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Number of subarrays having absolute sum greater than K | Set-2
  • Difficulty Level : Hard
  • Last Updated : 13 May, 2021

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

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++




// 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




// 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#




// 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




# 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

Javascript




<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>
Output: 
10

 

Time complexity : O(Nlog(N))
 

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