Sum of array elements that is first continuously increasing then decreasing
Last Updated :
03 May, 2023
Given an array where elements are first continuously increasing and after that its continuously decreasing unit first number is reached again. We want to add the elements of array. We may assume that there is no overflow in sum.
Examples:
Input : arr[] = {5, 6, 7, 6, 5}.
Output : 29
Input : arr[] = {10, 11, 12, 13, 12, 11, 10}
Output : 79
A simple solution is to traverse through n and add the elements of array.
Implementation:
C++
#include <iostream>
using namespace std;
int arraySum( int arr[], int n)
{
int sum = 0;
for ( int i = 0; i < n; i++)
sum = sum + arr[i];
return sum;
}
int main()
{
int arr[] = {10, 11, 12, 13, 12, 11, 10};
int n = sizeof (arr) / sizeof (arr[0]);
cout << arraySum(arr, n);
return 0;
}
|
Java
class GFG {
public static int arraySum( int arr[], int n)
{
int sum = 0 ;
for ( int i = 0 ; i < n; i++)
sum = sum + arr[i];
return sum;
}
public static void main(String[] args)
{
int arr[] = { 10 , 11 , 12 , 13 , 12 , 11 , 10 };
int n = arr.length;
System.out.print(arraySum(arr, n));
}
}
|
Python3
def arraySum( arr, n):
_sum = 0
for i in range (n):
_sum = _sum + arr[i]
return _sum
arr = [ 10 , 11 , 12 , 13 , 12 , 11 , 10 ]
n = len (arr)
print (arraySum(arr, n))
|
C#
using System;
class GFG {
public static int arraySum( int []arr, int n)
{
int sum = 0;
for ( int i = 0; i < n; i++)
sum = sum + arr[i];
return sum;
}
public static void Main()
{
int []arr = {10, 11, 12, 13, 12, 11, 10};
int n = arr.Length;
Console.WriteLine(arraySum(arr, n));
}
}
|
PHP
<?php
function arraySum( $arr , $n )
{
$sum = 0;
for ( $i = 0; $i < $n ; $i ++)
$sum = $sum + $arr [ $i ];
return $sum ;
}
$arr = array (10, 11, 12, 13,
12, 11, 10);
$n = sizeof( $arr );
echo (arraySum( $arr , $n ));
?>
|
Javascript
<script>
function arraySum(arr, n)
{
let sum = 0;
for (let i = 0; i < n; i++)
sum = sum + arr[i];
return sum;
}
let arr = [10, 11, 12, 13,
12, 11, 10];
let n = arr.length;
document.write(arraySum(arr, n));
</script>
|
Time Complexity : O(n)
Space Complexity : O(1)
An efficient solution is to apply below formula.
sum = (arr[0] - 1)*n + ?n/2?2
How does it work?
If we take a closer look, we can notice that the
sum can be written as.
(arr[0] - 1)*n + (1 + 2 + .. x + (x -1) + (x-2) + ..1)
Let us understand above result with example {10, 11,
12, 13, 12, 11, 10}. If we subtract 9 (arr[0]-1) from
this array, we get {1, 2, 3, 2, 1}.
Where x = ceil(n/2) [Half of array size]
As we know that 1 + 2 + 3 + . . . + x = x * (x + 1)/2.
And we have given
= 1 + 2 + 3 + . . . + x + (x - 1) + . . . + 3 + 2 + 1
= (1 + 2 + 3 + . . . + x) + ((x - 1) + . . . + 3 + 2 + 1)
= (x * (x + 1))/2 + ((x - 1) * x)/2
= (x2 + x)/2 + (n2 - x)/2
= (2 * x2)/2
= x2
Implementation:
C++
#include <iostream>
using namespace std;
int arraySum( int arr[], int n)
{
int x = (n+1)/2;
return (arr[0] - 1)*n + x*x;
}
int main()
{
int arr[] = {10, 11, 12, 13, 12, 11, 10};
int n = sizeof (arr) / sizeof (arr[0]);
cout << arraySum(arr, n);
return 0;
}
|
Java
class GFG {
public static int arraySum( int arr[], int n)
{
int x = (n + 1 ) / 2 ;
return (arr[ 0 ] - 1 ) * n + x * x;
}
public static void main(String[] args)
{
int arr[] = { 10 , 11 , 12 , 13 , 12 , 11 , 10 };
int n = arr.length;
System.out.print(arraySum(arr, n));
}
}
|
Python3
def arraySum( arr, n):
x = (n + 1 ) / 2
return (arr[ 0 ] - 1 ) * n + x * x
arr = [ 10 , 11 , 12 , 13 , 12 , 11 , 10 ]
n = len (arr)
print (arraySum(arr, n))
|
C#
using System;
class GFG {
public static int arraySum( int []arr, int n)
{
int x = (n + 1) / 2;
return (arr[0] - 1) * n + x * x;
}
public static void Main()
{
int []arr = {10, 11, 12, 13, 12, 11, 10};
int n = arr.Length;
Console.WriteLine(arraySum(arr, n));
}
}
|
PHP
<?php
function arraySum( $arr , $n )
{
$x = ( $n + 1) / 2;
return ( $arr [0] - 1) *
$n + $x * $x ;
}
$arr = array (10, 11, 12, 13,
12, 11, 10);
$n = sizeof( $arr );
echo (arraySum( $arr , $n ));
?>
|
Javascript
function arraySum(arr, n)
{
let x = (n + 1) / 2;
return (arr[0] - 1) *
n + x * x;
}
let arr = [10, 11, 12, 13,
12, 11, 10];
let n = arr.length;
document.write(arraySum(arr, n));
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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