Dada una array arr[] de enteros positivos y dos enteros L y R que definen el rango [L, R] . La tarea es imprimir los subarreglos que tienen una suma en el rango L a R .
Ejemplos:
Entrada: arr[] = {1, 4, 6}, L = 3, R = 8
Salida: {1, 4}, {4}, {6}.
Explicación: Todos los subarreglos posibles son los siguientes
{1] con suma 1.
{1, 4} con suma 5.
{1, 4, 6} con suma 11.
{4} con suma 4.
{4, 6} con suma 10.
{6} con suma 6.
Por lo tanto, los subarreglos {1, 4}, {4}, {6} tienen suma en el rango [3, 8].Entrada: arr[] = {2, 3, 5, 8}, L = 4, R = 13
Salida: {2, 3}, {2, 3, 5}, {3, 5}, {5}, { 5, 8}, {8}.
Enfoque: este problema se puede resolver haciendo fuerza bruta y verificando todos y cada uno de los subarreglos posibles usando dos bucles. A continuación se muestra la implementación del enfoque anterior.
C++
// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find subarrays in given range
void subArraySum(int arr[], int n,
int leftsum, int rightsum)
{
int curr_sum, i, j, res = 0;
// Pick a starting point
for (i = 0; i < n; i++) {
curr_sum = arr[i];
// Try all subarrays starting with 'i'
for (j = i + 1; j <= n; j++) {
if (curr_sum > leftsum
&& curr_sum < rightsum) {
cout << "{ ";
for (int k = i; k < j; k++)
cout << arr[k] << " ";
cout << "}\n";
}
if (curr_sum > rightsum || j == n)
break;
curr_sum = curr_sum + arr[j];
}
}
}
// Driver Code
int main()
{
int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 };
int N = sizeof(arr) / sizeof(arr[0]);
int L = 10, R = 23;
subArraySum(arr, N, L, R);
return 0;
}
Java
// Java code for the above approach
import java.io.*;
class GFG
{
// Function to find subarrays in given range
static void subArraySum(int arr[], int n, int leftsum,
int rightsum)
{
int curr_sum, i, j, res = 0;
// Pick a starting point
for (i = 0; i < n; i++) {
curr_sum = arr[i];
// Try all subarrays starting with 'i'
for (j = i + 1; j <= n; j++) {
if (curr_sum > leftsum
&& curr_sum < rightsum) {
System.out.print("{ ");
for (int k = i; k < j; k++)
System.out.print(arr[k] + " ");
System.out.println("}");
}
if (curr_sum > rightsum || j == n)
break;
curr_sum = curr_sum + arr[j];
}
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 };
int N = arr.length;
int L = 10, R = 23;
subArraySum(arr, N, L, R);
}
}
// This code is contributed by Potta Lokesh
Python3
# Python program for above approach
# Function to find subarrays in given range
def subArraySum (arr, n, leftsum, rightsum):
res = 0
# Pick a starting point
for i in range(n):
curr_sum = arr[i]
# Try all subarrays starting with 'i'
for j in range(i + 1, n + 1):
if (curr_sum > leftsum
and curr_sum < rightsum):
print("{ ", end="")
for k in range(i, j):
print(arr[k], end=" ")
print("}")
if (curr_sum > rightsum or j == n):
break
curr_sum = curr_sum + arr[j]
# Driver Code
arr = [15, 2, 4, 8, 9, 5, 10, 23]
N = len(arr)
L = 10
R = 23
subArraySum(arr, N, L, R)
# This code is contributed by Saurabh Jaiswal
C#
// C# code for the above approach
using System;
class GFG
{
// Function to find subarrays in given range
static void subArraySum(int []arr, int n, int leftsum,
int rightsum)
{
int curr_sum, i, j, res = 0;
// Pick a starting point
for (i = 0; i < n; i++) {
curr_sum = arr[i];
// Try all subarrays starting with 'i'
for (j = i + 1; j <= n; j++) {
if (curr_sum > leftsum
&& curr_sum < rightsum) {
Console.Write("{ ");
for (int k = i; k < j; k++)
Console.Write(arr[k] + " ");
Console.WriteLine("}");
}
if (curr_sum > rightsum || j == n)
break;
curr_sum = curr_sum + arr[j];
}
}
}
// Driver Code
public static void Main()
{
int []arr = { 15, 2, 4, 8, 9, 5, 10, 23 };
int N = arr.Length;
int L = 10, R = 23;
subArraySum(arr, N, L, R);
}
}
// This code is contributed by Samim Hossain Mondal.
Javascript
<script>
// JavaScript program for above approach
// Function to find subarrays in given range
const subArraySum = (arr, n, leftsum, rightsum) => {
let curr_sum, i, j, res = 0;
// Pick a starting point
for (i = 0; i < n; i++) {
curr_sum = arr[i];
// Try all subarrays starting with 'i'
for (j = i + 1; j <= n; j++) {
if (curr_sum > leftsum
&& curr_sum < rightsum) {
document.write("{ ");
for (let k = i; k < j; k++)
document.write(`${arr[k]} `);
document.write("}<br/>");
}
if (curr_sum > rightsum || j == n)
break;
curr_sum = curr_sum + arr[j];
}
}
}
// Driver Code
let arr = [15, 2, 4, 8, 9, 5, 10, 23];
let N = arr.length;
let L = 10, R = 23;
subArraySum(arr, N, L, R);
// This code is contributed by rakeshsahni
</script>
Producción
{ 15 }
{ 15 2 }
{ 15 2 4 }
{ 2 4 8 }
{ 4 8 }
{ 4 8 9 }
{ 8 9 }
{ 8 9 5 }
{ 9 5 }
{ 5 10 }
Complejidad de tiempo: O(N^3)
Espacio Auxiliar: O(1)