Given an array arr[] consisting of N integers, the task is to determine if the sum of array elements can be reduced to 0 by performing the following operations any number of times:
- Choose an element A[i] and reduce A[i] by i(1-based indexing), any number of times, possibly 0.
- If the sum can be reduced to 0, print “Yes“. Otherwise, print “No“.
Examples:
Input: arr[] = {2, 3, 1}
Output: Yes
Explanation:
Select A[2] = 3
Perform given operation 3 times to obtain the following result:
3 -> (3 -2) -> (3 – 2 – 2) -> (3 – 2 – 2 – 2)
Sum of the modified array = 2 + (-3) + 1 = 0.
Therefore, the required answer is 0.Input: arr[] = {-5, 3}
Output: No
Approach: This problem can be solved based on the following observations:
- If a positive number is repetitively subtracted from another positive number, then eventually, 0 can be obtained. But, repetitively subtracting a positive number from a negative number, 0 can never be obtained as it keeps on decreasing negatively.
- The task to reduce the sum to 0 by subtracting i from A[i].
- Therefore, on choosing an element and reducing the value of the element by i, the sum is being reduced by i.
Let the sum of the array be S.
S = A[1] + A[2] + …. + A[N]
After performing given operations, sum of the array modifies to
S2 = (A[i] – (i)) + (A[i+1] – (i+1)) ….
S2 = A[i] + A[i+1]…. – (i + i+1….)
S2 = S – (i + i + 1…..)
Therefore, after every operation, the original sum is reduced.
- Therefore, the task reduces to checking if the initial sum of the array is positive or 0. If found to be true, print “Yes“. Otherwise, print “No“.
Below is the implementation for the above approach:
// C++ Program for the above approach #include <iostream> using namespace std;
// Function to check if an array // sum can be reduced to zero or not bool isPossible( int arr[], int n)
{ // Stores the sum of the array
int S = 0;
for ( int i = 0; i < n; i++) {
// Update sum of the array
S = S + arr[i];
}
// If the sum is positive
if (S >= 0) {
// Array sum can be
// reduced to 0
return true ;
}
// Otherwise
else {
// Array sum cannot
// be reduced to 0
return false ;
}
} // Driver Code int main()
{ int arr[] = { -5, 3 };
int n = sizeof (arr) / sizeof ( int );
// Print the answer
if (isPossible(arr, n)) {
cout << "Yes" ;
}
else {
cout << "No" ;
}
} |
// Java program for the above approach import java.io.*;
class GFG {
// Function to check if array sum
// can be reduced to zero or not
static boolean isPossible( int [] arr, int n)
{
// Stores the sum of the array
int S = 0 ;
for ( int i = 0 ; i < n; i++) {
S = S + arr[i];
}
// If array sum is positive
if (S >= 0 )
// Array sum can be
// reduced to 0
return true ;
// Otherwise
else
// Array sum cannot
// be reduced to 0
return false ;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { - 5 , 3 };
int n = arr.length;
// Function call
if (isPossible(arr, n))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
} |
# Python program for the above approach # Function to check if an array # sum can be reduced to zero or not def isPossible(arr, n):
# Stores sum of the array
S = sum (arr)
# If sum is positive
if (S > = 0 ):
return true
# If sum is negative
else :
return false
# Driver Code if __name__ = = '__main__' :
arr = [ - 5 , 3 ]
n = len (arr)
# Function call
if (isPossible(arr, n)):
print ( "Yes" )
else :
print ( "No" )
|
// C# program for // the above approach using System;
class GFG{
// Function to check if array sum // can be reduced to zero or not static bool isPossible( int [] arr,
int n)
{ // Stores the sum
// of the array
int S = 0;
for ( int i = 0; i < n; i++)
{
S = S + arr[i];
}
// If array sum is positive
if (S >= 0)
// Array sum can be
// reduced to 0
return true ;
// Otherwise
else
// Array sum cannot
// be reduced to 0
return false ;
} // Driver Code public static void Main()
{ int [] arr = {-5, 3};
int n = arr.Length;
// Function call
if (isPossible(arr, n))
Console.Write( "Yes" );
else
Console.Write( "No" );
} } // This code is contributed by Chitranayal |
<script> // javascript program for the above approach // Function to check if array sum // can be reduced to zero or not
function isPossible(arr , n)
{
// Stores the sum of the array
var S = 0;
for (i = 0; i < n; i++) {
S = S + arr[i];
}
// If array sum is positive
if (S >= 0)
// Array sum can be
// reduced to 0
return true ;
// Otherwise
else
// Array sum cannot
// be reduced to 0
return false ;
}
// Driver Code
var arr = [ -5, 3 ];
var n = arr.length;
// Function call
if (isPossible(arr, n))
document.write( "Yes" );
else
document.write( "No" );
// This code is contributed by todaysgaurav </script> |
No
Time Complexity: O(N)
Auxiliary Space: O(1)