Given an array, arr[] of size N and a positive integer D, the task is to find the position i of an element in arr[] such that the prefix sum, arr[i] and suffix sum is in arithmetic progression with common difference D, i.e. arr[i] – sum(arr[0, . . ., i-1]) = sum(arr[i+1, . . ., N-1]) – arr[i] = D. If no such position exists return -1.
Examples:
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
Approach: The given problem can be solved by using Linear Search approach based on the below observation:
- If the size of array is less than 3 then no sequence is possible so simply return -1.
- Calculate sum of array arr[] and store it in Sum.
- if Sum % 3 != 0, then return 0.
- Initialize a variable say Mid = Sum / 3 to store the middle element of the AP series.
- Iterate arr[] from index i = 1 to N – 2:
- Calculate the prefix sum till i-1 (say temp)
- If temp is equal to Mid – D then suffix sum is Mid + D. So return the position i+1.
- Else return -1.
- If loop terminates and no element in arr[] is equal to mid then simply return -1.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> 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 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 Taranpreet |
#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# 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. |
<script> // JavaScript program for the above approach
// Function to check if there is
// an element forming A.P. series
// having common difference d
const checkArray = (arr, N, d) => {
// If size of array is less than
// three then return -1
if (N < 3)
return -1;
// Initialize the variables
let 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
let Mid = parseInt(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
let arr = [4, 6, 20, 10, 15, 5];
let N = arr.length;
let D = 10;
// Function call
document.write(checkArray(arr, N, D));
// This code is contributed by rakeshsahni </script> |
3
Time Complexity: O(N)
Auxiliary Space: O(1)