Dada una array arr[] de tamaño N y un entero K > 0 . La tarea es encontrar el número de subarreglos con una suma de al menos K .
Ejemplos:
Entrada: arr[] = {6, 1, 2, 7}, K = 10
Salida: 2
{6, 1, 2, 7} y {1, 2, 7} son los únicos subarreglos válidos.
Entrada: arr[] = {3, 3, 3}, K = 5
Salida: 3
Enfoque: para un índice izquierdo fijo (digamos l ), intente encontrar el primer índice a la derecha de l (digamos r ) tal que (arr[l] + arr[l + 1] + … + arr[r]) ≥ k _ Luego agregue N – r + 1 a la respuesta requerida. Repita este proceso para todos los índices de la izquierda.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the number of
// subarrays with sum atleast k
int k_sum(int a[], int n, int k)
{
// To store the right index
// and the current sum
int r = 0, sum = 0;
// To store the number of sub-arrays
int ans = 0;
// For all left indexes
for (int l = 0; l < n; l++) {
// Get elements till current sum
// is less than k
while (sum < k) {
if (r == n)
break;
else {
sum += a[r];
r++;
}
}
// No such subarray is possible
if (sum < k)
break;
// Add all possible subarrays
ans += n - r + 1;
// Remove the left most element
sum -= a[l];
}
// Return the required answer
return ans;
}
// Driver code
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
// Java implementation of the approach
class GFG
{
// Function to return the number of
// subarrays with sum atleast k
static int k_sum(int a[], int n, int k)
{
// To store the right index
// and the current sum
int r = 0, sum = 0;
// To store the number of sub-arrays
int ans = 0;
// For all left indexes
for (int l = 0; l < n; l++)
{
// Get elements till current sum
// is less than k
while (sum < k)
{
if (r == n)
break;
else
{
sum += a[r];
r++;
}
}
// No such subarray is possible
if (sum < k)
break;
// Add all possible subarrays
ans += n - r + 1;
// Remove the left most element
sum -= a[l];
}
// Return the required answer
return ans;
}
// Driver code
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));
}
}
// This code is contributed by kanugargng
Python3
# Python3 implementation of the approach # Function to return the number of # subarrays with sum atleast k def k_sum(a, n, k): # To store the right index # and the current sum r, sum = 0, 0; # To store the number of sub-arrays ans = 0; # For all left indexes for l in range(n): # Get elements till current sum # is less than k while (sum < k): if (r == n): break; else: sum += a[r]; r += 1; # No such subarray is possible if (sum < k): break; # Add all possible subarrays ans += n - r + 1; # Remove the left most element sum -= a[l]; # Return the required answer return ans; # Driver code a = [ 6, 1, 2, 7 ]; k = 10; n = len(a); print(k_sum(a, n, k)); # This code contributed by PrinciRaj1992
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the number of
// subarrays with sum atleast k
static int k_sum(int []a, int n, int k)
{
// To store the right index
// and the current sum
int r = 0, sum = 0;
// To store the number of sub-arrays
int ans = 0;
// For all left indexes
for (int l = 0; l < n; l++)
{
// Get elements till current sum
// is less than k
while (sum < k)
{
if (r == n)
break;
else
{
sum += a[r];
r++;
}
}
// No such subarray is possible
if (sum < k)
break;
// Add all possible subarrays
ans += n - r + 1;
// Remove the left most element
sum -= a[l];
}
// Return the required answer
return ans;
}
// Driver code
public static void Main()
{
int []a = { 6, 1, 2, 7 };
int k = 10;
int n = a.Length;
Console.WriteLine(k_sum(a, n, k));
}
}
// This code is contributed by AnkitRai01
Javascript
<script>
// Javascript implementation of the approach
// Function to return the number of
// subarrays with sum atleast k
function k_sum(a, n, k)
{
// To store the right index
// and the current sum
let r = 0, sum = 0;
// To store the number of sub-arrays
let ans = 0;
// For all left indexes
for (let l = 0; l < n; l++) {
// Get elements till current sum
// is less than k
while (sum < k) {
if (r == n)
break;
else {
sum += a[r];
r++;
}
}
// No such subarray is possible
if (sum < k)
break;
// Add all possible subarrays
ans += n - r + 1;
// Remove the left most element
sum -= a[l];
}
// Return the required answer
return ans;
}
// Driver code
let a = [6, 1, 2, 7], k = 10;
let n = a.length;
document.write(k_sum(a, n, k));
// This code is contributed by _saurabh_jaiswal.
</script>
Producción:
2
Complejidad temporal: O(r * k)
Espacio Auxiliar: O(1)
Tema relacionado: Subarrays, subsecuencias y subconjuntos en array
Publicación traducida automáticamente
Artículo escrito por pawan_asipu y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA