Find the index in a circular array from which prefix sum is always non-negative
Given a circular array arr[] consisting of N integers, the task is to find the starting index of the circular array such that the prefix sum from that index is always non-negative. If there exists no such index, then print “-1”.
Examples:
Input: arr[] = {3, -6, 7, -1, -4, 5, -1}
Output: 2
Explanation:
Consider the index 2 from where the prefix sum of the given circular array is being calculated, then the prefix sum is given by {9, 3, 7, 6, 2, 7, 6}
Input: arr[] = {3, -5, -1}
Output: -1
Approach: The given problem can be solved based on the following observations:
Follow the steps to solve the problem:
- Initialize a variable, say sum as 0, that stores the sum of array elements.
- Initialize a variable, say in as 0 that stores the starting index for the circular traversal.
- Initialize a variable, say min as INT_MAX that stores the minimum prefix sum of the array arr[].
- Traverse the given array and perform the following steps:
- Update the value of sum as the sum of sum and the current element arr[i].
- If the value of sum is less than min, then update the min as sum and in as (i + 1).
- If the sum of the array is negative, then print -1. Otherwise, print the value of (in % N) as the resultant possible index.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int startingPoint( int A[], int N)
{
int sum = 0;
int in = 0;
int min = INT_MAX;
for ( int i = 0; i < N; i++) {
sum += A[i];
if (sum < min) {
min = sum;
in = i + 1;
}
}
if (sum < 0) {
return -1;
}
return in % N;
}
int main()
{
int arr[] = { 3, -6, 7, -4, -4, 6, -1 };
int N = ( sizeof (arr) / sizeof (arr[0]));
cout << startingPoint(arr, N);
return 0;
}
|
Java
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
static int startingPoint( int A[], int N)
{
int sum = 0 ;
int in = 0 ;
int min = Integer.MAX_VALUE;
for ( int i = 0 ; i < N; i++)
{
sum += A[i];
if (sum < min)
{
min = sum;
in = i + 1 ;
}
}
if (sum < 0 )
{
return - 1 ;
}
return in % N;
}
public static void main(String[] args)
{
int arr[] = { 3 , - 6 , 7 , - 4 , - 4 , 6 , - 1 };
int N = arr.length;
System.out.print(startingPoint(arr, N));
}
}
|
Python3
import sys
def startingPoint(A, N):
sum = 0
startingindex = 0
min = sys.maxsize
for i in range ( 0 , N):
sum + = A[i]
if ( sum < min ):
min = sum
startingindex = i + 1
if ( sum < 0 ):
return - 1
return startingindex % N
arr = [ 3 , - 6 , 7 , - 4 , - 4 , 6 , - 1 ]
N = len (arr)
print (startingPoint(arr,N))
|
C#
using System;
class GFG{
static int startingPoint( int [] A, int N)
{
int sum = 0;
int ind = 0;
int min = Int32.MaxValue;
for ( int i = 0; i < N; i++)
{
sum += A[i];
if (sum < min)
{
min = sum;
ind = i + 1;
}
}
if (sum < 0)
{
return -1;
}
return ind % N;
}
public static void Main()
{
int [] arr = { 3, -6, 7, -4, -4, 6, -1 };
int N = arr.Length;
Console.Write(startingPoint(arr, N));
}
}
|
Javascript
<script>
function startingPoint(A , N) {
var sum = 0;
var it = 0;
var min = Number.MAX_VALUE;
for (i = 0; i < N; i++) {
sum += A[i];
if (sum < min) {
min = sum;
it = i + 1;
}
}
if (sum < 0) {
return -1;
}
return it % N;
}
var arr = [ 3, -6, 7, -4, -4, 6, -1 ];
var N = arr.length;
document.write(startingPoint(arr, N));
</script>
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Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
29 Apr, 2021
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