Dada una array arr[] de N enteros, la tarea es encontrar la suma de todos los elementos entre dos ceros en la array dada. Si es posible, imprima toda la suma, de lo contrario imprima «-1» .
Nota: No hay un cero continuo en la array dada.
Ejemplos:
Entrada: arr[] = { 1, 0, 3, 4, 0, 4, 4, 0, 2, 1, 4, 0, 3 } Salida: 7 8 7
Explicación :
La
suma de elementos entre cada cero es:
3 + 4 = 7
4 + 4 = 8
2 + 1 + 4 = 7Entrada: arr[] = { 1, 3, 4, 6, 0}
Salida: -1
Acercarse:
- Recorre la array dada arr[] y encuentra el primer índice con el elemento 0.
- Si aparece algún elemento con el valor cero, comience a almacenar la suma de los elementos que le siguen en un vector (por ejemplo, A[] ) hasta que aparezca el siguiente cero.
- Repita los pasos anteriores para cada cero que se produzca.
- Imprime los elementos almacenados en A[].
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include "bits/stdc++.h"
using namespace std;
// Function to find the sum between two
// zeros in the given array arr[]
void sumBetweenZero(int arr[], int N)
{
int i = 0;
// To store the sum of element
// between two zeros
vector<int> A;
// To store the sum
int sum = 0;
// Find first index of 0
for (i = 0; i < N; i++) {
if (arr[i] == 0) {
i++;
break;
}
}
// Traverse the given array arr[]
for (; i < N; i++) {
// If 0 occurs then add it to A[]
if (arr[i] == 0) {
A.push_back(sum);
sum = 0;
}
// Else add element to the sum
else {
sum += arr[i];
}
}
// Print all the sum stored in A
for (int i = 0; i < A.size(); i++) {
cout << A[i] << ' ';
}
// If there is no such element print -1
if (A.size() == 0)
cout << "-1";
}
// Driver Code
int main()
{
int arr[] = { 1, 0, 3, 4, 0, 4, 4,
0, 2, 1, 4, 0, 3 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function call
sumBetweenZero(arr, N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the sum between two
// zeros in the given array arr[]
static void sumBetweenZero(int arr[], int N)
{
int i = 0;
// To store the sum of element
// between two zeros
Vector<Integer> A = new Vector<Integer>();
// To store the sum
int sum = 0;
// Find first index of 0
for(i = 0; i < N; i++)
{
if (arr[i] == 0)
{
i++;
break;
}
}
// Traverse the given array arr[]
for(; i < N; i++)
{
// If 0 occurs then add it to A[]
if (arr[i] == 0)
{
A.add(sum);
sum = 0;
}
// Else add element to the sum
else
{
sum += arr[i];
}
}
// Print all the sum stored in A
for(int j = 0; j < A.size(); j++)
{
System.out.print(A.get(j) + " ");
}
// If there is no such element print -1
if (A.size() == 0)
System.out.print("-1");
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 0, 3, 4, 0, 4, 4,
0, 2, 1, 4, 0, 3 };
int N = arr.length;
// Function call
sumBetweenZero(arr, N);
}
}
// This code is contributed by gauravrajput1
Python3
#Python3 program for the above approach
# Function to find the sum between two
# zeros in the given array arr[]
def sumBetweenZero(arr, N):
i = 0
# To store the sum of the element
# between two zeros
A = []
# To store the sum
sum = 0
# Find first index of 0
for i in range(N):
if (arr[i] == 0):
i += 1
break
k = i
# Traverse the given array arr[]
for i in range(k, N, 1):
# If 0 occurs then add it to A[]
if (arr[i] == 0):
A.append(sum)
sum = 0
# Else add element to the sum
else:
sum += arr[i]
# Print all the sum stored in A
for i in range(len(A)):
print(A[i], end = ' ')
# If there is no such element print -1
if (len(A) == 0):
print("-1")
# Driver Code
if __name__ == '__main__':
arr = [ 1, 0, 3, 4, 0, 4, 4,
0, 2, 1, 4, 0, 3 ]
N = len(arr)
# Function call
sumBetweenZero(arr, N)
# This code is contributed by Bhupendra_Singh
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the sum between two
// zeros in the given array []arr
static void sumBetweenZero(int []arr, int N)
{
int i = 0;
// To store the sum of element
// between two zeros
List<int> A = new List<int>();
// To store the sum
int sum = 0;
// Find first index of 0
for(i = 0; i < N; i++)
{
if (arr[i] == 0)
{
i++;
break;
}
}
// Traverse the given array []arr
for(; i < N; i++)
{
// If 0 occurs then add it to []A
if (arr[i] == 0)
{
A.Add(sum);
sum = 0;
}
// Else add element to the sum
else
{
sum += arr[i];
}
}
// Print all the sum stored in A
for(int j = 0; j < A.Count; j++)
{
Console.Write(A[j] + " ");
}
// If there is no such element print -1
if (A.Count == 0)
Console.Write("-1");
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 0, 3, 4, 0, 4, 4,
0, 2, 1, 4, 0, 3 };
int N = arr.Length;
// Function call
sumBetweenZero(arr, N);
}
}
// This code is contributed by gauravrajput1
Javascript
<script>
// Javascript program for the above approach
// Function to find the sum between two
// zeros in the given array arr[]
function sumBetweenZero(arr, N) {
let i = 0;
// To store the sum of element
// between two zeros
let A = new Array();
// To store the sum
let sum = 0;
// Find first index of 0
for (i = 0; i < N; i++) {
if (arr[i] == 0) {
i++;
break;
}
}
// Traverse the given array arr[]
for (; i < N; i++) {
// If 0 occurs then add it to A[]
if (arr[i] == 0) {
A.push(sum);
sum = 0;
}
// Else add element to the sum
else {
sum += arr[i];
}
}
// Print all the sum stored in A
for (let i = 0; i < A.length; i++) {
document.write(A[i] + ' ');
}
// If there is no such element print -1
if (A.length == 0)
document.write("-1");
}
// Driver Code
let arr = [1, 0, 3, 4, 0, 4, 4,
0, 2, 1, 4, 0, 3];
let N = arr.length;
// Function call
sumBetweenZero(arr, N);
// This code is contributed by _saurabh_jaiswal
</script>
Producción:
7 8 7
Complejidad de tiempo: O(N) , donde N es la longitud de la array.
Publicación traducida automáticamente
Artículo escrito por ShivaTeja2 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA