计算给定总和的最大不重叠子数组

给定一个由N个整数和一个整数目标组成的数组arr [] ，任务是找到最大数量的非空不重叠子数组，以使每个子数组中的数组元素之和等于目标。

例子：

Input: arr[] = {2, -1, 4, 3, 6, 4, 5, 1}, target = 6
Output: 3
Explanation: 
Subarrays {-1, 4, 3}, {6} and {5, 1} have sum equal to target(= 6).

Input: arr[] = {2, 2, 2, 2, 2}, target = 4
Output: 2 

的方法：为了得到具有总和目标的最小的非重叠子阵列中，目标是使用前缀和技术。请按照以下步骤解决问题：

1. 将到目前为止计算出的所有总和存储在Map mp中，其中key为直到该索引的前缀总和，而value为具有该总和的子数组的结束索引。
2. 如果直到索引i的前缀和(即sum )等于target ，请检查map中是否存在sum – target 。
3. 如果map中存在sum – target ，并且mp [sum – target] = idx ，则表示[idx + 1，i]的子数组的sum等于target 。
4. 现在，对于不重叠的子数组，维护一个附加变量availIdx (最初设置为-1)，并且仅当mp [sum – target]≥availIdx时，才从[idx + 1，i]中获取子数组。
5. 只要找到这样的子数组，就增加答案并将availIdx的值更改为当前索引。
6. 同样，对于不重叠的子阵列，贪婪地采用尽可能小的子阵列总是有好处的。因此，对于找到的每个前缀和，即使其已经存在，也要更新其在Map中的索引。
7. 完成上述步骤后，打印计数值。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count maximum number
// of non-overlapping subarrays with
// sum equals to the target
int maximumSubarrays(int arr[], int N,
                     int target)
{
    // Stores the final count
    int ans = 0;
 
    // Next subarray should start
    // from index >= availIdx
    int availIdx = -1;
 
    // Tracks the prefix sum
    int cur_sum = 0;
 
    // Map to store the prefix sum
    // for respective indices
    unordered_map mp;
    mp[0] = -1;
 
    for (int i = 0; i < N; i++) {
 
        cur_sum += arr[i];
 
        // Check if cur_sum - target is
        // present in the array or not
        if (mp.find(cur_sum - target)
                != mp.end()
            && mp[cur_sum - target]
                   >= availIdx) {
 
            ans++;
            availIdx = i;
        }
 
        // Update the index of
        // current prefix sum
        mp[cur_sum] = i;
    }
 
    // Return the count of subarrays
    return ans;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 2, -1, 4, 3,
                  6, 4, 5, 1 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Given sum target
    int target = 6;
 
    // Function Call
    cout << maximumSubarrays(arr, N,
                             target);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to count maximum number
// of non-overlapping subarrays with
// sum equals to the target
static int maximumSubarrays(int arr[], int N,
                            int target)
{
     
    // Stores the final count
    int ans = 0;
 
    // Next subarray should start
    // from index >= availIdx
    int availIdx = -1;
 
    // Tracks the prefix sum
    int cur_sum = 0;
 
    // Map to store the prefix sum
    // for respective indices
    HashMap mp = new HashMap();
    mp.put(0, 1);
 
    for(int i = 0; i < N; i++)
    {
        cur_sum += arr[i];
 
        // Check if cur_sum - target is
        // present in the array or not
        if (mp.containsKey(cur_sum - target) &&
            mp.get(cur_sum - target) >= availIdx)
        {
            ans++;
            availIdx = i;
        }
         
        // Update the index of
        // current prefix sum
        mp.put(cur_sum, i);
    }
 
    // Return the count of subarrays
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = { 2, -1, 4, 3,
                  6, 4, 5, 1 };
 
    int N = arr.length;
 
    // Given sum target
    int target = 6;
 
    // Function call
    System.out.print(maximumSubarrays(arr, N,
                                      target));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Function to count maximum number
# of non-overlapping subarrays with
# sum equals to the target
def maximumSubarrays(arr, N, target):
     
    # Stores the final count
    ans = 0
 
    # Next subarray should start
    # from index >= availIdx
    availIdx = -1
 
    # Tracks the prefix sum
    cur_sum = 0
 
    # Map to store the prefix sum
    # for respective indices
    mp = {}
    mp[0] = -1
 
    for i in range(N):
        cur_sum += arr[i]
 
        # Check if cur_sum - target is
        # present in the array or not
        if ((cur_sum - target) in mp and
          mp[cur_sum - target] >= availIdx):
            ans += 1
            availIdx = i
 
        # Update the index of
        # current prefix sum
        mp[cur_sum] = i
 
    # Return the count of subarrays
    return ans
 
# Driver Code
if __name__ == '__main__':
     
    # Given array arr[]
    arr = [ 2, -1, 4, 3,
            6, 4, 5, 1 ]
 
    N = len(arr)
 
    # Given sum target
    target = 6
 
    # Function call
    print(maximumSubarrays(arr, N, target))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to count maximum number
// of non-overlapping subarrays with
// sum equals to the target
static int maximumSubarrays(int []arr, int N,
                            int target)
{
  // Stores the readonly count
  int ans = 0;
 
  // Next subarray should start
  // from index >= availIdx
  int availIdx = -1;
 
  // Tracks the prefix sum
  int cur_sum = 0;
 
  // Map to store the prefix sum
  // for respective indices
  Dictionary mp = new Dictionary();
  mp.Add(0, 1);
 
  for(int i = 0; i < N; i++)
  {
    cur_sum += arr[i];
 
    // Check if cur_sum - target is
    // present in the array or not
    if (mp.ContainsKey(cur_sum - target) &&
        mp[cur_sum - target] >= availIdx)
    {
      ans++;
      availIdx = i;
    }
 
    // Update the index of
    // current prefix sum
    if(mp.ContainsKey(cur_sum))
      mp[cur_sum] = i;
    else
      mp.Add(cur_sum, i);
  }
 
  // Return the count of subarrays
  return ans;
}
 
// Driver Code
public static void Main(String[] args)
{   
  // Given array []arr
  int []arr = {2, -1, 4, 3,
               6, 4, 5, 1};
   
  int N = arr.Length;
 
  // Given sum target
  int target = 6;
 
  // Function call
  Console.Write(maximumSubarrays(arr, N,
                                 target));
}
}
 
// This code is contributed by Princi Singh

输出：

3

时间复杂度： O(N)
辅助空间： O(N)