Number of subarrays having absolute sum greater than K | Set-2

Given an integer array arr[] of length N consisting of both positive and negative integers, the task is to find the number of sub-arrays with the absolute value of sum greater than a given positive number K. 

Examples:  

Input : arr[] = {-1, 0, 1}, K = 0 
Output : 4 
All possible sub-arrays and there total sum: 
{-1} = -1 
{0} = 0 
{1} = 1 
{-1, 0} = -1 
{0, 1} = 1 
{-1, 0, 1} = 0 
Thus, 4 sub-arrays have absolute 
value of sum greater than 0.

Input : arr[] = {2, 3, 4}, K = 4 
Output : 3 

Approach: A similar approach that works on a positive integer array is discussed here.
In this article, we will look at an algorithm that solves this problem for both positive and negative integers.  

1. Create a prefix-sum array of the given array.
2. Sort the prefix-sum array.
3. Create variable ans, find the number of elements in the prefix-sum array with value lesser than -K or greater than K, and initialize ans with this value.
4. Now, iterate the sorted prefix-sum array and for every index i, find the index of the first element with a value greater than arr[i] + K. Let's say this index is j.

Then ans can be updated as ans += N - j as the number of elements in the prefix-sum array larger than the value of arr[i]+K will be equal to N - j. 
To find the index j, perform binary search on prefix-sum array. Specifically, find the upper-bound on the value of prefix-sum[i] + k. 

Below is the implementation of the above approach: 

// C++ implementation of the above approach
#include <bits/stdc++.h>
#define maxLen 30
using namespace std;

// Function to find required value
int findCnt(int arr[], int n, int k)
{
    // Variable to store final answer
    int ans = 0;

    // Loop to find prefix-sum
    for (int i = 1; i < n; i++) {
        arr[i] += arr[i - 1];
        if (arr[i] > k or arr[i] < -1 * k)
            ans++;
    }

    if (arr[0] > k || arr[0] < -1 * k)
        ans++;

    // Sorting prefix-sum array
    sort(arr, arr + n);

    // Loop to find upper_bound
    // for each element
    for (int i = 0; i < n; i++)
        ans += n - 
       (upper_bound(arr, arr + n, arr[i] + k) - arr);

    // Returning final answer
    return ans;
}

// Driver code
int main()
{
    int arr[] = { -1, 4, -5, 6 };
    int n = sizeof(arr) / sizeof(int);
    int k = 0;

    // Function to find required value
    cout << findCnt(arr, n, k);
}

// Java implementation of the approach
import java.util.*;

class GFG 
{
    
static int maxLen = 30;


// Function to find required value
static int findCnt(int arr[], int n, int k)
{
    // Variable to store final answer
    int ans = 0;

    // Loop to find prefix-sum
    for (int i = 1; i < n; i++)
    {
        arr[i] += arr[i - 1];
        if (arr[i] > k || arr[i] < -1 * k)
            ans++;
    }

    if (arr[0] > k || arr[0] < -1 * k)
        ans++;

    // Sorting prefix-sum array
    Arrays.sort(arr);

    // Loop to find upper_bound
    // for each element
    for (int i = 0; i < n; i++)
        ans += n - upper_bound(arr, 0, n, arr[i] + k);

    // Returning final answer
    return ans;
}

static int upper_bound(int[] a, int low, 
                    int high, int element)
{
    while(low < high)
    {
        int middle = low + (high - low)/2;
        if(a[middle] > element)
            high = middle;
        else
            low = middle + 1;
    }
    return low;
} 

// Driver code
public static void main(String[] args) 
{
    int arr[] = { -1, 4, -5, 6 };
    int n = arr.length;
    int k = 0;

    // Function to find required value
    System.out.println(findCnt(arr, n, k));
}
}

// This code is contributed by 29AjayKumar

// C# implementation of the approach
using System;

class GFG 
{
    // Function to find required value
    static int findCnt(int []arr, int n, int k)
    {
        // Variable to store final answer
        int ans = 0;
    
        // Loop to find prefix-sum
        for (int i = 1; i < n; i++)
        {
            arr[i] += arr[i - 1];
            if (arr[i] > k || arr[i] < -1 * k)
                ans++;
        }
    
        if (arr[0] > k || arr[0] < -1 * k)
            ans++;
    
        // Sorting prefix-sum array
        Array.Sort(arr);
    
        // Loop to find upper_bound
        // for each element
        for (int i = 0; i < n; i++)
            ans += n - upper_bound(arr, 0, n, arr[i] + k);
    
        // Returning final answer
        return ans;
    }
    
    static int upper_bound(int[] a, int low, 
                        int high, int element)
    {
        while(low < high)
        {
            int middle = low + (high - low)/2;
            if(a[middle] > element)
                high = middle;
            else
                low = middle + 1;
        }
        return low;
    } 
    
    // Driver code
    public static void Main() 
    {
        int []arr = { -1, 4, -5, 6 };
        int n = arr.Length;
        int k = 0;
    
        // Function to find required value
        Console.WriteLine(findCnt(arr, n, k));
    }
}

// This code is contributed by AnkitRai01

# Python3 implementation of the above approach
from bisect import bisect as upper_bound

maxLen=30

# Function to find required value
def findCnt(arr, n, k):

    # Variable to store final answer
    ans = 0

    # Loop to find prefix-sum
    for i in range(1,n):
        arr[i] += arr[i - 1]
        if (arr[i] > k or arr[i] < -1 * k):
            ans+=1

    if (arr[0] > k or arr[0] < -1 * k):
        ans+=1

    # Sorting prefix-sum array
    arr=sorted(arr)

    # Loop to find upper_bound
    # for each element
    for i in range(n):
        ans += n - upper_bound(arr,arr[i] + k)

    # Returning final answer
    return ans


# Driver code

arr = [-1, 4, -5, 6]
n = len(arr)
k = 0

# Function to find required value
print(findCnt(arr, n, k))

# This code is contributed by mohit kumar 29

<script>

// Javascript implementation of the above approach
var maxLen = 30;

function upper_bound(a, low, high, element)
    {
        while(low < high)
        {
            var middle = low + parseInt((high - low)/2);
            if(a[middle] > element)
                high = middle;
            else
                low = middle + 1;
        }
        return low;
    } 

// Function to find required value
function findCnt(arr, n, k)
{
    // Variable to store final answer
    var ans = 0;

    // Loop to find prefix-sum
    for (var i = 1; i < n; i++) {
        arr[i] += arr[i - 1];
        if (arr[i] > k || arr[i] < -1 * k)
            ans++;
    }

    if (arr[0] > k || arr[0] < -1 * k)
        ans++;

    // Sorting prefix-sum array
    arr.sort((a,b)=>a-b)

    // Loop to find upper_bound
    // for each element
    for (var i = 0; i < n; i++)
        ans += (n - upper_bound(arr, 0, n, arr[i] + k));

    // Returning final answer
    return ans;
}

// Driver code
var arr = [ -1, 4, -5, 6 ];
var n = arr.length;
var k = 0;
// Function to find required value
document.write( findCnt(arr, n, k));


</script>  

10

Time complexity : O(Nlog(N))
Auxiliary Space: O(1), no extra space is required, so it is a constant