by Samim Hossain Mondal.Javascript
输出
3时间复杂度： O(N)
辅助空间： O(1)