Count of subarrays with sum at least K
Last Updated :
22 Feb, 2023
Given an array arr[] of size N and an integer K > 0. The task is to find the number of subarrays with sum at least K.
Examples:
Input: arr[] = {6, 1, 2, 7}, K = 10
Output: 2
{6, 1, 2, 7} and {1, 2, 7} are the only valid subarrays.
Input: arr[] = {3, 3, 3}, K = 5
Output: 3
Approach: For a fixed left index (say l), try to find the first index on the right of l (say r) such that (arr[l] + arr[l + 1] + … + arr[r]) ? K. Then add N – r + 1 to the required answer. Repeat this process for all the left indices.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int k_sum( int a[], int n, int k)
{
int r = 0, sum = 0;
int ans = 0;
for ( int l = 0; l < n; l++) {
while (sum < k) {
if (r == n)
break ;
else {
sum += a[r];
r++;
}
}
if (sum < k)
break ;
ans += n - r + 1;
sum -= a[l];
}
return ans;
}
int main()
{
int a[] = { 6, 1, 2, 7 }, k = 10;
int n = sizeof (a) / sizeof (a[0]);
cout << k_sum(a, n, k);
return 0;
}
|
Java
class GFG
{
static int k_sum( int a[], int n, int k)
{
int r = 0 , sum = 0 ;
int ans = 0 ;
for ( int l = 0 ; l < n; l++)
{
while (sum < k)
{
if (r == n)
break ;
else
{
sum += a[r];
r++;
}
}
if (sum < k)
break ;
ans += n - r + 1 ;
sum -= a[l];
}
return ans;
}
public static void main (String[] args)
{
int a[] = { 6 , 1 , 2 , 7 }, k = 10 ;
int n = a.length;
System.out.println(k_sum(a, n, k));
}
}
|
Python3
def k_sum(a, n, k):
r, sum = 0 , 0 ;
ans = 0 ;
for l in range (n):
while ( sum < k):
if (r = = n):
break ;
else :
sum + = a[r];
r + = 1 ;
if ( sum < k):
break ;
ans + = n - r + 1 ;
sum - = a[l];
return ans;
a = [ 6 , 1 , 2 , 7 ]; k = 10 ;
n = len (a);
print (k_sum(a, n, k));
|
C#
using System;
class GFG
{
static int k_sum( int []a, int n, int k)
{
int r = 0, sum = 0;
int ans = 0;
for ( int l = 0; l < n; l++)
{
while (sum < k)
{
if (r == n)
break ;
else
{
sum += a[r];
r++;
}
}
if (sum < k)
break ;
ans += n - r + 1;
sum -= a[l];
}
return ans;
}
public static void Main()
{
int []a = { 6, 1, 2, 7 };
int k = 10;
int n = a.Length;
Console.WriteLine(k_sum(a, n, k));
}
}
|
Javascript
<script>
function k_sum(a, n, k)
{
let r = 0, sum = 0;
let ans = 0;
for (let l = 0; l < n; l++) {
while (sum < k) {
if (r == n)
break ;
else {
sum += a[r];
r++;
}
}
if (sum < k)
break ;
ans += n - r + 1;
sum -= a[l];
}
return ans;
}
let a = [6, 1, 2, 7], k = 10;
let n = a.length;
document.write(k_sum(a, n, k));
</script>
|
Time Complexity: O(r * k)
Auxiliary Space: O(1)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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