Given a set of integers, check whether there is a subsequence with odd sum and if yes, then finding the maximum odd sum. If no subsequence contains odd sum, return -1.
Examples :
Input : arr[] = {2, 5, -4, 3, -1};
Output : 9
The subsequence with maximum odd
sum is 2, 5, 3 and -1.
Input : arr[] = {4, -3, 3, -5}
Output : 7
The subsequence with maximum odd
sum is 4 and 3
Input : arr[] = {2, 4, 6}
Output : -1
There is no subsequence with odd sum.A simple solution to generate all subsequences and find the maximum sum of all subsequences with odd sums. Time complexity of this solution would be exponential.
An efficient solution can work in O(n) time. The idea is based on the following facts.
- Odd sum is not possible if all numbers are even. Otherwise, we always find an answer.
- If odd sum is possible, we find sum of all positive integers. If sum is odd, we return it as this is the maximum overall positive sum. If sum is even, we subtract the odd number with the smallest absolute value from sum. This step can be justified by the fact that the smallest absolute odd value number becomes part of result if it is negative, and removed from the result when it is positive.
Implementation:
C++
// C++ program to find maximum sum of odd // subsequence if it exists. #include<bits/stdc++.h>using namespace std;
// Returns maximum sum odd subsequence if exists// Else returns -1int findMaxOddSubarraySum(int arr[], int n)
{ // Here min_odd is the minimum odd number (in
// absolute terms). Initializing with max value
// of int .
int min_odd = INT_MAX;
// To check if there is al-least one odd number.
bool isOdd = false;
int sum = 0; // To store sum of all positive elements
for (int i=0 ; i<n ; i++)
{
// Adding positive number would increase
// the sum.
if (arr[i] > 0)
sum = sum + arr[i];
// To find the minimum odd number(absolute)
// in the array.
if (arr[i]%2 != 0)
{
isOdd = true;
if (min_odd> abs(arr[i]))
min_odd = abs(arr[i]);
}
}
// If there was no odd number
if (isOdd == false)
return -1;
// Now, sum will be either odd or even.
// If even, changing it to odd. As, even - odd = odd.
// since m is the minimum odd number(absolute).
if (sum%2 == 0)
sum = sum - min_odd;
return sum;
}// Driver codeint main()
{ int arr[] = {2, -3, 5, -1, 4};
int n = sizeof(arr)/sizeof(arr[0]);
cout << findMaxOddSubarraySum(arr, n);
return 0;
} |
Java
// Java program to find maximum sum // of odd subsequence if it exists.import java.io.*;
class GFG {
// Returns maximum sum odd subsequence,// if exists Else returns -1static int findMaxOddSubarraySum(int arr[], int n)
{ // Here min_odd is the minimum odd number
// (in absolute terms). Initializing with
// max value of int .
int min_odd = Integer.MAX_VALUE;
// To check if there is al-least
// one odd number.
boolean isOdd = false;
// To store sum of all positive elements
int sum = 0;
for (int i = 0 ; i < n ; i++)
{
// Adding positive number would
// increase the sum.
if (arr[i] > 0)
sum = sum + arr[i];
// To find the minimum odd number
// (absolute) in the array.
if (arr[i] % 2 != 0)
{
isOdd = true;
if (min_odd > Math.abs(arr[i]))
min_odd = Math.abs(arr[i]);
}
}
// If there was no odd number
if (isOdd == false)
return -1;
// Now, sum will be either odd or even.
// If even, changing it to odd.
// As, even - odd = odd.
// since m is the minimum odd
// number(absolute).
if (sum % 2 == 0)
sum = sum - min_odd;
return sum;
}// Driver codepublic static void main (String[] args)
{ int arr[] = {2, -3, 5, -1, 4};
int n = arr.length;
System.out.println(findMaxOddSubarraySum(arr, n));
}}// This code is contributed by vt_m |
Python3
# Python program to find# maximum sum of odd # subsequence if it exists. # Returns maximum sum odd# subsequence if exists# Else returns -1def findMaxOddSubarraySum(arr, n):
# Here min_odd is the
# minimum odd number (in
# absolute terms).
# Initializing with max value
# of int .
min_odd = +2147483647
# To check if there is
# at-least one odd number.
isOdd = False
# To store sum of
# all positive elements
sum = 0 for i in range(n):
# Adding positive number
# would increase
# the sum.
if (arr[i] > 0):
sum = sum + arr[i]
# To find the minimum
# odd number(absolute)
# in the array.
if (arr[i]%2 != 0):
isOdd = True
if (min_odd > abs(arr[i])):
min_odd = abs(arr[i])
# If there was no odd number
if (isOdd == False):
return -1
# Now, sum will be
# either odd or even.
# If even, changing it to
# odd. As, even - odd = odd.
# since m is the minimum
# odd number(absolute).
if (sum%2 == 0):
sum = sum - min_odd
return sum
# Driver codearr = [2, -3, 5, -1, 4]
n =len(arr)
print(findMaxOddSubarraySum(arr, n))
# This code is contributed# by Anant Agarwal. |
C#
// C# program to find maximum sum // of odd subsequence if it exists.using System;
class GFG {
// Returns maximum sum odd subsequence,
// if exists Else returns -1
static int findMaxOddSubarraySum(int []arr, int n)
{
// Here min_odd is the minimum odd number
// (in absolute terms). Initializing with
// max value of int .
int min_odd = int.MaxValue;
// To check if there is al-least
// one odd number.
bool isOdd = false;
// To store sum of all positive elements
int sum = 0;
for (int i = 0 ; i < n ; i++)
{
// Adding positive number would
// increase the sum.
if (arr[i] > 0)
sum = sum + arr[i];
// To find the minimum odd number
// (absolute) in the array.
if (arr[i] % 2 != 0)
{
isOdd = true;
if (min_odd > Math.Abs(arr[i]))
min_odd = Math.Abs(arr[i]);
}
}
// If there was no odd number
if (isOdd == false)
return -1;
// Now, sum will be either odd or even.
// If even, changing it to odd.
// As, even - odd = odd.
// since m is the minimum odd
// number(absolute).
if (sum % 2 == 0)
sum = sum - min_odd;
return sum;
}
// Driver code
public static void Main ()
{
int []arr = {2, -3, 5, -1, 4};
int n = arr.Length;
Console.Write(findMaxOddSubarraySum(arr, n));
}
}// This code is contributed by nitin mittal. |
PHP
<?php// PHP program to find maximum // sum of odd subsequence if // it exists. // Returns maximum sum odd // subsequence if exists Else// returns -1function findMaxOddSubarraySum( $arr, $n)
{ // Here min_odd is the minimum
// odd number (in absolute terms).
// Initializing with max value
// of int .
$min_odd = PHP_INT_MAX;
// To check if there is
// at-least one odd number.
$isOdd = false;
// To store sum of all
// positive elements
$sum = 0;
for ( $i = 0; $i < $n; $i++)
{
// Adding positive number
// would increase the sum.
if ($arr[$i] > 0)
$sum = $sum + $arr[$i];
// To find the minimum odd
// number(absolute) in the array.
if ($arr[$i] % 2 != 0)
{
$isOdd = true;
if ($min_odd > abs($arr[$i]))
$min_odd = abs($arr[$i]);
}
}
// If there was no odd number
if ($isOdd == false)
return -1;
// Now, sum will be either
// odd or even. If even,
// changing it to odd. As,
// even - odd = odd. since
// m is the minimum odd
// number(absolute).
if ($sum % 2 == 0)
$sum = $sum - $min_odd;
return $sum;
}// Driver code$arr = array(2, -3, 5, -1, 4);
$n = count($arr);
echo findMaxOddSubarraySum($arr, $n);
// This code is contributed by anuj_67.?> |
Javascript
<script> // Javascript program to find maximum sum
// of odd subsequence if it exists.
// Returns maximum sum odd subsequence,
// if exists Else returns -1
function findMaxOddSubarraySum(arr, n)
{
// Here min_odd is the minimum odd number
// (in absolute terms). Initializing with
// max value of int .
let min_odd = Number.MAX_VALUE;
// To check if there is al-least
// one odd number.
let isOdd = false;
// To store sum of all positive elements
let sum = 0;
for (let i = 0 ; i < n ; i++)
{
// Adding positive number would
// increase the sum.
if (arr[i] > 0)
sum = sum + arr[i];
// To find the minimum odd number
// (absolute) in the array.
if (arr[i] % 2 != 0)
{
isOdd = true;
if (min_odd > Math.abs(arr[i]))
min_odd = Math.abs(arr[i]);
}
}
// If there was no odd number
if (isOdd == false)
return -1;
// Now, sum will be either odd or even.
// If even, changing it to odd.
// As, even - odd = odd.
// since m is the minimum odd
// number(absolute).
if (sum % 2 == 0)
sum = sum - min_odd;
return sum;
}
let arr = [2, -3, 5, -1, 4];
let n = arr.length;
document.write(findMaxOddSubarraySum(arr, n));
// This code is contributed by rameshtravel07.
</script> |
Output
11
Time Complexity : O(n)
Auxiliary Space : O(1)
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