计算给定数组中总和非零的子数组

给定大小为N的数组arr [] ，任务是计算给定数组arr []的子数组的总数，该子数组具有非零和。
例子：

Input: arr[] = {-2, 2, -3} 
Output: 4 
Explanation: 
The subarrays with non zero sum are: [-2], [2], [2, -3], [-3]. All possible subarray of the given input array are [-2], [2], [-3], [2, -2], [2, -3], [-2, 2, -3]. Out of these [2, -2] is not included in the count because 2+(-2) = 0 and [-2, 2, -3] is not selected because one the subarray [2, -2] of this array has a zero sum of elements.
Input: arr[] = {1, 3, -2, 4, -1} 
Output: 15 
Explanation: 
There are 15 subarray for the given array {1, 3, -2, 4, -1} which has a non zero sum.

方法：
解决上述问题的主要思想是使用“前缀和数组”和“地图数据结构”。

• 首先，构建给定数组的Prefix sum数组，并使用以下公式检查子数组是否有0个元素和。

Sum of Subarray[L, R] = Prefix[R] – Prefix[L – 1]. So, If Sum of Subarray[L, R] = 0
Then, Prefix[R] – Prefix[L – 1] = 0. Hence, Prefix[R] = Prefix[L – 1] 
 

• 现在，从1迭代到N，并保留一个Hash表以存储该元素先前出现的索引和一个变量(例如last) ，并将其初始化为0。
• 检查Prefix [i]是否已存在于哈希中。如果是，则将last更新为last = max(last，hash [Prefix [i]] +1) 。否则，在答案前添加i-最后更新哈希表。

下面是上述方法的实现：

C++

// C++ program to Count the total number of
// subarrays for a given array such that its
// subarray should have non zero sum
 
#include 
using namespace std;
 
// Function to build the Prefix sum array
vector PrefixSumArray(int arr[], int n)
{
    vector prefix(n);
 
    // Store prefix of the first position
    prefix[0] = arr[0];
 
    for (int i = 1; i < n; i++)
        prefix[i] = prefix[i - 1] + arr[i];
 
    return prefix;
}
 
// Function to return the Count of
// the total number of subarrays
int CountSubarray(int arr[], int n)
{
    vector Prefix(n);
 
    // Calculating the prefix array
    Prefix = PrefixSumArray(arr, n);
 
    int last = 0, ans = 0;
 
    map Hash;
 
    Hash[0] = -1;
 
    for (int i = 0; i <= n; i++) {
        // Check if the element already exists
        if (Hash.count(Prefix[i]))
            last = max(last, Hash[Prefix[i]] + 1);
 
        ans += max(0, i - last);
 
        // Mark the element
        Hash[Prefix[i]] = i;
    }
 
    // Return the final answer
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 3, -2, 4, -1 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << CountSubarray(arr, N);
}

Java

// Java program to count the total number of
// subarrays for a given array such that its
// subarray should have non zero sum
import java.util.*;
 
class GFG{
 
// Function to build the Prefix sum array
static int[] PrefixSumArray(int arr[], int n)
{
    int []prefix = new int[n];
     
    // Store prefix of the first position
    prefix[0] = arr[0];
 
    for(int i = 1; i < n; i++)
        prefix[i] = prefix[i - 1] + arr[i];
 
    return prefix;
}
 
// Function to return the Count of
// the total number of subarrays
static int CountSubarray(int arr[], int n)
{
    int []Prefix = new int[n];
 
    // Calculating the prefix array
    Prefix = PrefixSumArray(arr, n);
 
    int last = 0, ans = 0;
 
    HashMap Hash = new HashMap();
 
    Hash.put(0, -1);
 
    for(int i = 0; i <= n; i++)
    {
         
        // Check if the element already exists
        if (i < n && Hash.containsKey(Prefix[i]))
            last = Math.max(last,
                            Hash.get(Prefix[i]) + 1);
 
        ans += Math.max(0, i - last);
 
        // Mark the element
        if (i < n)
        Hash.put(Prefix[i], i);
    }
 
    // Return the final answer
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 3, -2, 4, -1 };
 
    int N = arr.length;
 
    System.out.print(CountSubarray(arr, N));
}
}
 
// This code is contributed by amal kumar choubey

Python3

# Python3 program to count the total number 
# of subarrays for a given array such that
# its subarray should have non zero sum
 
# Function to build the prefix sum array
def PrefixSumArray(arr, n):
 
    prefix = [0] * (n + 1);
 
    # Store prefix of the first position
    prefix[0] = arr[0];
 
    for i in range(1, n):
        prefix[i] = prefix[i - 1] + arr[i];
         
    return prefix;
 
# Function to return the count of
# the total number of subarrays
def CountSubarray(arr, n):
 
    Prefix = [0] * (n + 1);
 
    # Calculating the prefix array
    Prefix = PrefixSumArray(arr, n);
 
    last = 0; ans = 0;
 
    Hash = {};
 
    Hash[0] = -1;
 
    for i in range(n + 1):
         
        # Check if the element already exists
        if (Prefix[i] in Hash):
            last = max(last, Hash[Prefix[i]] + 1);
 
        ans += max(0, i - last);
 
        # Mark the element
        Hash[Prefix[i]] = i;
 
    # Return the final answer
    return ans;
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 1, 3, -2, 4, -1 ];
    N = len(arr);
 
    print(CountSubarray(arr, N));
     
# This code is contributed by AnkitRai01

C#

// C# program to count the total number of
// subarrays for a given array such that its
// subarray should have non zero sum
using System;
using System.Collections.Generic;
class GFG{
 
// Function to build the Prefix sum array
static int[] PrefixSumArray(int []arr, int n)
{
    int []prefix = new int[n];
     
    // Store prefix of the first position
    prefix[0] = arr[0];
 
    for(int i = 1; i < n; i++)
        prefix[i] = prefix[i - 1] + arr[i];
    return prefix;
}
 
// Function to return the Count of
// the total number of subarrays
static int CountSubarray(int []arr, int n)
{
    int []Prefix = new int[n];
 
    // Calculating the prefix array
    Prefix = PrefixSumArray(arr, n);
 
    int last = 0, ans = 0;
    Dictionary Hash = new Dictionary();
    Hash.Add(0, -1);
    for(int i = 0; i <= n; i++)
    {       
        // Check if the element already exists
        if (i < n && Hash.ContainsKey(Prefix[i]))
            last = Math.Max(last,
                            Hash[Prefix[i]] + 1);
 
        ans += Math.Max(0, i - last);
 
        // Mark the element
        if (i < n)
        Hash.Add(Prefix[i], i);
    }
 
    // Return the readonly answer
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = {1, 3, -2, 4, -1};
    int N = arr.Length;
    Console.Write(CountSubarray(arr, N));
}
}
 
// This code is contributed by shikhasingrajput

输出：

15

时间复杂度： O(N)