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:

• If the sum of the array elements is negative, then there exists no such index whose prefix sum from that point is non-negative.
• Otherwise, the index is the index after the index having the Minimum Prefix Sum.

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++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the starting index
// of the given circular array s.t.
// prefix sum array is non negative
int startingPoint(int A[], int N)
{
    // Stores the sum of the array
    int sum = 0;

    // Stores the starting index
    int in = 0;

    // Stores the minimum prefix
    // sum of A[0..i]
    int min = INT_MAX;

    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {

        // Update the value of sum
        sum += A[i];

        // If sum is less than min
        if (sum < min) {

            // Update the min as the
            // value of prefix sum
            min = sum;

            // Update in
            in = i + 1;
        }
    }

    // Otherwise, no such index is
    // possible
    if (sum < 0) {
        return -1;
    }

    return in % N;
}

// Driver Code
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

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG{

// Function to find the starting index
// of the given circular array s.t.
// prefix sum array is non negative
static int startingPoint(int A[], int N)
{

    // Stores the sum of the array
    int sum = 0;

    // Stores the starting index
    int in = 0;

    // Stores the minimum prefix
    // sum of A[0..i]
    int min = Integer.MAX_VALUE;

    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {

        // Update the value of sum
        sum += A[i];

        // If sum is less than min
        if (sum < min)
        {

            // Update the min as the
            // value of prefix sum
            min = sum;

            // Update in
            in = i + 1;
        }
    }

    // Otherwise, no such index is
    // possible
    if (sum < 0)
    {
        return -1;
    }

    return in % N;
}

// Driver Code
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));
}
}

// This code is contributed by Kingash

Python3

# Python3 program for the above approach

# Function to find the starting index
# of the given circular array
# prefix sum array is non negative
import sys

def startingPoint(A, N):

    # Stores the sum of the array
    sum = 0

    # Stores the starting index
    startingindex = 0

    # Stores the minimum prefix
    # sum of A[0..i]
    min = sys.maxsize

    # Traverse the array
    for i in range(0, N):

        # Update the value of sum
        sum += A[i]

        # If sum is less than minimum
        if (sum < min):

            # Update the min as
            # the value of prefix sum
            min = sum

            # Update starting index
            startingindex = i + 1

    # Otherwise no such index is possible
    if (sum < 0):
        return -1

    return startingindex % N

# Driver code
arr = [ 3, -6, 7, -4, -4, 6, -1 ]
N = len(arr)

print(startingPoint(arr,N))

# This code is contributed by Virusbuddah

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the starting index
// of the given circular array s.t.
// prefix sum array is non negative
static int startingPoint(int[] A, int N)
{

    // Stores the sum of the array
    int sum = 0;

    // Stores the starting index
    int ind = 0;

    // Stores the minimum prefix
    // sum of A[0..i]
    int min = Int32.MaxValue;

    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {

        // Update the value of sum
        sum += A[i];

        // If sum is less than min
        if (sum < min)
        {

            // Update the min as the
            // value of prefix sum
            min = sum;

            // Update in
            ind = i + 1;
        }
    }

    // Otherwise, no such index is
    // possible
    if (sum < 0)
    {
        return -1;
    }

    return ind % N;
}

// Driver Code
public static void Main()
{
    int[] arr = { 3, -6, 7, -4, -4, 6, -1 };
    int N = arr.Length;

    Console.Write(startingPoint(arr, N));
}
}

// This code is contributed by ukasp

Javascript

<script>
// javascript program for the above approach
    // Function to find the starting index
    // of the given circular array s.t.
    // prefix sum array is non negative
    function startingPoint(A , N) {

        // Stores the sum of the array
        var sum = 0;

        // Stores the starting index
        var it = 0;

        // Stores the minimum prefix
        // sum of A[0..i]
        var min = Number.MAX_VALUE;

        // Traverse the array arr
        for (i = 0; i < N; i++) {

            // Update the value of sum
            sum += A[i];

            // If sum is less than min
            if (sum < min) {

                // Update the min as the
                // value of prefix sum
                min = sum;

                // Update in
                it = i + 1;
            }
        }

        // Otherwise, no such index is
        // possible
        if (sum < 0) {
            return -1;
        }

        return it % N;
    }

    // Driver Code

        var arr = [ 3, -6, 7, -4, -4, 6, -1 ];
        var N = arr.length;

        document.write(startingPoint(arr, N));

// This code contributed by umadevi9616
</script>

Output:

5

Time Complexity: O(N) 
Auxiliary Space: O(1)